🧠:[<think>Okay, so I need to find the last three digits of the number 2003^(2002^2001). That's a really big exponent, so calculating it directly is impossible. I remember that to find the last three digits of a number, we need to compute the number modulo 1000. So, essentially, I need to compute 2003^(2002^2001) mod 1000. First, let's simplify the base modulo 1000. The base is 2003, and when we divide 2003 by 1000, the remainder is 3. So, 2003 ≡ 3 mod 1000. Therefore, the problem reduces to finding 3^(2002^2001) mod 1000. Now, to compute a large exponent modulo 1000, Euler's theorem might be helpful. Euler's theorem states that if a and n are coprime, then a^φ(n) ≡ 1 mod n, where φ(n) is Euler's totient function. Since 3 and 1000 are coprime (their GCD is 1), we can use this theorem. First, let's compute φ(1000). Since 1000 = 2^3 * 5^3, φ(1000) = 1000*(1 - 1/2)*(1 - 1/5) = 1000*(1/2)*(4/5) = 1000*(2/5) = 400. So φ(1000) = 400. Therefore, 3^400 ≡ 1 mod 1000. Therefore, 3^k mod 1000 can be simplified by taking k mod 400. So, 3^(2002^2001) mod 1000 is equal to 3^( (2002^2001 mod 400) ) mod 1000. So, the next step is to compute 2002^2001 mod 400. Let's work on that. First, simplify 2002 mod 400. 2000 is divisible by 400 (since 400*5=2000), so 2002 mod 400 = 2. Therefore, 2002 ≡ 2 mod 400. Therefore, 2002^2001 ≡ 2^2001 mod 400. So now, we need to compute 2^2001 mod 400. Hmm, calculating 2^2001 mod 400. Maybe we can use Euler's theorem here again. But first check if 2 and 400 are coprime. Well, 400 is 2^4*5^2, so GCD(2,400)=2, which is not 1. Therefore, Euler's theorem doesn't apply here. Alternatively, perhaps we can use the Chinese Remainder Theorem (CRT) by splitting modulo 16 and modulo 25, since 16*25=400 and they are coprime. Then compute 2^2001 mod 16 and mod 25, then combine the results. Let's try that. First, compute 2^2001 mod 16. Since 16 is 2^4, and the exponent 2001 is large. For powers of 2 modulo 16, note that 2^4 = 16 ≡ 0 mod 16. Therefore, any higher power of 2 will also be 0 mod 16. So, 2^2001 mod 16 = 0. Now compute 2^2001 mod 25. Since 25 is coprime with 2, we can use Euler's theorem here. φ(25) = 25*(1 - 1/5) = 20. Therefore, 2^20 ≡ 1 mod 25. Therefore, 2^2001 = 2^(20*100 + 1) = (2^20)^100 * 2^1 ≡ 1^100 * 2 ≡ 2 mod 25. So now, we have 2^2001 ≡ 0 mod 16 and 2^2001 ≡ 2 mod 25. We need to find a number x such that x ≡ 0 mod 16 and x ≡ 2 mod 25. To solve this system using CRT. Let x = 16k. Then 16k ≡ 2 mod 25 => 16k ≡ 2 mod 25. Multiply both sides by the modular inverse of 16 mod 25. Let's find the inverse of 16 modulo 25. We need an integer m such that 16m ≡ 1 mod 25. Testing m=11: 16*11=176. 176 mod 25: 25*7=175, so 176-175=1. So 16*11 ≡ 1 mod 25. So inverse of 16 mod 25 is 11. Therefore, k ≡ 2*11 mod 25 => k ≡ 22 mod 25. Therefore, k=25m +22 for some integer m. Therefore, x=16k=16*(25m +22)=400m + 352. Therefore, the smallest positive solution is x=352 mod 400. Therefore, 2^2001 ≡ 352 mod 400. So going back, we had 2002^2001 ≡ 2^2001 ≡ 352 mod 400. Therefore, the exponent in the original problem is 352. Therefore, we need to compute 3^352 mod 1000. Now, compute 3^352 mod 1000. This is still a large exponent. Let's see if we can find a pattern or use exponentiation by squaring. First, note that φ(1000)=400, and since 3 and 1000 are coprime, 3^400 ≡ 1 mod 1000. But 352 is less than 400, so maybe we can compute 3^352 directly using exponentiation by squaring. Alternatively, we can compute 3^352 mod 8 and mod 125, then use CRT to combine the results. Since 1000=8*125, and 8 and 125 are coprime. Let's try that. First, compute 3^352 mod 8. Since 3 mod 8 is 3. 3^1=3 mod8, 3^2=1 mod8, 3^3=3 mod8, etc. So the cycle is every 2 exponents. Since 352 is even, 3^352=(3^2)^176 ≡ 1^176 ≡1 mod8. Next, compute 3^352 mod 125. This is more involved. Since 125 is 5^3, and 3 and 5 are coprime, Euler's theorem applies. φ(125)=125*(1 -1/5)=100. Therefore, 3^100 ≡1 mod125. So 3^352=3^(100*3 +52)= (3^100)^3 *3^52 ≡1^3 *3^52≡3^52 mod125. So compute 3^52 mod125. Let's compute this step by step using exponentiation by squaring. Compute 3^1=3 mod1253^2=9 mod1253^4=(3^2)^2=81 mod1253^8=(3^4)^2=81^2=6561 mod125. Let's divide 6561 by 125: 125*52=6500, 6561-6500=61. So 3^8≡61 mod1253^16=(3^8)^2=61^2=3721 mod125. 125*29=3625, 3721-3625=96. So 3^16≡96 mod1253^32=(3^16)^2=96^2=9216 mod125. 125*73=9125, 9216-9125=91. So 3^32≡91 mod125Now, 3^52=3^32 * 3^16 *3^4. Let's compute that:3^32=91, 3^16=96, 3^4=81So 91*96=8736 mod125. 125*69=8625, 8736-8625=111. So 91*96≡111 mod125Then multiply by 81: 111*81=8991 mod125. 125*71=8875, 8991-8875=116. So 3^52≡116 mod125. So, putting it together, 3^352 ≡116 mod125. Now, we have 3^352 ≡1 mod8 and 3^352≡116 mod125. We need to find a number x such that x≡1 mod8 and x≡116 mod125. Let x=125k +116. Then this must satisfy 125k +116 ≡1 mod8. Compute 125 mod8: 125 /8=15*8=120, remainder 5. So 125 ≡5 mod8. 116 mod8: 116/8=14*8=112, remainder 4. So:125k +116 ≡5k +4 ≡1 mod8 =>5k ≡1 -4= -3 ≡5 mod8. Therefore, 5k≡5 mod8. Multiply both sides by inverse of 5 mod8. The inverse of 5 mod8 is 5, since 5*5=25≡1 mod8. Therefore, k≡5*5=25≡1 mod8. Therefore, k≡1 mod8. So k=8m +1 for some integer m. Therefore, x=125*(8m +1) +116=1000m +125 +116=1000m +241. Therefore, the smallest positive solution is x=241 mod1000. Therefore, 3^352 ≡241 mod1000. Therefore, the last three digits of 2003^(2002^2001) are 241. Wait, but let me double-check all the steps to ensure no miscalculations. First, confirming 2003 mod1000 is 3, correct. Then, φ(1000)=400, correct. Then, exponent 2002^2001 mod400. 2002 mod400=2, so 2^2001 mod400. Then using CRT on mod16 and mod25. 2^2001 mod16: since 2^4=16≡0, any higher exponent is 0 mod16. Correct. 2^2001 mod25: φ(25)=20, so exponent 2001 mod20=1, so 2^1=2 mod25. Correct. Then solving x≡0 mod16 and x≡2 mod25 gives x=352 mod400. Correct. Then, 3^352 mod1000. Splitting into mod8 and mod125. 3^352 mod8: since 3^2=1 mod8, even exponent gives 1. Correct. 3^352 mod125: 3^100≡1 mod125, so 3^352=3^(3*100 +52)=3^52 mod125. Then, computing 3^52 step by step. Let me check the exponentials again:3^1=33^2=93^4=813^8=81^2=6561 mod125: 6561 - 125*52=6561 -6500=61, correct.3^16=61^2=3721 mod125: 3721 -125*29=3721 -3625=96, correct.3^32=96^2=9216 mod125: 9216 -125*73=9216 -9125=91, correct.3^52=3^32*3^16*3^4=91*96*81. First 91*96=8736 mod125: 8736 -125*69=8736 -8625=111, correct. Then 111*81=8991 mod125: 8991 -125*71=8991 -8875=116, correct. So 3^52≡116 mod125. Correct.Then, solving x≡1 mod8 and x≡116 mod125. x=125k +116. Then 125k +116 ≡5k +4 ≡1 mod8 =>5k≡-3≡5 mod8. So k≡1 mod8. Therefore, x=125*(8m +1)+116=1000m +241. So x≡241 mod1000. Hence, the final result is 241. Wait, but let me verify 3^352 mod1000 using another method, just to be sure. Maybe using exponentiation by squaring directly. Compute 3^352 mod1000. Let's compute this step by step. First, note that 3^1=33^2=93^4=813^8=81^2=6561 mod1000=5613^16=(561)^2=314721 mod1000=7213^32=(721)^2=519841 mod1000=8413^64=(841)^2=707281 mod1000=2813^128=(281)^2=789^61 mod1000= 78961 mod1000=9613^256=(961)^2=923521 mod1000=521Now, 3^352=3^256 *3^64 *3^32. Let's compute each term:3^256=5213^64=2813^32=841First, multiply 521 and 281:521*281: Let's compute step by step. 500*281=140500, 21*281=5901. Sum:140500 +5901=146401. 146401 mod1000=401So 521*281≡401 mod1000Next, multiply this by 841:401*841. Let's compute 400*841=336,400. 1*841=841. Sum:336,400 +841=337,241. 337,241 mod1000=241.Therefore, 3^352 mod1000=241. This matches the previous result. So that's reassuring. Therefore, the last three digits of 2003^(2002^2001) are 241. Final Answerboxed{241}</think>To find the last three digits of the number (2003^{2002^{2001}}), we need to compute (2003^{2002^{2001}} mod 1000). First, simplify the base modulo 1000:[2003 equiv 3 mod 1000]Thus, we need to compute (3^{2002^{2001}} mod 1000).Using Euler's theorem, since (3) and (1000) are coprime, we have:[3^{phi(1000)} equiv 1 mod 1000]where (phi(1000) = 1000 left(1 - frac{1}{2}right)left(1 - frac{1}{5}right) = 400). Therefore, we need the exponent (2002^{2001} mod 400).Simplify (2002 mod 400):[2002 equiv 2 mod 400]Thus, we need (2^{2001} mod 400). Using the Chinese Remainder Theorem, we split this into modulo 16 and 25:- (2^{2001} equiv 0 mod 16)- (2^{2001} equiv 2 mod 25) (using Euler's theorem and solving the exponent)Combining these results using CRT, we find:[2^{2001} equiv 352 mod 400]Now, compute (3^{352} mod 1000). Using CRT again, we split this into modulo 8 and 125:- (3^{352} equiv 1 mod 8)- (3^{352} equiv 116 mod 125) (using exponentiation by squaring)Combining these results using CRT, we find:[3^{352} equiv 241 mod 1000]Thus, the last three digits of (2003^{2002^{2001}}) are (boxed{241}).,<think>Okay, so I need to find the last three digits of the number 2003 raised to the power of 2002 raised to the power of 2001. That is, compute 2003^(2002^2001) mod 1000. Hmm, this seems like a modular exponentiation problem. Let me recall how to approach such problems.First, to find the last three digits of a number, we need to compute the number modulo 1000. So, the task is to compute 2003^2002^2001 mod 1000. Since 2003 mod 1000 is 3, because 2000 is divisible by 1000 and 2000 + 3 mod 1000 is 3. Therefore, the problem reduces to finding 3^(2002^2001) mod 1000.Now, computing such a huge exponent directly is impossible, so I need to find a way to simplify this. I remember that Euler's theorem could be useful here. Euler's theorem states that if a and n are coprime, then a^φ(n) ≡ 1 mod n, where φ(n) is Euler's totient function. So, perhaps I can use this theorem to reduce the exponent.First, check if 3 and 1000 are coprime. Since 3 is a prime number and 3 does not divide 1000 (which factors into 2^3 * 5^3), they are indeed coprime. Therefore, φ(1000) = 1000 * (1 - 1/2) * (1 - 1/5) = 1000 * 1/2 * 4/5 = 1000 * 2/5 = 400. So φ(1000) = 400. Therefore, 3^400 ≡ 1 mod 1000.That means that 3^k ≡ 3^(k mod 400) mod 1000. Therefore, to compute 3^(2002^2001) mod 1000, we can first compute the exponent 2002^2001 mod 400, then take 3 raised to that result modulo 1000.So now the problem reduces to finding 2002^2001 mod 400. Let's compute that.First, note that 2002 mod 400. Since 2000 is divisible by 400 (as 2000 / 400 = 5), 2002 mod 400 = 2002 - 5*400 = 2002 - 2000 = 2. Therefore, 2002 ≡ 2 mod 400. Therefore, 2002^2001 ≡ 2^2001 mod 400.So now we need to compute 2^2001 mod 400. Hmm, 2^2001 is a huge exponent. Let's see if we can find a pattern or use Euler's theorem again here. But first, check if 2 and 400 are coprime. However, 400 factors into 2^4 * 5^2, and since 2 divides 400, they are not coprime. Therefore, Euler's theorem does not apply here. Alternatively, we can use the Chinese Remainder Theorem by splitting modulus 400 into prime power components.But perhaps we can compute 2^2001 mod 16 and 2^2001 mod 25, then combine them using the Chinese Remainder Theorem. Since 400 = 16 * 25, and 16 and 25 are coprime.First, compute 2^2001 mod 16:Note that 2^4 = 16 ≡ 0 mod 16. Therefore, for exponents ≥ 4, 2^n ≡ 0 mod 16. Since 2001 is much larger than 4, 2^2001 ≡ 0 mod 16.Now compute 2^2001 mod 25. Here, 2 and 25 are coprime, so Euler's theorem applies. φ(25) = 25 * (1 - 1/5) = 20. Therefore, 2^20 ≡ 1 mod 25. Therefore, 2^2001 = 2^(20*100 + 1) = (2^20)^100 * 2^1 ≡ 1^100 * 2 ≡ 2 mod 25.Therefore, 2^2001 ≡ 0 mod 16 and 2^2001 ≡ 2 mod 25. Now, we need to find a number x such that:x ≡ 0 mod 16x ≡ 2 mod 25Let x = 16k. Then, 16k ≡ 2 mod 25. So 16k ≡ 2 mod 25. Multiply both sides by the inverse of 16 mod 25. Let's find the inverse of 16 mod 25. 16 * 11 = 176 ≡ 176 - 7*25 = 176 - 175 = 1 mod 25. So inverse of 16 mod 25 is 11. Therefore, k ≡ 2*11 ≡ 22 mod 25. Therefore, k = 25m + 22. Therefore, x = 16*(25m + 22) = 400m + 352. Therefore, the smallest positive solution is x = 352 mod 400. Therefore, 2^2001 ≡ 352 mod 400.Therefore, the exponent 2002^2001 ≡ 2^2001 ≡ 352 mod 400. Therefore, going back to the original problem, 3^(2002^2001) ≡ 3^352 mod 1000.So now we need to compute 3^352 mod 1000. Again, this is a large exponent, but perhaps we can compute it using exponentiation by squaring or break it down using modulus properties.Alternatively, since 1000 = 8 * 125, and 8 and 125 are coprime, we can compute 3^352 mod 8 and 3^352 mod 125, then combine them via Chinese Remainder Theorem.First, compute 3^352 mod 8:3 mod 8 = 3. 3^1 = 3 mod 8, 3^2 = 9 ≡ 1 mod 8. So 3^2 ≡ 1 mod 8. Therefore, 3^352 = (3^2)^176 ≡ 1^176 ≡ 1 mod 8.Next, compute 3^352 mod 125. This is more involved. Since 3 and 125 are coprime, Euler's theorem applies. φ(125) = 125 * (1 - 1/5) = 100. Therefore, 3^100 ≡ 1 mod 125. Therefore, 3^352 = 3^(100*3 + 52) = (3^100)^3 * 3^52 ≡ 1^3 * 3^52 ≡ 3^52 mod 125.So need to compute 3^52 mod 125. Let's compute this step by step using exponentiation by squaring:First, compute 3^1 = 3 mod 1253^2 = 9 mod 1253^4 = (3^2)^2 = 81 mod 1253^8 = (3^4)^2 = 81^2 = 6561 mod 125. Let's compute 6561 / 125: 125*52 = 6500, so 6561 - 6500 = 61. Therefore, 3^8 ≡ 61 mod 125.3^16 = (3^8)^2 = 61^2 = 3721 mod 125. 125*29 = 3625. 3721 - 3625 = 96. Therefore, 3^16 ≡ 96 mod 125.3^32 = (3^16)^2 = 96^2 = 9216 mod 125. 125*73 = 9125, so 9216 - 9125 = 91. Therefore, 3^32 ≡ 91 mod 125.Now, 3^52 = 3^32 * 3^16 * 3^4. Let's compute each part:3^32 ≡ 91 mod 1253^16 ≡ 96 mod 1253^4 ≡ 81 mod 125Multiply them step by step:First, 91 * 96 mod 125. 91*96 = (90*96) + (1*96) = 8640 + 96 = 8736. 8736 mod 125. 125*69 = 8625. 8736 - 8625 = 111. So 91*96 ≡ 111 mod 125.Then, 111 * 81 mod 125. 111*80 = 8880, 111*1 = 111. So total 8880 + 111 = 8991. 8991 mod 125. 125*71 = 8875. 8991 - 8875 = 116. So 111*81 ≡ 116 mod 125.Therefore, 3^52 ≡ 116 mod 125.Therefore, 3^352 ≡ 116 mod 125.So now we have:3^352 ≡ 1 mod 83^352 ≡ 116 mod 125We need to find x such that x ≡ 1 mod 8 and x ≡ 116 mod 125. Then x ≡ ... mod 1000.Let’s solve this system of congruences. Let x = 125k + 116. Then, x ≡ 125k + 116 ≡ 5k + 4 mod 8 (since 125 mod 8 is 5 and 116 mod 8 is 4). We need 5k + 4 ≡ 1 mod 8. So, 5k ≡ -3 ≡ 5 mod 8. Therefore, 5k ≡ 5 mod 8. Multiply both sides by inverse of 5 mod 8. Since 5*5=25≡1 mod8, so inverse is 5. Therefore, k ≡5*5 ≡25≡1 mod8. Therefore, k =8m +1. Thus, x =125*(8m +1)+116=1000m +125 +116=1000m +241. Therefore, the smallest positive solution is x=241 mod1000. Therefore, 3^352 ≡241 mod1000.Therefore, the last three digits of 2003^2002^2001 are 241.Wait, but let me verify this step by step because I might have made a mistake somewhere. Let me check the calculations again.First, 2003 mod 1000 is indeed 3. Then, we need to compute 3^(2002^2001) mod 1000. So we compute 2002^2001 mod φ(1000)=400. 2002 mod400=2, so exponent is 2^2001 mod400. Then, we found that 2^2001 mod400=352. Then, 3^352 mod1000. Then, split into mod8 and mod125. 3^352 mod8=1, which is correct because 3^2≡1 mod8, so even exponents would be 1. Then mod125, 3^352=3^(3*100 +52)= (3^100)^3 *3^52≡1^3 *3^52≡3^52 mod125. Then 3^52 computed as 3^32 *3^16 *3^4: 91*96*81 mod125. Then 91*96=8736 mod125=8736-69*125=8736-8625=111 mod125. Then 111*81=8991 mod125=8991-71*125=8991-8875=116 mod125. So that's correct.Then solving x≡1 mod8 and 116 mod125. x=125k +116. Then 125k +116≡5k +4≡1 mod8. 5k≡-3≡5 mod8. Multiply by 5 inverse which is 5, so k≡25≡1 mod8. Hence k=8m+1. x=125*(8m+1)+116=1000m+241. So x≡241 mod1000. Therefore, the last three digits are 241.Hmm, but let me check if 3^352 is indeed 241 mod1000. Let me compute 3^10=59049. 59049 mod1000=49. Then 3^20=(3^10)^2=49^2=2401 mod1000=401. 3^40=(3^20)^2=401^2=160801 mod1000=801. 3^80=(3^40)^2=801^2=641601 mod1000=601. 3^160=(3^80)^2=601^2=361201 mod1000=201. 3^320=(3^160)^2=201^2=40401 mod1000=401.Now, 3^352=3^320 *3^32. 3^32= (3^16)^2. Let's compute 3^16:3^1=33^2=93^4=813^8=81^2=6561 mod1000=5613^16=561^2. 561^2: 500^2=250000, 2*500*61=61000, 61^2=3721. Total=250000+61000+3721=314,721. 314721 mod1000=721. So 3^16=721 mod1000. Then 3^32=(3^16)^2=721^2=519,841 mod1000=841. So 3^32=841 mod1000.Therefore, 3^352=3^320 *3^32=401 *841 mod1000. Let's compute 401*841. 400*800=320,000; 400*41=16,400; 1*800=800; 1*41=41. So total is 320,000 +16,400 +800 +41=337,241. 337,241 mod1000=241. Yes, so 3^352 mod1000=241. Therefore, the answer is 241. So the last three digits are 241.Therefore, after verifying step by step, I think the answer is indeed 241. I might have made a mistake earlier in the calculation of 3^352 mod125, but cross-checking via modulus 1000 using exponentiation by squaring shows that 3^352 mod1000 is 241. Therefore, the last three digits of 2003^(2002^2001) are 241. So I think the answer is boxed{241}.Final Answerboxed{241}</think>To find the last three digits of the number (2003^{2002^{2001}}), we need to compute (2003^{2002^{2001}} mod 1000). First, simplify the base modulo 1000:[2003 equiv 3 mod 1000]Thus, the problem reduces to finding (3^{2002^{2001}} mod 1000).Using Euler's theorem, we know that if (a) and (n) are coprime, then (a^{phi(n)} equiv 1 mod n). Here, (phi(1000) = 400), so:[3^{400} equiv 1 mod 1000]We need to compute the exponent (2002^{2001} mod 400). First, simplify the base modulo 400:[2002 equiv 2 mod 400]Thus, we need to compute (2^{2001} mod 400). Using the Chinese Remainder Theorem, we split this into modulo 16 and modulo 25:- (2^{2001} equiv 0 mod 16) (since (2^4 = 16))- (2^{2001} mod 25): Using Euler's theorem, (phi(25) = 20), so (2^{20} equiv 1 mod 25). Therefore, (2^{2001} = 2^{20 cdot 100 + 1} equiv 2 mod 25).Combining these results using the Chinese Remainder Theorem, we solve:[x equiv 0 mod 16 quad text{and} quad x equiv 2 mod 25]The solution is (x equiv 352 mod 400). Thus, (2002^{2001} equiv 352 mod 400).Now, we need to compute (3^{352} mod 1000). Using the Chinese Remainder Theorem again, we split this into modulo 8 and modulo 125:- (3^{352} equiv 1 mod 8) (since (3^2 equiv 1 mod 8))- (3^{352} mod 125): Using Euler's theorem, (phi(125) = 100), so (3^{100} equiv 1 mod 125). Therefore, (3^{352} = 3^{100 cdot 3 + 52} equiv 3^{52} mod 125). Through step-by-step computation, we find (3^{52} equiv 116 mod 125).Combining these results using the Chinese Remainder Theorem, we solve:[x equiv 1 mod 8 quad text{and} quad x equiv 116 mod 125]The solution is (x equiv 241 mod 1000).Thus, the last three digits of (2003^{2002^{2001}}) are (boxed{241}).]

🔑:We must find the remainder when 2003^{2002^{2001}} is divided by 1000, which will be the same as the remainder when 3^{2002^{2001}} is divided by 1000, since 2003 equiv 3(bmod 1000). To do this we will first find a positive integer n such that 3^{n} equiv 1(bmod 1000) and then try to express 2002^{2001} in the form n k+r, so that2003^{2002^{2001}} equiv 3^{n k+r} equivleft(3^{n}right)^{k} cdot 3^{r} equiv 1^{k} cdot 3^{r} equiv 3^{r}(bmod 1000)Since 3^{2}=10-1, we can evaluate 3^{2 m} using the binomial theorem:3^{2 m}=(10-1)^{m}=(-1)^{m}+10 m(-1)^{m-1}+100 frac{m(m-1)}{2}(-1)^{m-2}+cdots+10^{m} .After the first 3 terms of this expansion, all remaining terms are divisible by 1000, so letting m=2 q, we have that3^{4 q} equiv 1-20 q+100 q(2 q-1)(bmod 1000)Using this, we can check that 3^{100} equiv 1(bmod 1000) and now we wish to find the remainder when 2002^{2001} is divided by 100.Now 2002^{2001} equiv 2^{2001}(bmod 100) equiv 4 cdot 2^{1999}(bmod 4 cdot 25), so we'll investigate powers of 2 modulo 25. Noting that 2^{10}=1024 equiv-1(bmod 25), we have2^{1999}=left(2^{10}right)^{199} cdot 2^{9} equiv(-1)^{199} cdot 512 equiv-12 equiv 13(bmod 25) .Thus 2^{2001} equiv 4 cdot 13=52(bmod 100). Therefore 2002^{2001} can be written in the form 100 k+52 for some integer k, so2003^{2002^{2001}} equiv 3^{52}(bmod 1000) equiv 1-20 cdot 13+1300 cdot 25 equiv 241(bmod 1000)using equation (1). So the last 3 digits of 2003^{2002^{2001}} are 241.

🧠:[<think>Okay, so I have this inequality to solve: 9.254 times the logarithm base 3 of (logarithm base 0.2 of (logarithm base 32 of ((x-1)/(x+5))) is greater than 0. Hmm, that looks a bit complicated with multiple logarithms nested inside each other. Let me break it down step by step.First, I need to remember the properties of logarithms and inequalities. Since logarithms can be tricky, especially with different bases and nested logs, I should start from the outermost logarithm and work my way inwards. Also, I need to consider the domains of each logarithm because the arguments must be positive.Let me rewrite the inequality for clarity:9.254 * log₃ [ log₀.₂ ( log₃₂ ( (x - 1)/(x + 5) ) ) ] > 0.Since 9.254 is a positive constant, I can divide both sides by 9.254 without changing the inequality direction. That simplifies it to:log₃ [ log₀.₂ ( log₃₂ ( (x - 1)/(x + 5) ) ) ] > 0.Now, log base 3 of something is greater than 0. Remember that log_b(a) > 0 when a > 1 if b > 1. Since the base here is 3, which is greater than 1, the argument inside the log must be greater than 1. So:log₀.₂ ( log₃₂ ( (x - 1)/(x + 5) ) ) > 1.Wait, log base 0.2. Hmm, base 0.2 is the same as 1/5, which is a fraction between 0 and 1. So logarithm with a base between 0 and 1 behaves differently. For log_b(a) > c when 0 < b < 1, the inequality flips when we exponentiate. Specifically, log_b(a) > c implies that a < b^c. Let me confirm that.Yes, because logarithm with a base less than 1 is a decreasing function. So if log_b(a) > c, then a < b^c. So applying that here:log₀.₂ ( log₃₂ ( (x - 1)/(x + 5) ) ) > 1implies that the argument of log₀.₂ is less than (0.2)^1 = 0.2.Therefore:log₃₂ ( (x - 1)/(x + 5) ) < 0.2.Wait, 0.2 is the same as 1/5. But base 32. Let me check. log base 32 of something is less than 1/5. So again, since 32 is greater than 1, the logarithm function is increasing. Therefore, log₃₂(A) < 1/5 implies that A < 32^(1/5).Calculating 32^(1/5). Since 32 is 2^5, so 32^(1/5) = 2. So:log₃₂ ( (x - 1)/(x + 5) ) < 1/5implies:(x - 1)/(x + 5) < 32^(1/5) = 2.So now, the inequality becomes:(x - 1)/(x + 5) < 2.But before solving that, I need to check the domain of each logarithm. Starting from the innermost one:1. The argument of log₃₂ must be positive: (x - 1)/(x + 5) > 0.2. The argument of log₀.₂ must be positive: log₃₂( (x -1)/(x + 5) ) > 0.3. The argument of log₃ must be positive: log₀.₂( log₃₂( (x -1)/(x + 5) ) ) > 0.But maybe I should handle the domain step by step as I go through each logarithm.First, the innermost logarithm: log₃₂( (x -1)/(x +5) ). For this to be defined, (x -1)/(x +5) must be > 0. Let's solve this inequality.(x -1)/(x +5) > 0.This occurs when both numerator and denominator are positive or both are negative.Case 1: x -1 > 0 and x +5 > 0. So x >1 and x > -5. Which simplifies to x >1.Case 2: x -1 <0 and x +5 <0. So x <1 and x < -5. Which simplifies to x < -5.So the domain here is x ∈ (-∞, -5) ∪ (1, ∞).Next, the result of log₃₂( (x -1)/(x +5) ) must be positive for the next logarithm log₀.₂ to be defined. Wait, actually, the next logarithm is log₀.₂ of that result. So log₀.₂( log₃₂( (x -1)/(x +5) ) ). For log₀.₂(arg) to be defined, arg must be positive. So log₃₂( (x -1)/(x +5) ) > 0.But log base 32 of something is >0 when the argument is >1, because log_b(a) >0 when a>1 if b>1. Since 32>1, log₃₂(A) >0 implies A>1. Therefore:(x -1)/(x +5) >1.Solve this inequality:(x -1)/(x +5) >1Subtract 1 from both sides:(x -1)/(x +5) -1 >0Combine terms:[(x -1) - (x +5)]/(x +5) >0Simplify numerator:x -1 -x -5 = -6So -6/(x +5) >0Which is equivalent to:-6/(x +5) >0Multiply both sides by -1 (inequality flips):6/(x +5) <0Which implies that x +5 <0 => x < -5.So the domain for log₀.₂( log₃₂( (x -1)/(x +5) ) ) to be defined is x < -5. But wait, original domain from the first step was x ∈ (-∞, -5) ∪ (1, ∞). But here, it requires x < -5. So the domain here is x < -5.But wait, let me check. If x < -5, then (x -1)/(x +5) is positive, as we saw in the first domain step. But now, log₃₂( (x -1)/(x +5) ) >1. Because log₃₂(A) >0 requires A >1. So A = (x -1)/(x +5) >1. Which led us to x < -5. So yes, combining both conditions, the domain here is x < -5.Therefore, the domain up to this point is x < -5.Now, moving up, the next logarithm is log₃ of something. The argument must be positive. So log₀.₂( log₃₂( (x -1)/(x +5) ) ) >0. Wait, no. Wait, the entire expression inside log₃ is log₀.₂( log₃₂( (x -1)/(x +5) ) ). For log₃(arg) to be defined, arg must be positive. But since log₀.₂(...) is the argument, log₀.₂(...) must be positive. However, log₀.₂(...) is positive when its argument is between 0 and 1, because log base 0.2 (which is less than 1) is positive when the argument is between 0 and 1. Let me verify that.For log_b(a) where 0 < b <1, log_b(a) >0 when 0 < a <1. Because log base 0.2 of a number between 0 and1 is positive. For example, log_0.2(0.5) is positive since 0.2^k =0.5 would require k positive. Because 0.2^k decreases as k increases. Wait, maybe it's better to recall the graph. When the base is between 0 and1, the logarithm function is decreasing. So log_b(a) is positive when a <1. So if log_b(a) >0, then a <1 because log_b(1)=0, and since it's decreasing, for a <1, log_b(a) >0.Therefore, log₀.₂( log₃₂( (x -1)/(x +5) ) ) >0 implies that log₃₂( (x -1)/(x +5) ) <1.But wait, we already had from the previous step that log₃₂( (x -1)/(x +5) ) >1 (from ensuring log₀.₂'s argument is positive). Wait, now I might be confused.Wait, let's recap:First, the innermost log requires (x -1)/(x +5) >0, which gives x < -5 or x >1.Then, log₃₂( (x -1)/(x +5) ) must be >0 for log₀.₂ to be defined. log₃₂(A) >0 implies A >1, so (x -1)/(x +5) >1, leading to x < -5.So the domain is x < -5.Then, log₀.₂( log₃₂( (x -1)/(x +5) ) ) must be the argument of log₃. For log₃ to be defined, its argument must be positive. So log₀.₂( log₃₂( (x -1)/(x +5) ) ) >0.But since base 0.2 is less than1, log₀.₂(B) >0 implies B <1. So log₃₂( (x -1)/(x +5) ) <1.But earlier, we had log₃₂( (x -1)/(x +5) ) >1 to make log₀.₂'s argument positive. Wait, this seems conflicting.Wait, let me go through this again carefully.First, log₃₂( (x -1)/(x +5) ) must be positive because it's the argument of log₀.₂. Since log₀.₂ requires its argument to be positive. Therefore, log₃₂( (x -1)/(x +5) ) >0. But log base32 (which is >1) of something is >0 when that something is >1. So (x -1)/(x +5) >1. As before, which gives x < -5.But then, the next layer is log₀.₂( log₃₂(...) ). Since log₀.₂'s argument is log₃₂(...) which is >1, but log₀.₂ is defined for positive arguments, which is satisfied here. Then, log₀.₂(B) where B >1. Since base 0.2 is less than1, log₀.₂(B) is negative if B >1. Because log_b(B) for B >1 and 0 <b <1 is negative. For example, log_0.2(2) is negative because 0.2^k =2 implies k is negative.Therefore, log₀.₂( log₃₂(...) ) would be negative here. But then, this negative result is the argument of log₃. But log₃ requires its argument to be positive. Therefore, log₀.₂( log₃₂(...) ) must be positive. But if log₀.₂(B) is positive, then B must be <1. However, earlier, we required B = log₃₂(...) >1. There's a contradiction here. Which suggests that there is no solution?Wait, this is a problem. Let me check again:Original inequality:9.254 log₃ [ log₀.₂ ( log₃₂ ( (x - 1)/(x + 5) ) ) ] >0.For the entire expression to be defined:1. (x -1)/(x +5) >0 ⇒ x < -5 or x >1.2. log₃₂( (x -1)/(x +5) ) >0 ⇒ (x -1)/(x +5) >1 ⇒ x < -5.3. log₀.₂( log₃₂( (x -1)/(x +5) ) ) >0 ⇒ log₃₂( (x -1)/(x +5) ) <1 (since base 0.2 <1).But from step 2, log₃₂( (x -1)/(x +5) ) >1. So steps 2 and 3 are conflicting. Therefore, there is no x that satisfies both log₃₂(...) >1 and log₃₂(...) <1. Hence, the domain is empty. Therefore, the original inequality has no solution.But wait, the original problem states that the inequality is greater than 0, but maybe there's a mistake in my reasoning. Let me verify each step again.Starting over:1. The innermost expression: (x -1)/(x +5) >0 ⇒ x < -5 or x >1.2. Next layer: log₃₂( (x -1)/(x +5) ) must be defined and positive. Defined when (x -1)/(x +5) >0 (already satisfied). Positive when (x -1)/(x +5) >1 ⇒ x < -5.3. Next layer: log₀.₂( log₃₂( (x -1)/(x +5) ) ) must be defined and positive. Defined when log₃₂(...) >0 (already satisfied). Positive when log₃₂(...) <1, since base 0.2 is less than1. Therefore, log₀.₂(B) >0 ⇒ B <1.But log₃₂( (x -1)/(x +5) ) <1 ⇒ (x -1)/(x +5) <32^1=32.But (x -1)/(x +5) <32. Let's solve this inequality.(x -1)/(x +5) <32.Multiply both sides by (x +5). But x < -5, so x +5 is negative. Therefore, multiplying by a negative number reverses the inequality:x -1 >32(x +5).Solve for x:x -1 >32x +160x -32x >160 +1-31x >161Divide both sides by -31, which flips the inequality:x < -161/31 ≈ -5.1935.But from step 2, we already have x < -5. So combining x < -5 and x < -161/31 ≈ -5.1935, the domain is x < -161/31.But wait, 161 divided by31: 31*5=155, 161-155=6, so 161/31=5 +6/31≈5.1935. So -161/31≈-5.1935.So the domain now is x < -5.1935.But then, we also have the requirement from step3 that log₀.₂(B) >0 implies B <1. Wait, but in step2, B = log₃₂(...) >1, but step3 requires B <1. Contradiction?Wait, no. Wait, let's clarify:Wait, step2: log₃₂( (x -1)/(x +5) ) >1 implies (x -1)/(x +5) >32^1=32. Therefore, (x -1)/(x +5) >32. But solving this:(x -1)/(x +5) >32.Since x < -5 (from step1), x +5 <0. Multiply both sides:x -1 <32(x +5)x -1 <32x +160-31x <161x > -161/31 ≈-5.1935.But x < -5 from step1. Therefore, combining x < -5 and x > -5.1935, we have x ∈ (-5.1935, -5).But wait, this contradicts the previous conclusion that (x -1)/(x +5) >1 leads to x < -5. But perhaps I made a mistake here.Wait, let me redo step2:log₃₂( (x -1)/(x +5) ) >0 ⇒ (x -1)/(x +5) >1.Solving (x -1)/(x +5) >1:Subtract 1: [(x -1) - (x +5)]/(x +5) >0 ⇒ (-6)/(x +5) >0.This is equivalent to (x +5) <0 ⇒ x < -5. So step2 gives x < -5.Then step3: log₀.₂( log₃₂( (x -1)/(x +5) ) ) >0 ⇒ log₃₂( (x -1)/(x +5) ) <1.Which implies (x -1)/(x +5) <32^1=32.Solve (x -1)/(x +5) <32, given that x < -5.Multiply both sides by (x +5), which is negative, so inequality flips:x -1 >32(x +5)x -1 >32x +160-31x >161x < -161/31 ≈-5.1935.So combining x < -5 (from step2) and x < -5.1935 (from step3), the domain is x < -5.1935.Therefore, the domain for this step is x < -161/31 ≈-5.1935.Now, moving up, the argument of log₃ is log₀.₂(...) which must be positive (already handled in step3). Then, the original inequality after dividing by 9.254:log₃ [ log₀.₂ ( log₃₂ ( (x - 1)/(x +5) ) ) ] >0.Since log base3 of something >0 implies that something >1 (because base3>1). So:log₀.₂ ( log₃₂ ( (x -1)/(x +5) ) ) >1.But log₀.₂(B) >1. Since base0.2 is less than1, log₀.₂(B) >1 implies B <0.2^1=0.2.Because log_b(a) >c when b<1 implies a <b^c.Therefore, log₀.₂(B) >1 ⇒ B <0.2.Here, B is log₃₂( (x -1)/(x +5) ).Therefore:log₃₂( (x -1)/(x +5) ) <0.2.Now, log base32 of something <0.2. Since base32>1, log₃₂(A) <0.2 implies A <32^0.2.Calculate 32^0.2: since 32=2^5, 32^0.2=2^(5*0.2)=2^1=2.Therefore, log₃₂( (x -1)/(x +5) ) <0.2 ⇒ (x -1)/(x +5) <2.Solve this inequality with x < -161/31 ≈-5.1935.Again, (x -1)/(x +5) <2.Multiply both sides by (x +5), which is negative (since x < -5.1935), so inequality flips:x -1 >2(x +5)x -1 >2x +10- x >11x < -11.So combining with previous domain x < -5.1935, now x must be < -11.But wait, let's check.(x -1)/(x +5) <2.For x < -5.1935, x +5 is negative.So let's solve:(x -1)/(x +5) <2.Subtract 2:(x -1)/(x +5) -2 <0Combine terms:(x -1 -2x -10)/(x +5) <0(-x -11)/(x +5) <0Multiply numerator and denominator by -1 (inequality flips):(x +11)/(x +5) >0So this inequality holds when both numerator and denominator are positive or both negative.Numerator x +11 >0 ⇒x >-11.Denominator x +5 >0 ⇒x >-5.So both positive when x >-5, but our domain is x < -5.1935, so this case doesn't apply.Both negative: x +11 <0 and x +5 <0 ⇒x < -11 and x < -5. Which simplifies to x < -11.So the solution is x < -11.Therefore, combining with the previous domain x < -5.1935, the solution is x < -11.But wait, x < -11 is already less than -5.1935 ≈-5.1935, so the solution is x < -11.Therefore, the final solution is x ∈ (-∞, -11).But we need to check if this satisfies all previous domain restrictions.Check x < -11.1. (x -1)/(x +5): numerator x -1 < -12, denominator x +5 < -6. So (negative)/(negative)=positive. Okay.2. log₃₂( (x -1)/(x +5) ) >0: (x -1)/(x +5) >1. Let's check with x < -11:Take x = -12: ( -12 -1 )/( -12 +5 ) = (-13)/(-7) ≈1.857 >1. Okay.3. log₀.₂( log₃₂( (x -1)/(x +5) ) ) >0: log₃₂(...) <1. For x = -12, log₃₂(13/7) ≈ log₃₂(1.857). Since 32^0.2=2, so log₃₂(2)=0.2. Since 1.857 <2, log₃₂(1.857) <0.2. Therefore, log₀.₂( <0.2 ). Since base0.2, log₀.₂(0.2) =1. But if it's less than0.2, log₀.₂( something <0.2 ) >1. For example, log₀.₂(0.1)= log_{1/5}(1/10)= ln(1/10)/ln(1/5)= ( -ln10 )/( -ln5 )= ln10/ln5 ≈1.4307>1. So yes, log₀.₂( <0.2 )>1. Therefore, log₀.₂( log₃₂(...) ) >1.Then, log₃( something >1 ) >0. Since log base3 of something >1 is log₃( >1 ), which is positive. Hence, 9.254 times a positive number is positive. Therefore, the inequality holds.Therefore, the solution is x < -11.But wait, let's check another point, say x = -12.(x -1)/(x +5) = (-13)/(-7) ≈1.857.log₃₂(1.857) ≈0.2 * (log2(1.857)/log2(32^(0.2))) ). Wait, 32 is 2^5, so log base32 of A is (log2(A))/5.So log₃₂(1.857)= (log2(1.857))/5 ≈(0.906)/5≈0.1812.Then, log₀.₂(0.1812)= ln(0.1812)/ln(0.2)≈(-1.707)/(-1.609)≈1.06.Then, log₃(1.06)= ln(1.06)/ln3≈0.058/1.0986≈0.0528.Multiply by9.254≈0.0528*9.254≈0.489. Which is greater than0. So the inequality holds for x=-12.Now check x=-11:(x-1)/(x+5)=(-12)/(-6)=2.log₃₂(2)=0.2, because 32^0.2=2.Then log₀.₂(0.2)=1.log₃(1)=0.9.254*0=0. So the inequality is 0>0, which is false. So x=-11 is not included.Check x=-10.5 (but x=-10.5 is greater than -5.1935, so not in the domain). But let's see:Wait, x=-10.5 is not less than -5.1935≈-5.1935. So it's not in the domain. So the solution is x < -11.Thus, the final answer is x ∈ (-∞, -11).</think>9.254 log _{3} log _{0.2} log _{32} frac{x-1}{x+5}>0.Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.9.254 log _{3} log _{0.2} log _{32} frac{x-1}{x+5}>0. ---Solution:The inequality to solve is: 9.254 log _{3} log _{0.2} log _{32} frac{x-1}{x+5}>0.1. Simplify the inequality: Divide both sides by 9.254 (positive, so inequality direction remains): log _{3} log _{0.2} log _{32} frac{x-1}{x+5} > 0.2. Outermost logarithm condition: log_{3}(A) > 0 implies A > 1 (since base 3 > 1): log _{0.2} log _{32} frac{x-1}{x+5} > 1.3. Handle log_{0.2} (base <1): log_{0.2}(B) > 1 implies B < 0.2^1 = 0.2 (inequality flips): log _{32} frac{x-1}{x+5} < 0.2.4. Solve log_{32}(C) < 0.2: Convert to exponential form: C < 32^{0.2}. Since 32 = 2^5, we have 32^{0.2} = 2^{5 cdot 0.2} = 2^1 = 2: frac{x-1}{x+5} < 2.5. Solve frac{x-1}{x+5} < 2 with domain constraints: - Domain for log_{32}frac{x-1}{x+5}: frac{x-1}{x+5} > 0 implies x < -5 or x > 1. - Domain for log_{0.2}log_{32}frac{x-1}{x+5}: log_{32}frac{x-1}{x+5} > 1 implies frac{x-1}{x+5} > 32 implies x < -frac{161}{31} approx -5.1935. - Combining domains: x < -5.1935. Solve frac{x-1}{x+5} < 2 for x < -5.1935: frac{x-1}{x+5} < 2 implies x < -11.6. Final domain intersection: x < -11 satisfies all constraints. Answer: The solution is boxed{(-infty, -11)}.,<think>Okay, let's tackle this inequality step by step. The problem is: 9.254 times the logarithm base 3 of (the logarithm base 0.2 of (the logarithm base 32 of ((x-1)/(x+5)))) is greater than 0. So, written out, it's 9.254 log₃(log₀.₂(log₃₂((x-1)/(x+5)))) > 0.First, since 9.254 is a positive constant, I can divide both sides by 9.254 without changing the inequality direction. That simplifies things to log₃(log₀.₂(log₃₂((x-1)/(x+5)))) > 0. Now, for a logarithm base 3 to be greater than 0, the argument of the log must be greater than 1 because log_b(a) > 0 when a > 1 if b > 1. Since 3 is greater than 1, this means log₀.₂(log₃₂((x-1)/(x+5))) > 3^0 = 1. Wait, no. Wait, log₃(something) > 0 implies that something > 3^0 = 1. So yes, log₀.₂(log₃₂((x-1)/(x+5))) > 1.But here, the next logarithm is base 0.2, which is less than 1. So when dealing with log base less than 1, the inequality flips when we exponentiate. Let me recall: If we have log_b(a) > c where b < 1, then a < b^c. Because the log function is decreasing when the base is between 0 and 1. So in this case, log₀.₂(Y) > 1 implies that Y < 0.2^1 = 0.2. So Y must be less than 0.2. Here, Y is log₃₂((x-1)/(x+5)). So log₃₂((x-1)/(x+5)) < 0.2.But wait, log base 32 of ((x-1)/(x+5)) < 0.2. Now, 32 is 2^5, which is greater than 1. So log base 32 of Z < 0.2. To solve this, we can rewrite the inequality. Since the base is greater than 1, log_b(Z) < c implies that Z < b^c. So Z < 32^0.2. What's 32^0.2? Since 32 is 2^5, 32^0.2 = (2^5)^(1/5) = 2^(5*(1/5)) = 2^1 = 2. Therefore, (x-1)/(x+5) < 2.So now, we need to solve (x-1)/(x+5) < 2. But before that, we also need to consider the domains of all the logarithms. Let's backtrack and make sure each logarithm is defined.Starting from the innermost logarithm: log₃₂((x-1)/(x+5)). For this to be defined, (x-1)/(x+5) must be positive. So (x-1)/(x+5) > 0. That happens when both numerator and denominator are positive or both are negative. Case 1: x - 1 > 0 and x + 5 > 0. So x > 1 and x > -5. Since x > 1 already satisfies x > -5, this case is x > 1.Case 2: x - 1 < 0 and x + 5 < 0. So x < 1 and x < -5. Here, x < -5. So the domain for the innermost log is x > 1 or x < -5.Next, log₀.₂(log₃₂((x-1)/(x+5))). The argument here is log₃₂((x-1)/(x+5)) which we have already considered, but also for log base 0.2 to be defined, log₃₂((x-1)/(x+5)) must be positive. Because logarithm is only defined for positive numbers. So log₃₂((x-1)/(x+5)) > 0. Which implies (x-1)/(x+5) > 32^0 = 1. So (x-1)/(x+5) > 1.Solving (x-1)/(x+5) > 1. Let's subtract 1: [(x-1)/(x+5) - 1] > 0. [(x -1 - (x +5))/(x+5)] > 0 => (-6)/(x+5) > 0. So the numerator is -6, which is negative. Therefore, -6/(x+5) > 0 implies that (x+5) must be negative. So x +5 < 0 => x < -5. Therefore, the argument of log₀.₂ must be positive, so we need log₃₂((x-1)/(x+5)) > 0, which gives x < -5. But previously, the domain for the innermost log was x >1 or x < -5. So combining these, the domain for the next logarithm (log₀.₂) is x < -5.Moving up, log₀.₂(log₃₂((x-1)/(x+5))) must be defined and then we had the inequality log₀.₂(...) >1. But since x < -5 is required for log₀.₂ to be defined, we have to remember that.Earlier, we deduced that log₀.₂(Y) >1 implies Y < 0.2, where Y is log₃₂((x-1)/(x+5)). But Y has to be positive as well because it's inside log₀.₂. So log₃₂((x-1)/(x+5)) >0, which we already have, and log₃₂((x-1)/(x+5)) < 0.2.But log₃₂((x-1)/(x+5)) < 0.2. As 32^0.2 =2, so (x-1)/(x+5) < 2. But x < -5 here. Let's solve (x-1)/(x+5) < 2 with x < -5.Multiply both sides by (x+5). But x+5 is negative because x < -5. So multiplying by a negative number reverses the inequality.Therefore, (x -1) > 2(x +5). Let's solve that:x -1 > 2x +10Subtract x from both sides: -1 > x +10Subtract 10: -11 > xSo x < -11.But we had x < -5 from the domain. So combining these, x < -11.But we need to check if this is consistent.Wait, let's verify. (x-1)/(x+5) < 2 with x < -5.Let me pick x = -12, which is less than -11.Calculate (x-1)/(x+5) = (-12 -1)/(-12 +5) = (-13)/(-7) = 13/7 ≈1.857, which is less than 2. Okay.Another check: x = -10, which is between -11 and -5. But x < -11 is required. If x = -10, then (x-1)/(x+5) = (-11)/(-5) = 11/5 = 2.2, which is greater than 2. So indeed, x < -11 is necessary to make (x-1)/(x+5) < 2.Therefore, this gives us x < -11.But we need to check if all the previous conditions are satisfied. The original domains:1. (x-1)/(x+5) >0: x >1 or x < -5. But due to log₀.₂ requiring log₃₂(...) >0, which gives x < -5. So domain here is x < -5.2. log₃₂((x-1)/(x+5)) >0: same as above, x < -5.3. log₀.₂(...) >1: led us to x < -11.But we also need to ensure that log₀.₂(log₃₂(...)) is defined. Which requires log₃₂(...) >0, which we already have as x < -5. So the final solution is x < -11.But let's verify with x < -11:Take x = -12. Then:(x-1)/(x+5) = (-13)/(-7) ≈1.857. Then log base32 of that is log32(1.857). Since 32^0.2 =2, so 32^0.2 =2, so log32(2) =0.2. So log32(1.857) is slightly less than 0.2. Then log base0.2 of that number. Since 0.2 is 1/5, log base 1/5 of a number less than 0.2. Let's say log32(1.857)=c≈0.198 (since 32^0.198 ≈1.857). Then log_0.2(c) = log_{1/5}(c) = ln(c)/ln(1/5) = ln(c)/(-ln5) = -ln(c)/ln5. Since c <0.2, ln(c) is negative (since c <1). So -ln(c)/ln5 is positive. So log_0.2(c) is positive. Then log base3 of that positive number. Then 9.254 times that is positive. Wait, but the inequality is 9.254 log3(...) >0. So we need log3(...) >0, which requires the argument of log3 to be >1. So let's go step by step.Wait, let's double-check:If x < -11, then (x-1)/(x+5) is positive and less than 2. So log32 of that is less than 0.2, as 32^0.2=2. So log32((x-1)/(x+5)) <0.2. Then log_0.2 of that is log_{1/5}(something <0.2). Since the base is less than 1, and the argument is less than the base^1, which is 0.2. Wait, base is 0.2, so log_{0.2}(Y) where Y <0.2. For example, log_{0.2}(0.1) would be greater than 1 because 0.2^2 =0.04, 0.2^3=0.008, but wait, log_{0.2}(Y) is equal to ln(Y)/ln(0.2). Since ln(0.2) is negative, and if Y <0.2, ln(Y) is also negative (since Y <1). So ln(Y)/ln(0.2) is positive. Let's compute log_{0.2}(0.1):log_{0.2}(0.1) = ln(0.1)/ln(0.2) ≈ (-2.3026)/(-1.6094) ≈1.4307. Which is greater than 1. So if Y <0.2, then log_{0.2}(Y) >1. Wait, but earlier, when we had log_{0.2}(Y) >1, we concluded that Y <0.2. But actually, for log base 0.2 (which is less than 1), log_b(Y) > c implies Y < b^c when b <1. Let's check with Y=0.1, log_{0.2}(0.1)≈1.43>1. So yes, Y=0.1 <0.2^1=0.2. So log_{0.2}(Y) >1 implies Y <0.2. But when we have Y <0.2, log_{0.2}(Y) can be greater than 1. For example, Y approaching 0, log_{0.2}(Y) approaches infinity. Wait, no. Wait, if Y approaches 0 from the right, log_{0.2}(Y) approaches infinity because 0.2^k approaches 0 as k approaches infinity. So log_{0.2}(Y) increases without bound as Y approaches 0. So if Y <0.2, then log_{0.2}(Y) >1. Wait, that seems contradictory to previous step.Wait, let's clarify:For log_b(a) > c where 0 < b <1.We know that log_b(a) = ln(a)/ln(b). Since ln(b) is negative, the sign of log_b(a) depends on ln(a). If a >1, ln(a) is positive, so log_b(a) is negative. If a=1, log_b(a)=0. If a <1, ln(a) is negative, so log_b(a) is positive.So log_{0.2}(a) >1 implies that a <0.2^1=0.2. Because when a <1, log_{0.2}(a) is positive, and increasing as a decreases. So log_{0.2}(a) >1 => a <0.2.Therefore, our previous step is correct: log_{0.2}(Y) >1 implies Y <0.2. But Y must be positive. So Y ∈ (0,0.2).But in our case, Y = log32((x-1)/(x+5)). We had Y <0.2, which is correct. But also, Y must be positive because it's the argument of log_{0.2}. So Y >0. So combining, 0 < Y <0.2.But Y = log32((x-1)/(x+5)). So 0 < log32((x-1)/(x+5)) <0.2. Which implies that 32^0 < (x-1)/(x+5) <32^0.2. 32^0 =1 and 32^0.2=2. So 1 < (x-1)/(x+5) <2. But wait, earlier we solved for (x-1)/(x+5) <2 and got x < -11. But this seems contradictory because (x-1)/(x+5) must be between 1 and 2?Wait, no. Wait, log32((x-1)/(x+5)) >0 implies (x-1)/(x+5) >1, which as we solved earlier gives x < -5. But now, we also have log32((x-1)/(x+5)) <0.2 implies (x-1)/(x+5) <32^0.2=2. So combined, 1 < (x-1)/(x+5) <2, and x < -5.But solving 1 < (x-1)/(x+5) <2 with x < -5.First, let's write 1 < (x-1)/(x+5) <2. We can split this into two inequalities:1 < (x-1)/(x+5) and (x-1)/(x+5) <2.We already solved (x-1)/(x+5) >1 and found x < -5.Then (x-1)/(x+5) <2 gives x < -11 as found earlier.So combining both, x < -11.Therefore, the solution is x < -11.But let's check if (x-1)/(x+5) is between 1 and 2 when x < -11.Take x = -12:(x-1)/(x+5) = (-13)/(-7) ≈1.857 which is between 1 and 2. Correct.Take x = -20:(-20 -1)/(-20 +5) = (-21)/(-15) = 1.4, which is between 1 and 2.So indeed, for x < -11, (x-1)/(x+5) is between 1 and 2.Therefore, log32 of that is between 0 and 0.2 (since 32^0=1 and 32^0.2=2). So log32((x-1)/(x+5)) is between 0 and 0.2.Then log_0.2 of that (which is log base 1/5) of a number between 0 and 0.2. As we saw earlier, log_{0.2}(0.1)≈1.43 which is greater than 1, and log_{0.2}(0.19)=ln(0.19)/ln(0.2)≈(-1.660)/(-1.609)≈1.032>1. So even if Y approaches 0.2 from below, log_{0.2}(Y) approaches 1 from above. Therefore, log_{0.2}(Y) >1.Then log base3 of that number (which is greater than1) is positive, since log3(something >1) is positive. Therefore, 9.254 times a positive number is positive, which satisfies the original inequality.Therefore, the solution is x < -11.But to confirm, let's test x = -12:Compute innermost log: log32((x-1)/(x+5)) = log32(13/7) ≈ log32(1.857). Since 32^0.2 =2, and 1.857 <2, so log32(1.857) ≈0.198 (since 32^0.198 ≈1.857). Then log_0.2(0.198) = ln(0.198)/ln(0.2) ≈ (-1.619)/(-1.609) ≈1.006. Then log3(1.006) ≈0.009 (since 3^0.009≈1.006). Multiply by 9.254 gives ≈0.083, which is greater than 0. So the inequality holds.Another test at x = -11. But wait, x < -11. Let's take x approaching -11 from the left.x approaches -11 from the left, so (x-1)/(x+5) approaches (-12)/(-6) =2. So log32(2)=0.2. Then log_0.2(0.2)=1. Then log3(1)=0. So 9.254*0=0. The inequality is >0, so x=-11 is not included. Thus, the solution is x < -11.So final answer: x ∈ (-∞, -11). So in boxed notation, boxed{(-infty, -11)}.</think>9.254 log _{3} log _{0.2} log _{32} frac{x-1}{x+5}>0.Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.9.254 log _{3} log _{0.2} log _{32} frac{x-1}{x+5}>0.The solution involves analyzing the nested logarithms step-by-step, considering the domains and properties of logarithmic functions. The key steps are:1. Simplify the inequality by dividing both sides by 9.254 (positive constant).2. Solve the resulting logarithmic inequality by ensuring each logarithm's argument is valid and applying logarithmic properties.3. Determine the domain restrictions for each logarithm to ensure all expressions are defined.4. Solve the rational inequality ((x-1)/(x+5) < 2) under the domain constraints.5. Combine all conditions to find the final solution interval.After detailed analysis, the solution to the inequality is (boxed{(-infty, -11)}).]

🔑:Solution. The original inequality is equivalent to the following:log _{0.2} log _{32} frac{x-1}{x+5}>1 Leftrightarrowleft{begin{array}{l}log _{32} frac{x-1}{x+5}>0, log _{32} frac{x-1}{x+5}<1, frac{x-1}{x+5}>0end{array} Leftrightarrow x<-11right.right.right..Answer: quad x in(-infty ;-11).

🧠:[<think>Okay, let's see. The problem is asking for the smallest real root 'a' of the equation sqrt(x(x+1)(x+2)(x+3) + 1) = 71. Hmm, alright. So first, I need to solve this equation for x and then find the smallest real root. Let me start by understanding the equation step by step.The left side is a square root of a product of four consecutive terms: x, x+1, x+2, x+3, all multiplied together, then add 1. Then the square root of that equals 71. So, the equation inside the square root must be 71 squared, right? Because sqrt(something) = 71 implies that something is 71^2. Let me write that down.So, sqrt(x(x+1)(x+2)(x+3) + 1) = 71.If I square both sides, that should eliminate the square root:x(x+1)(x+2)(x+3) + 1 = 71².Calculating 71 squared, 71*71. Let me do that. 70 squared is 4900, so 71 squared is 70² + 2*70 +1 = 4900 + 140 +1 = 5041. So, 71² = 5041.Therefore, the equation becomes:x(x+1)(x+2)(x+3) + 1 = 5041.Subtract 1 from both sides:x(x+1)(x+2)(x+3) = 5040.Hmm, okay. Now, the left-hand side is x multiplied by x+1, x+2, x+3. That seems like a product of four consecutive integers, except x isn't necessarily an integer here. Maybe there's a way to simplify this product. Let me think.I remember that when you have four consecutive terms, like x, x+1, x+2, x+3, their product can be related to (x^2 + 3x)(x^2 + 3x + 2). Let me check that:(x)(x+3) = x^2 + 3x.(x+1)(x+2) = x^2 + 3x + 2.So, the product x(x+1)(x+2)(x+3) is equal to (x^2 + 3x)(x^2 + 3x + 2). Let me verify:Multiply x and x+3: x^2 + 3x.Multiply x+1 and x+2: x^2 + 3x + 2.Then multiply those two quadratics: (x^2 + 3x)(x^2 + 3x + 2) = (x^2 + 3x)^2 + 2(x^2 + 3x).So, that's (x^2 + 3x)^2 + 2(x^2 + 3x). Let me denote y = x^2 + 3x, then the product becomes y^2 + 2y. Which is y(y + 2). So, that might be helpful. Alternatively, maybe completing the square.Alternatively, maybe let’s set t = x^2 + 3x + 1. Let me see. Let me think of the original equation.We have x(x+1)(x+2)(x+3) + 1 = 5041, but x(x+1)(x+2)(x+3) +1 is given under the square root. Wait, the problem had sqrt(x(x+1)(x+2)(x+3) + 1) = 71, which we squared to get 5041. So, the expression inside the square root is 5041. So, maybe x(x+1)(x+2)(x+3) +1 is a perfect square? Let me check.From the earlier substitution: x(x+1)(x+2)(x+3) = (x^2 + 3x)(x^2 + 3x + 2) = y(y + 2) where y = x^2 + 3x. So, adding 1 gives y(y + 2) + 1 = y^2 + 2y + 1 = (y + 1)^2. Oh! That's a perfect square. So, the left-hand side of the original equation is sqrt((y + 1)^2) = |y + 1|. Since we're dealing with square roots, we take the principal (non-negative) root, so |y + 1| = 71. Therefore, y + 1 = ±71, but since the square root is non-negative, |y + 1| = 71 implies that y +1 = 71 or y +1 = -71. But since sqrt(...) is 71, which is positive, but the expression inside sqrt is (y +1)^2, so regardless of y +1 being positive or negative, the square is positive, and the sqrt is the absolute value. Therefore, we have |y +1| = 71, so y +1 = 71 or y +1 = -71. Therefore, y = 70 or y = -72.But since y = x^2 + 3x, we can set up two equations:1) x^2 + 3x = 702) x^2 + 3x = -72So, solving these quadratics:First equation: x^2 + 3x - 70 = 0Second equation: x^2 + 3x + 72 = 0Let me solve the first equation: x^2 + 3x - 70 = 0Using quadratic formula: x = [-3 ± sqrt(9 + 280)] / 2 = [-3 ± sqrt(289)] / 2 = [-3 ±17]/2So, sqrt(289) is 17. Therefore,x = (-3 +17)/2 = 14/2 = 7x = (-3 -17)/2 = (-20)/2 = -10So, solutions are 7 and -10.Second equation: x^2 + 3x +72 =0Discriminant: 9 - 288 = -279, which is negative, so no real roots.Therefore, the real roots come from the first equation: 7 and -10.But the problem says "a is the smallest real root", so between 7 and -10, the smallest is -10. Therefore, a = -10.Wait, but hold on. Let me check if these solutions actually satisfy the original equation, because sometimes squaring both sides can introduce extraneous solutions.Original equation: sqrt(x(x+1)(x+2)(x+3) +1) =71.Let me check x =7.Compute x(x+1)(x+2)(x+3) +1. Let's compute 7*8*9*10 +1. 7*8=56, 56*9=504, 504*10=5040. 5040 +1=5041. sqrt(5041)=71. Correct. So x=7 is a valid solution.Now x=-10. Let's compute x(x+1)(x+2)(x+3)+1. For x=-10:(-10)(-9)(-8)(-7) +1. Multiply first two terms: (-10)(-9)=90. Next two: (-8)(-7)=56. Then 90*56=5040. 5040 +1=5041. sqrt(5041)=71. Correct. So x=-10 is also valid.Therefore, both roots are valid, and the smallest real root is -10.But let me just double-check. The original equation is sqrt(...)=71. So even if x is negative, but the product inside the sqrt is positive. Let's confirm that when x=-10, the product inside the sqrt is positive. Indeed, (-10)(-9)(-8)(-7) is positive because there are four negative terms, multiplying two at a time: (-10)(-9)=90, (-8)(-7)=56, then 90*56=5040, so positive. Then +1, so 5041. So sqrt(5041)=71. Correct. So x=-10 is indeed a valid solution.Therefore, the answer should be a=-10.Wait, but before finalizing, let me see if there's another way to approach this problem. Sometimes problems like this can be transformed by substitution. For example, maybe let’s set t = x + 1.5, to center the terms around the middle. Let me see.But maybe not necessary here. Since we already found the roots and checked them. Alternatively, maybe the original quartic equation can be written as (x^2 + 3x +1)^2 =71^2, which would lead to x^2 +3x +1 = ±71. But let's check that.Wait, earlier we saw that x(x+1)(x+2)(x+3) +1 = (x^2 +3x +1)^2. Let me verify that.From before: x(x+1)(x+2)(x+3) = (x^2 +3x)(x^2 +3x +2). Let's compute (x^2 +3x)(x^2 +3x +2):Multiply out: x^2*(x^2 +3x +2) +3x*(x^2 +3x +2)= x^4 +3x^3 +2x^2 +3x^3 +9x^2 +6xCombine like terms:x^4 + (3x^3 +3x^3) + (2x^2 +9x^2) +6x= x^4 +6x^3 +11x^2 +6xNow, (x^2 +3x +1)^2 = x^4 +6x^3 +11x^2 +6x +1. Therefore, x(x+1)(x+2)(x+3) +1 = (x^2 +3x +1)^2. Therefore, indeed, sqrt((x^2 +3x +1)^2) = |x^2 +3x +1| =71. Therefore, x^2 +3x +1= ±71.Hence, x^2 +3x +1=71 => x^2 +3x -70=0, which is the same as before, leading to x=7 and x=-10.Or x^2 +3x +1= -71 => x^2 +3x +72=0, which has discriminant 9 -288= -279, no real roots. So same result. So the solutions are x=7 and x=-10.Therefore, the smallest real root is -10. So a=-10.I think that's thorough. So the answer is boxed{-10}.Final Answerboxed{-10}</think>Given the equation (sqrt{x(x+1)(x+2)(x+3) + 1} = 71), we start by squaring both sides to eliminate the square root:[x(x+1)(x+2)(x+3) + 1 = 71^2]Calculating (71^2) gives (5041), so the equation becomes:[x(x+1)(x+2)(x+3) + 1 = 5041]Subtracting 1 from both sides, we get:[x(x+1)(x+2)(x+3) = 5040]We recognize that the product (x(x+1)(x+2)(x+3)) can be rewritten as ((x^2 + 3x)(x^2 + 3x + 2)). Let (y = x^2 + 3x), then the equation becomes:[y(y + 2) = 5040]This simplifies to:[y^2 + 2y - 5040 = 0]Solving this quadratic equation using the quadratic formula (y = frac{-b pm sqrt{b^2 - 4ac}}{2a}) where (a = 1), (b = 2), and (c = -5040):[y = frac{-2 pm sqrt{4 + 20160}}{2} = frac{-2 pm sqrt{20164}}{2} = frac{-2 pm 142}{2}]This gives the solutions (y = 70) and (y = -72). Substituting back (y = x^2 + 3x), we get two quadratic equations:1. (x^2 + 3x - 70 = 0)2. (x^2 + 3x + 72 = 0)For the first equation (x^2 + 3x - 70 = 0):[x = frac{-3 pm sqrt{9 + 280}}{2} = frac{-3 pm 17}{2}]This gives the solutions (x = 7) and (x = -10).The second equation (x^2 + 3x + 72 = 0) has a negative discriminant, indicating no real solutions.Checking both solutions (x = 7) and (x = -10) in the original equation confirms they are valid. The smallest real root is (-10).Thus, the value of (a) is (boxed{-10}).,<think>Okay, so I have this equation to solve: the square root of x(x+1)(x+2)(x+3) plus 1 equals 71. And I need to find the smallest real root, which is called 'a'. Hmm, let me try to unpack this step by step.First, the equation is √[x(x+1)(x+2)(x+3) + 1] = 71. So, maybe I should start by squaring both sides to eliminate the square root. That would give me x(x+1)(x+2)(x+3) + 1 = 71². Calculating 71 squared, that's 5041. So, the equation simplifies to x(x+1)(x+2)(x+3) + 1 = 5041. Then subtract 1 from both sides: x(x+1)(x+2)(x+3) = 5040.Now, I need to handle the product x(x+1)(x+2)(x+3). That looks like four consecutive terms multiplied together. Maybe there's a way to simplify this expression. Let me think. If I pair the terms: x and (x+3), and (x+1) and (x+2). Let's try that:First pair: x(x+3) = x² + 3xSecond pair: (x+1)(x+2) = x² + 3x + 2So, the product becomes (x² + 3x)(x² + 3x + 2). Hmm, that's interesting. Let me set y = x² + 3x. Then the product is y(y + 2) = y² + 2y. So, the equation becomes y² + 2y = 5040.So, substituting back, we have (x² + 3x)² + 2(x² + 3x) - 5040 = 0. Let me check that. Yes, expanding (x² + 3x)(x² + 3x + 2) gives (x² + 3x)² + 2(x² + 3x), so that's correct.So, let me set y = x² + 3x. Then the equation is y² + 2y - 5040 = 0. Let's solve this quadratic equation for y. Using the quadratic formula:y = [-2 ± √(4 + 4*5040)] / 2First, compute the discriminant: 4 + 4*5040 = 4(1 + 5040) = 4*5041. Since 5041 is 71², right? Because 71*71 is 5041. So, discriminant becomes 4*71². Therefore, the square root of the discriminant is 2*71 = 142.So, y = [-2 ± 142]/2. Let's compute both possibilities:First solution: (-2 + 142)/2 = 140/2 = 70Second solution: (-2 - 142)/2 = -144/2 = -72So, y = 70 or y = -72.But remember, y = x² + 3x. So now, we have two equations:1) x² + 3x = 702) x² + 3x = -72Let's solve both equations.Starting with the first equation: x² + 3x - 70 = 0Using the quadratic formula here: x = [-3 ± √(9 + 280)] / 2 = [-3 ± √289]/2√289 is 17, so:x = (-3 +17)/2 = 14/2 = 7x = (-3 -17)/2 = (-20)/2 = -10So, solutions here are x = 7 and x = -10.Now, the second equation: x² + 3x + 72 = 0Again, using the quadratic formula: x = [-3 ± √(9 - 288)] / 2 = [-3 ± √(-279)] / 2Since the discriminant is negative, there are no real solutions here. So, the only real solutions are x = 7 and x = -10.But the problem mentions that 'a' is the smallest real root. So, between 7 and -10, the smaller one is -10. Therefore, a = -10.But wait, hold on. Let me check if these solutions actually satisfy the original equation. Sometimes when we square both sides, we might introduce extraneous solutions.Let's check x = 7 first:Left-hand side: √[7*8*9*10 + 1] = √[5040 + 1] = √5041 = 71. Which matches the right-hand side. So, that's valid.Now, x = -10:Left-hand side: √[-10*(-9)*(-8)*(-7) + 1]. Let's compute the product inside:First, multiply the terms:-10 * (-9) = 9090 * (-8) = -720-720 * (-7) = 5040So, inside the sqrt: 5040 + 1 = 5041. So, sqrt(5041) = 71. Which again matches. So, x = -10 is also a valid solution.Therefore, both roots are valid. Since the problem asks for the smallest real root, which is -10. So, a = -10.But let me think again. Is there any other possible real roots? The equation reduced to two quadratics, one with real roots and one without. So, only x = 7 and x = -10. So, -10 is the smallest.Therefore, the answer should be boxed{-10}Final Answerboxed{-10}</think>Given the equation (sqrt{x(x+1)(x+2)(x+3) + 1} = 71), we start by squaring both sides to eliminate the square root:[x(x+1)(x+2)(x+3) + 1 = 71^2]Calculating (71^2 = 5041), we get:[x(x+1)(x+2)(x+3) + 1 = 5041]Subtracting 1 from both sides:[x(x+1)(x+2)(x+3) = 5040]We pair the terms (x(x+3)) and ((x+1)(x+2)):[x(x+3) = x^2 + 3x quad text{and} quad (x+1)(x+2) = x^2 + 3x + 2]Let (y = x^2 + 3x). The product becomes:[y(y + 2) = 5040]This simplifies to the quadratic equation:[y^2 + 2y - 5040 = 0]Solving this quadratic equation using the quadratic formula (y = frac{-b pm sqrt{b^2 - 4ac}}{2a}):[y = frac{-2 pm sqrt{4 + 4 cdot 5040}}{2} = frac{-2 pm sqrt{4 cdot 5041}}{2} = frac{-2 pm 142}{2}]This gives the solutions:[y = frac{140}{2} = 70 quad text{and} quad y = frac{-144}{2} = -72]For (y = 70):[x^2 + 3x - 70 = 0]Solving this quadratic equation:[x = frac{-3 pm sqrt{9 + 280}}{2} = frac{-3 pm 17}{2}]This gives the solutions (x = 7) and (x = -10).For (y = -72):[x^2 + 3x + 72 = 0]The discriminant is negative, so there are no real solutions.Checking the solutions (x = 7) and (x = -10) in the original equation confirms both are valid. The smallest real root is (-10).Thus, the value of (a) is (boxed{-10}).]

🔑:Let t=x+1.5, then the equation becomes sqrt{(t-1.5)(t-0.5)(t+0.5)(t+1.5)+1}=71begin{array}{l}sqrt{left(t^{2}-frac{9}{4}right)left(t^{2}-frac{1}{4}right)+1}=71 sqrt{t^{4}-frac{5}{2} t^{2}+frac{9}{16}+1}=71 Rightarrow sqrt{t^{4}-frac{5}{2} t^{2}+frac{25}{16}}=71 Rightarrow sqrt{left(t^{2}-frac{5}{4}right)^{2}}=71 t^{2}-frac{5}{4}=71 Rightarrow t^{2}=frac{289}{4} Rightarrow t=frac{17}{2} text { or } t=-frac{17}{2} x=t-1.5= pm frac{17}{2}-frac{3}{2}=7 text { or }-10 a=text { the smallest root }=-10end{array}

🧠:[<think>Okay, let's see. The problem gives a piecewise function f(x) and asks which interval cannot have a zero. The options are A: (1,4), B: (3,7), C: (8,13), D: (11,18). Hmm, so I need to check each interval to see if the function could cross zero there. If it's always positive or always negative in that interval, then there's no zero. First, let me understand the function. It's defined in two parts. From x=0 to x=10, it's 3*2^x -24. Then from x=10 to x=20, it's -2^{x-5} +126. So, two different exponential functions. I need to check continuity at x=10 as well, maybe. Wait, does the function have a value at x=10 in both cases? Let me check. For the first part, when x=10, it's 3*2^10 -24. For the second part, when x=10, it's -2^{10-5} +126. Let me compute both.First part at x=10: 3*1024 -24 = 3072 -24 = 3048.Second part at x=10: -2^5 +126 = -32 +126 = 94. Wait, that's a huge discrepancy. So the function is not continuous at x=10? That's odd, but maybe that's how it's defined. So at x=10, the first part gives 3048, and the second part gives 94. So there's a jump discontinuity there. So when considering the intervals, we have to remember that the function jumps down from 3048 to 94 at x=10.Okay, now let's analyze each piece separately.First part: f(x) = 3*2^x -24 for 0 <= x <=10.Second part: f(x) = -2^{x-5} +126 for 10 < x <=20.So to find zeros, we need to solve 3*2^x -24=0 for x in [0,10], and -2^{x-5} +126=0 for x in (10,20].Let's solve each equation.First equation: 3*2^x -24=0 => 3*2^x=24 => 2^x=8 => x=3. Because 2^3=8. So x=3 is a zero in the first part. So that would be in interval A: (1,4) and B: (3,7). Since x=3 is the exact point, but intervals are open. So in interval A (1,4), since x=3 is within (1,4), the function crosses zero there. But since the interval is open, does x=3 count? Wait, the zero is at x=3, which is the boundary of the interval B: (3,7). So the interval B is (3,7), which does not include 3. So does the function cross zero in (3,7)? Well, right after x=3, like x=3.0001, the function is increasing because it's an exponential function. Let's confirm.The first part, 3*2^x -24. The derivative is 3*ln2*2^x, which is always positive, so the function is strictly increasing from x=0 to x=10. Therefore, the only zero in the first part is at x=3. So in the interval (1,4), since the function is increasing and crosses zero at x=3, which is inside (1,4), so there is a zero in interval A. But interval A is (1,4), which includes x=3. Wait, (1,4) is open, so x=3 is in (1,4). Wait, 3 is between 1 and 4, so yes, open interval (1,4) includes x=3. Wait, open intervals don't include endpoints, but 3 is not an endpoint of (1,4). The endpoints are 1 and 4. So (1,4) includes all x such that 1 < x < 4. So 3 is inside that interval. Therefore, the zero at x=3 is inside interval A. Therefore, interval A does contain a zero.But wait, the problem is asking which interval cannot have a zero. So if interval A contains a zero, then it's not the answer. Similarly, interval B is (3,7). The zero at x=3 is not included in (3,7), but in the interval (3,7), the function is increasing from x=3 onwards. Wait, at x=3, the function is zero, then for x>3, since it's increasing, the function becomes positive. So in the interval (3,7), the function is always positive. Therefore, there's no zero in (3,7). Because after x=3, f(x) is positive, and since it's increasing, it remains positive. Therefore, interval B: (3,7) cannot have a zero. Hmm, but let me double-check.Wait, at x=3, f(x)=0, but x=3 is not in (3,7). So in (3,7), the function is positive. So no zeros there. So interval B cannot have a zero. Then why is the problem giving four options, and maybe that's the answer. Wait, but let me check other intervals as well.Next, interval C: (8,13). Let's see. For x in (8,10], the function is still the first part, 3*2^x -24. At x=8, f(8)=3*256 -24=768-24=744. At x=10, as calculated before, it's 3048. So from x=8 to x=10, the function is increasing from 744 to 3048, so all positive. Then for x>10, up to 13, it's the second part: -2^{x-5} +126. Let's compute f(10). Wait, at x=10, the first part is 3048, and the second part is 94. So from x=10 onwards, the function drops to 94 and then continues with the second part. Let's compute the second part at x=10: -2^{10-5} +126= -32 +126=94. Then as x increases beyond 10, the term 2^{x-5} increases, so -2^{x-5} decreases, so the function -2^{x-5} +126 is decreasing. Let's check when it reaches zero.Set -2^{x-5} +126=0 => 2^{x-5}=126 => x-5=log2(126). Since 2^6=64, 2^7=128. So log2(126) is a little less than 7. Precisely, log2(126)=ln(126)/ln(2)≈4.836/0.693≈6.977. So x≈5+6.977≈11.977. So approximately x≈12. So the zero for the second part is around x≈11.977, which is approximately 12. So in the interval (10,20], the function -2^{x-5}+126 decreases from 94 to negative values, crossing zero at around x≈12. Therefore, in the interval (8,13), from x=8 to x=10, the function is positive, then at x=10 it drops to 94, and continues decreasing to zero at x≈12, so between x=10 and x≈12, the function goes from 94 to 0. Therefore, there is a zero in (10,12), which is within (8,13). So interval C: (8,13) includes x≈12 where the zero occurs. So interval C does contain a zero.Then interval D: (11,18). The zero occurs at x≈12, so (11,18) includes x≈12, so there is a zero there. Therefore, interval D can have a zero.Therefore, the intervals that cannot have a zero are B: (3,7). Because after x=3, the function is positive and increasing, so in (3,7), it's always positive. So no zero. The other intervals either contain a zero or cross zero within them. So the answer should be B.Wait, but let me check again. Let's make sure.First part: f(x)=3*2^x -24. We found the zero at x=3. Since the function is increasing, left of x=3, it's negative, right of x=3, positive. So in (1,4), it crosses zero at x=3, so interval A contains a zero. Interval B is (3,7). From x=3 onwards, the function is positive. So in (3,7), f(x) is positive. Therefore, no zero in B.For the second part, the zero is around x≈12, which is in interval C: (8,13) and D: (11,18). So both C and D contain the zero. Therefore, only interval B cannot have a zero. So answer is B.But wait, the options were A,B,C,D, and the question says "cannot have a zero in which of the following intervals". So the answer is B. But let me confirm once more.Wait, interval C is (8,13). From x=8 to x=10, the function is the first part, which is positive. At x=10, it's 3048. Then from x=10 to 13, it's the second part, which starts at 94 and decreases to f(13)= -2^{13-5} +126= -2^8 +126= -256 +126= -130. So between x=10 and x=13, the function goes from 94 to -130, crossing zero at x≈12. Therefore, in interval C, (8,13), there is a zero at x≈12. So interval C does contain a zero.Similarly, interval D: (11,18). The zero is at x≈12, which is within (11,18). So interval D also contains a zero.Therefore, only interval B: (3,7) cannot have a zero. So the answer should be B. But wait, let me check the options again. The problem might have tricked me.Wait, original function: first part is 0<=x<=10, second part is 10<x<=20. So at x=10, the function is 3*2^10 -24=3048. Then for x>10, it's the second function. So between x=10 and x=20, the function is decreasing from 94 to -2^{15}+126= -32768 +126= -32642, which is way negative. So the second part starts at 94 when x=10 (though technically x=10 is included in the first part), and then decreases. So between x=10 and x≈12, the function crosses zero. So in interval C: (8,13), which includes x=10 to x=13, the function crosses zero. So in that interval, there is a zero. Similarly, interval D: (11,18) includes x≈12. So yes, zeros there.But maybe there's another zero? Wait, the first part is 3*2^x -24. Since it's strictly increasing, only one zero at x=3. The second part is -2^{x-5} +126, which is strictly decreasing, so only one zero at x≈12. Therefore, the function has two zeros: x=3 and x≈12. Therefore, intervals that include these points would have zeros. Intervals that don't include these points and where the function is entirely positive or negative would not have zeros.So interval A: (1,4). Includes x=3. So has a zero. Interval B: (3,7). Starts just above 3, where the function is positive and increasing. So from x=3 to x=7, the function is positive. So no zero. Interval C: (8,13). Includes x≈12. So has a zero. Interval D: (11,18). Also includes x≈12. So has a zero. Therefore, the answer is B.But let me check once more. Let's compute f(4). For x=4, first part: 3*2^4 -24= 3*16 -24=48-24=24. So f(4)=24. At x=3.5: 3*2^3.5 -24. 2^3.5= sqrt(2^7)=sqrt(128)=approx 11.3137. So 3*11.3137≈33.941 -24≈9.941. So positive. At x=3.1: 3*2^3.1 -24. 2^3.1≈8.574. 3*8.574≈25.722 -24≈1.722. Positive. So even just above 3, the function is positive, so between 3 and 7, it's positive. Therefore, interval B: (3,7) has no zero.Therefore, the answer is B.Final Answerboxed{B}</think>To determine which interval cannot have a zero for the given piecewise function ( f(x) ), we analyze each part of the function and check for zeros in the given intervals.The function is defined as:[f(x) = begin{cases} 3 times 2^{x} - 24, & 0 leqslant x leqslant 10 -2^{x-5} + 126, & 10 < x leqslant 20 end{cases}]1. First part analysis ( ( 0 leqslant x leqslant 10 ) ): - Solve ( 3 times 2^x - 24 = 0 ): [ 3 times 2^x = 24 implies 2^x = 8 implies x = 3 ] - The function ( 3 times 2^x - 24 ) is strictly increasing, so it crosses zero at ( x = 3 ). This zero is within the interval ( (1, 4) ).2. Second part analysis ( ( 10 < x leqslant 20 ) ): - Solve ( -2^{x-5} + 126 = 0 ): [ 2^{x-5} = 126 implies x - 5 = log_2(126) implies x approx 5 + 6.977 approx 11.977 ] - The function ( -2^{x-5} + 126 ) is strictly decreasing, so it crosses zero at ( x approx 12 ). This zero is within the intervals ( (8, 13) ) and ( (11, 18) ).3. Checking intervals: - Interval A: ( (1, 4) ) includes ( x = 3 ), so it has a zero. - Interval B: ( (3, 7) ) starts just above 3 where the function is positive and remains positive, so it has no zero. - Interval C: ( (8, 13) ) includes ( x approx 12 ), so it has a zero. - Interval D: ( (11, 18) ) includes ( x approx 12 ), so it has a zero.Thus, the interval that cannot have a zero is (boxed{B}).,<think>Okay, let me try to figure out this problem. So, we have a piecewise function f(x) defined as two different exponential functions on two intervals: from 0 to 10, it's 3 times 2 to the x minus 24, and from 10 to 20, it's negative 2 to the (x minus 5) plus 126. The question is asking which of the given intervals cannot have a zero of the function. The options are A: (1,4), B: (3,7), C: (8,13), D: (11,18). First, I need to understand what it means for a function to have a zero in an interval. A zero is a point where the function crosses the x-axis, meaning f(x) = 0. So, if the function changes sign over an interval, by the Intermediate Value Theorem, there must be at least one zero in that interval. Conversely, if the function doesn't change sign in an interval, there can't be a zero there unless the function touches the x-axis but doesn't cross it. But since these are exponential functions, which are either always increasing or always decreasing, depending on their form, it's more likely that if the sign changes, there's exactly one zero. So, maybe the strategy here is to check for sign changes in each interval for both pieces of the function.Let's start by analyzing the first piece of the function: f(x) = 3*2^x - 24 for 0 ≤ x ≤ 10. Let's see where this part could have zeros. To find a zero, set 3*2^x - 24 = 0. Solving for x: 3*2^x = 24 ⇒ 2^x = 8 ⇒ x = 3. So, the first part of the function has a zero at x=3. Therefore, in the interval (3,7), which is option B, part of this interval is covered by the first piece of the function. Since the zero is at x=3, which is the lower bound of interval B, but the interval is open (3,7). However, the function might cross from negative to positive or vice versa around that point, so we need to check the behavior around x=3.Wait, let's actually compute the function at x=3. At x=3, f(3) = 3*2^3 -24 = 3*8 -24 = 24 -24 = 0. So x=3 is a zero. Now, let's check the sign around x=3. For x slightly less than 3, say x=2: f(2) = 3*4 -24 = 12 -24 = -12. For x slightly more than 3, say x=4: f(4) = 3*16 -24 = 48 -24 = 24. So, the function crosses from negative to positive at x=3. Therefore, the interval (3,7) includes x=3, but since the interval is open at 3, does that matter? Wait, the zero is exactly at x=3, which is not included in the interval (3,7). However, just to the right of 3, the function is positive. So, in the interval (3,7), the function is positive at x=4 (24) and continues to increase because 2^x is an increasing function. Therefore, in (3,7), the first piece of the function is always positive. But wait, x=7 is still within the first piece of the function (since 0 ≤x ≤10). Let's check f(7): 3*2^7 -24 = 3*128 -24 = 384 -24 = 360. So, it's still positive. Therefore, on the interval (3,7), the first piece of the function is entirely positive. Therefore, there can't be a zero in (3,7), except at x=3, which is excluded. So, maybe option B is the interval where there cannot be a zero? Wait, but hold on, let me check the second piece of the function as well. Wait, no, the second piece is for x >10, so in (3,7), it's still the first piece. So, if the first piece is positive throughout (3,7), then there's no zero in that interval. But the problem is, the options are about the function as a whole. So, even though the first piece doesn't have a zero in (3,7), maybe the second piece could? But no, the second piece is only for x>10. So, in (3,7), it's entirely the first piece, which is positive. Therefore, there cannot be a zero in (3,7). So, maybe B is the answer? But let's check other intervals as well to make sure.Now, let's check the other intervals. Option A is (1,4). The first piece applies here. Let's check the function at x=1: f(1) = 3*2 -24 = 6 -24 = -18. At x=4: f(4) = 24, as before. So, the function goes from negative to positive between x=1 and x=4, crossing zero at x=3. Therefore, by IVT, there is a zero in (1,4), so interval A does contain a zero, so it's not the answer.Option C is (8,13). Let's break this into two parts because x=10 is the boundary between the two pieces. So, from x=8 to x=10, it's the first piece, and from x=10 to x=13, it's the second piece. Let's check the first piece at x=8: f(8) = 3*2^8 -24 = 3*256 -24 = 768 -24 = 744. At x=10: f(10) = 3*2^10 -24 = 3*1024 -24 = 3072 -24 = 3048. So, in the interval (8,10], the first piece is very large and positive. Then, for x>10 up to 13, we have the second piece: f(x) = -2^{x-5} + 126. Let's compute f(10): but wait, at x=10, do we use the first or the second piece? The problem says 0 ≤x ≤10 uses the first, and 10 <x ≤20 uses the second. So, at x=10, it's the first piece. So, f(10) = 3048 as above. Then, just above 10, say x=11: f(11) = -2^{11-5} + 126 = -2^6 + 126 = -64 + 126 = 62. Then, x=13: f(13) = -2^{13-5} + 126 = -2^8 + 126 = -256 + 126 = -130. So, between x=10 and x=13, the second piece goes from 62 at x=11 to -130 at x=13. Therefore, since it's a continuous function (both pieces are continuous, and at x=10, the first piece is 3048 and the second piece at x=10 would be -2^{10-5} + 126 = -32 + 126 = 94, but wait, that's a problem. Wait, hold on, maybe I made a mistake here.Wait, no, the second piece is defined for x >10, so at x=10, it's still the first piece. But the second piece when x approaches 10 from the right would be -2^{10 -5} +126 = -32 + 126 = 94. But the first piece at x=10 is 3048. Therefore, there is a discontinuity at x=10. The left limit is 3048, and the right limit is 94. Therefore, the function is not continuous at x=10. Hmm, that's important. So, between x=10 and x=13, the second piece starts at 94 (right after x=10) and decreases to -130 at x=13. So, since the function is decreasing (because the exponent x-5 increases, so 2^{x-5} increases, making -2^{x-5} decrease), so from x=10 to x=13, the function goes from 94 to -130, crossing zero somewhere in between. Therefore, there must be a zero in (10,13). Therefore, in the interval (8,13), even though from 8 to 10 the function is positive, from 10 to 13 it goes from positive to negative, crossing zero. Therefore, there is a zero in (10,13). So, since (8,13) includes part of the second piece where the function crosses zero, interval C does contain a zero. Hence, option C is not the answer.Option D is (11,18). Let's check the second piece here. Let's compute f(11): as above, f(11)=62, f(13)=-130, f(18)= -2^{18-5} +126 = -2^{13} +126 = -8192 +126 = -8066. So, the function at x=11 is 62, at x=13 is -130, and continues decreasing to -8066 at x=18. Since the function is decreasing (as the coefficient of the exponential is negative), it crosses zero once between x=10 and x=13. Specifically, between x=10 and x=11, it was still positive (x=10: first piece 3048, x=11: 62). Then, it crosses zero somewhere between x=11 and x=13. Wait, at x=11, it's 62, and at x=13, it's -130. So, since it's continuous on (10,20), by IVT, there must be a zero between 11 and 13. Therefore, in the interval (11,18), which includes (11,13), there is a zero. However, after x=13, the function continues to decrease, so once it crosses zero, it stays negative. Therefore, there is exactly one zero in (10,13), which is within (11,18) only if the zero is between 11 and 13. But since f(11) is 62 and f(13) is -130, the zero is somewhere between 11 and 13. Therefore, the interval (11,18) does include that zero, so there is a zero in (11,18). Therefore, D is not the answer.Wait, but let me confirm. Let's check f(11)=62, f(12)= -2^{12-5} +126= -128 +126= -2. So, at x=12, f(12)= -2. Therefore, between x=11 and x=12, the function goes from 62 to -2. So, crossing zero between 11 and 12. Therefore, the zero is in (11,12), which is part of (11,18). Therefore, D includes a zero. So D is not the answer.So, recapping: A has a zero at x=3, so (1,4) includes a zero. B: (3,7) is entirely in the first piece where the function is positive, so no zero. C: (8,13) includes a zero in (10,13). D: (11,18) includes a zero in (11,12). Therefore, the interval that cannot have a zero is B: (3,7). Therefore, the answer is B.Wait, but let me double-check. The first piece: from x=0 to x=10. At x=3, it's zero. Then from x=3 onwards, the first piece is increasing because 2^x is increasing. So, from x=3 to x=10, f(x) goes from 0 to 3048. So, in (3,7), the function is positive. So, in that interval, f(x) is always positive, hence no zero. Therefore, B is the interval where there cannot be a zero. Therefore, the answer should be B.But let me check the options again. The problem says "cannot have a zero in which of the following intervals?" So, the answer is B. But let me make sure I didn't make a mistake with the other intervals.For interval C: (8,13). From x=8 to x=10, the first piece is positive. From x=10 to x=13, the second piece goes from 3048 (at x=10 left limit) to 94 (at x=10 right limit), but actually, at x=10, the first piece is 3048, then the second piece at x=10+ is 94, then decreasing to -130 at x=13. So, there's a discontinuity at x=10, but the second piece is continuous from x=10 to x=20. Wait, hold on, the function is defined as 3*2^x -24 for 0 ≤x ≤10, and -2^{x-5} +126 for 10 <x ≤20. So, at x=10, it's still the first piece. Then, just above x=10, it's the second piece. Therefore, the second piece is continuous on (10,20], but there's a jump discontinuity at x=10. So, from the left at x=10, the function is 3048; from the right, approaching x=10 from above, it's approaching 94. Therefore, the function isn't continuous at x=10. Therefore, when considering the interval (8,13), the function is positive from 8 to 10, then drops to 94 at x=10+ and continues decreasing to -130 at x=13. Since the second piece is continuous on (10,20], by IVT, since it goes from 94 to -130, it must cross zero somewhere between x=10 and x=13. Therefore, even though there's a discontinuity at x=10, the second piece on its own crosses zero in (10,13). Therefore, the entire function f(x) has a zero in (10,13), which is part of interval C: (8,13). Therefore, C does have a zero.Similarly, interval D: (11,18). The zero occurs between x=10 and x=13, specifically between x=11 and x=12 as we saw earlier. Therefore, (11,18) includes that zero.Interval A: (1,4). The zero is at x=3, so (1,4) includes x=3, but since it's open interval, x=3 is not included. Wait, but hold on, x=3 is part of the domain (0 ≤x ≤10), so in the first piece. But the interval (1,4) is open, so x=3 is not in the interval. Wait, but wait, the function has a zero exactly at x=3. But the interval (1,4) is open, so it doesn't include x=3. However, does the function cross zero within the interval (1,4)? At x=1, the function is -18, at x=4 it's 24. Therefore, even though the zero is at x=3, which is not included in the open interval (1,4), the function still crosses from negative to positive within that interval. Wait, actually, the zero is at x=3, which is within [0,10], but the interval (1,4) is open. But the question is about the interval (1,4). So, even if the zero is at x=3, which is in (1,4) because 1 < 3 <4. Wait, (1,4) is all real numbers between 1 and 4, not including 1 and 4. Since 3 is between 1 and 4, x=3 is in (1,4). Wait, no: hold on, open interval (1,4) includes all x such that 1 < x <4. Since 3 is between 1 and 4, 1 <3 <4, so x=3 is in (1,4). Therefore, the zero is at x=3, which is in the interval (1,4). Therefore, the function does have a zero in interval A. Therefore, interval A does contain a zero.Wait, but hold on, the options are intervals where the function "cannot have a zero". So, if the zero is at x=3, which is inside interval A: (1,4), then interval A does have a zero. So, A is not the answer.But earlier, I thought interval B: (3,7) is entirely in the first piece where the function is positive. Wait, but (3,7) is from x=3 to x=7. Since x=3 is part of the domain, but (3,7) is open, so x=3 is not included. But the function at x=3 is zero. However, in the open interval (3,7), x=3 is excluded. So, from x just above 3 to x=7, the function is positive because f(x) is 3*2^x -24. Since at x=3, it's zero, and 2^x is increasing, so for x >3, 2^x >8, so 3*2^x -24 >0. Therefore, in (3,7), the function is always positive. Therefore, there's no zero in (3,7). Therefore, the answer is B.Yes, that seems correct. So, the answer is B.Final Answerboxed{B}</think>To determine which interval cannot contain a zero of the piecewise function ( f(x) ), we analyze each piece of the function and check for sign changes within the given intervals.1. First piece ( 3 times 2^x - 24 ) for ( 0 leqslant x leqslant 10 ): - Solving ( 3 times 2^x - 24 = 0 ) gives ( x = 3 ). This zero is at ( x = 3 ). - For ( x > 3 ), the function is positive and increasing. In the interval ( (3, 7) ), the function remains positive as ( 3 times 2^x - 24 ) increases from 0 to 360.2. Second piece ( -2^{x-5} + 126 ) for ( 10 < x leqslant 20 ): - At ( x = 10 ), the first piece evaluates to 3048, and the second piece starts at 94 when approaching from the right. - The second piece decreases from 94 at ( x = 10 ) to -130 at ( x = 13 ), crossing zero between ( x = 11 ) and ( x = 12 ).3. Checking intervals: - Interval A: (1, 4) contains the zero at ( x = 3 ). - Interval B: (3, 7) is within the first piece where the function is positive, so no zero exists here. - Interval C: (8, 13) includes the zero between ( x = 10 ) and ( x = 13 ). - Interval D: (11, 18) includes the zero between ( x = 11 ) and ( x = 12 ).Thus, the interval that cannot have a zero is (boxed{B}).]

🔑:When 0 leqslant x leqslant 10, f(x) = 3 times 2^{x} - 24 is an increasing function. We have f(1) = 3 times 2 - 24 = -18 0. By the Intermediate Value Theorem, f(x) has a zero in (1, 4).Next, f(3) = 3 times 8 - 24 = 0 and f(7) = 3 times 2^{7} - 24 = 364 > 0. Since f(x) is increasing on (0, 10), it has no zero in (3, 7).When 10 0 and f(13) = 126 - 2^{8} 0 and f(18) = 126 - 2^{13} < 0. Since f(x) is decreasing on (11, 18), it has a zero in (11, 18).Therefore, f(x) cannot have a zero in (3, 7). The answer is boxed{B}.This problem requires understanding the concept of a function's zero and how to apply the Intermediate Value Theorem. It also tests computational skills and the ability to make judgments based on a function's monotonicity. This is a basic-level problem.

🧠:[<think>Okay, let's try to tackle this geometry problem. So, we have a right triangular pyramid A₁-ABC. AA₁ is perpendicular to the base ABC. BC is perpendicular to A₁B, and AA₁ equals AC equals 2. We need to find the surface area of the circumscribed sphere around this pyramid. Hmm, okay.First, let me visualize the pyramid. A right triangular pyramid means that one of the edges (AA₁) is perpendicular to the base ABC. The base is a triangle, ABC. Since AA₁ is perpendicular to ABC, A₁ is the apex of the pyramid, and AA₁ is the height from the apex to the base.Given that BC is perpendicular to A₁B. That seems important. So BC and A₁B are perpendicular. Also, AA₁ = AC = 2. So the length from A to A₁ is 2, and the length from A to C is also 2. Let me try to sketch this mentally.Let me start by setting up a coordinate system. Since AA₁ is perpendicular to the base ABC, let me place point A at the origin (0,0,0). Then, since AA₁ is vertical, A₁ will be at (0,0,2). The base ABC lies in the z=0 plane.Given that AC = 2. So point C is somewhere such that the distance from A (0,0,0) to C is 2. Let's say C is at (2,0,0), lying along the x-axis. But maybe that's not the case. Wait, but we need BC to be perpendicular to A₁B. Let me not assume coordinates yet.Alternatively, maybe assign coordinates step by step. Let's set point A at (0,0,0). Then, since AA₁ is vertical, A₁ is at (0,0,2). The base ABC is in the plane z=0. Let me denote coordinates for B and C as (x₁,y₁,0) and (x₂,y₂,0), respectively.We know AC = 2. So the distance from A (0,0,0) to C (x₂,y₂,0) is 2, so sqrt(x₂² + y₂²) = 2. Therefore, x₂² + y₂² = 4.Now, BC is perpendicular to A₁B. Let's parse this. BC is the vector from B to C, which is (x₂ - x₁, y₂ - y₁, 0). A₁B is the vector from A₁ to B, which is (x₁ - 0, y₁ - 0, 0 - 2) = (x₁, y₁, -2). For these vectors to be perpendicular, their dot product should be zero.So, (x₂ - x₁)(x₁) + (y₂ - y₁)(y₁) + (0)(-2) = 0. Therefore, (x₂ - x₁)x₁ + (y₂ - y₁)y₁ = 0.Simplifying, x₂x₁ - x₁² + y₂y₁ - y₁² = 0. So, x₁x₂ + y₁y₂ = x₁² + y₁².Alternatively, that's equivalent to x₁x₂ + y₁y₂ = |B|², where B is (x₁, y₁, 0). Hmm.Also, since ABC is a triangle in the z=0 plane. Maybe we can find coordinates for B and C such that the conditions are satisfied.Alternatively, maybe there's a way to choose coordinates so that some axes align with edges. Let me think.Since AA₁ is perpendicular to ABC, and we need BC perpendicular to A₁B. Maybe if we let B lie along the y-axis and C lie in the x-y plane.Wait, but AC = 2. Let's assume point C is (2,0,0), so AC is along the x-axis. Then, point B can be somewhere else. Let me try that.So, set A at (0,0,0), A₁ at (0,0,2), C at (2,0,0). Then, B is (x, y, 0). Then, BC is the vector from B to C: (2 - x, -y, 0). A₁B is the vector from A₁ to B: (x, y, -2). The dot product of BC and A₁B should be zero.Dot product: (2 - x)x + (-y)y + 0*(-2) = 0. So:x(2 - x) - y² = 0.That simplifies to 2x - x² - y² = 0.Additionally, since ABC is a triangle, points A, B, C are not colinear, so B cannot be on the line AC. Since AC is from (0,0,0) to (2,0,0), so if B is not on the x-axis.Let me also note that the coordinates of B are (x, y, 0), and we have 2x - x² - y² = 0.This is equivalent to x² + y² = 2x, which can be rewritten as (x - 1)^2 + y² = 1. So, point B lies on a circle of radius 1 centered at (1, 0, 0). That's interesting.So, possible coordinates for B are (1 + cosθ, sinθ, 0), where θ is some angle. But maybe we can pick a specific θ to simplify the problem?Alternatively, since we need to compute the surface area of the circumscribed sphere, which requires knowing the radius of the sphere that passes through all the vertices of the pyramid A₁-ABC. To find the radius, we need to find the circumradius of the pyramid, which is the radius of the sphere passing through all five points: A₁, A, B, C, and the other points. Wait, actually, a triangular pyramid (tetrahedron) has four vertices: A₁, A, B, C. So, the circumscribed sphere is the one passing through these four points. The surface area would then be 4πR², where R is the circumradius.Wait, the problem says "surface area of the circumscribed sphere". So, once we find R, the radius, then the surface area is 4πR². So, our main task is to find R.To find the circumradius of a tetrahedron, we can use the formula:R = frac{|vec{AB} cdot (vec{AC} times vec{AA₁})|}{6V}Wait, but maybe another formula. Alternatively, the circumradius can be found by solving the equation of the sphere passing through all four points. Since all four points lie on the sphere, we can set up equations based on their coordinates.Alternatively, given the coordinates of the four points, we can find the center (x₀, y₀, z₀) of the sphere such that the distance from (x₀, y₀, z₀) to each of the four points is equal to R. Then, solve the system of equations.But to do that, we need the coordinates of all four points. So, let's try to assign coordinates step by step.We have A at (0,0,0), A₁ at (0,0,2), C at (2,0,0). Let me denote point B as (x, y, 0). From earlier, we have the condition that 2x - x² - y² = 0, so x² + y² = 2x. Therefore, point B is somewhere on the circle (x - 1)^2 + y² = 1 in the z=0 plane.But to determine the exact coordinates of B, we might need another condition. Wait, are there any other given conditions? The problem states it's a right triangular pyramid. Wait, a right triangular pyramid usually means that the apex is directly over the right angle of the base. But in our case, the base is ABC, and AA₁ is perpendicular to the base. If ABC is a right triangle, then it's a right triangular pyramid. Wait, but the problem says "right triangular pyramid A₁-ABC", so maybe the base ABC is a right triangle? The term is a bit ambiguous.Wait, perhaps the pyramid is right in the sense that the apex is directly over the base's vertex, which is the case here since AA₁ is perpendicular to ABC, so A₁ is over A. But whether the base ABC is a right triangle or not is not explicitly stated. However, given that BC is perpendicular to A₁B, which might imply some right angles in the base.Alternatively, maybe the pyramid has some right triangular faces. Hmm.But perhaps we can proceed with coordinates. Let's assume C is at (2,0,0) as before. Then, point B is (x, y, 0) on the circle (x - 1)^2 + y² = 1. Let's parameterize B.Let me set θ as the angle parameter. So, coordinates of B can be (1 + cosθ, sinθ, 0). Then, let's compute the coordinates of B as (1 + cosθ, sinθ, 0). Then, we can proceed to find the circumradius.But maybe there's a specific θ that satisfies other conditions? Wait, perhaps not. Maybe the circumradius is the same regardless of θ? But that seems unlikely. Wait, but the problem doesn't give more constraints on B. Wait, but maybe there is another condition.Wait, in the problem statement, is there any missing information? Let's check again: In the right triangular pyramid A₁-ABC, AA₁ is perpendicular to the base ABC, BC is perpendicular to A₁B, and AA₁=AC=2. Find the surface area of the circumscribed sphere.So, we have AA₁=2, AC=2. So AC=2 is given, which we already considered. But BC is perpendicular to A₁B. So, given these conditions, perhaps the position of B is uniquely determined.Wait, in our earlier setup, with A at (0,0,0), A₁ at (0,0,2), C at (2,0,0), then B is on the circle (x - 1)^2 + y² =1. But is there another condition that can fix θ?Alternatively, maybe the triangle ABC must be such that BC is perpendicular to A₁B. Wait, but BC is in the base plane (z=0), and A₁B is a vector going from A₁(0,0,2) to B(x, y, 0). So their vectors are BC = (2 - x, -y, 0) and A₁B = (x, y, -2). Their dot product is zero, which gives 2x - x² - y² = 0, which we already have. So that equation is satisfied for all points B on that circle. So, maybe B can be any point on that circle, but the problem doesn't specify any further constraints. Hmm, that seems odd. Maybe there is a unique solution?Wait, but if B can be anywhere on that circle, then the resulting tetrahedron would have different shapes, and thus different circumradii. But the problem asks for a specific surface area, implying that the answer is unique. Therefore, there must be an additional constraint that we're missing.Wait, let me check the problem statement again: "right triangular pyramid A₁-ABC". The term "right triangular pyramid" typically means that the apex is directly over the right angle of the base triangle. So, if ABC is a right triangle with the right angle at A, then the apex A₁ is over A. But in that case, AA₁ is perpendicular to ABC, which matches the given condition.But the problem doesn't state that ABC is a right triangle, but if it's a right triangular pyramid, then the base must be a right triangle. So perhaps ABC is a right triangle with right angle at A. If that's the case, then AB and AC are perpendicular. Since AC is given as 2, and AA₁ is 2. Then, if ABC is a right triangle at A, then AB is perpendicular to AC.But in our earlier setup, if AC is along the x-axis from (0,0,0) to (2,0,0), and if ABC has a right angle at A, then AB would be along the y-axis. So, point B would be at (0, b, 0). But then, the vector BC would be (2 - 0, 0 - b, 0) = (2, -b, 0). The vector A₁B would be (0 - 0, b - 0, 0 - 2) = (0, b, -2). The dot product of BC and A₁B would be 2*0 + (-b)*b + 0*(-2) = -b². For them to be perpendicular, this should be zero, which implies b = 0. But if b = 0, then point B would coincide with point A, which is impossible. Therefore, this leads to a contradiction. Hence, ABC cannot be a right triangle at A.Therefore, maybe the right triangular pyramid refers to another configuration. Alternatively, maybe the term "right" here refers to the apex being directly over the base, which is a triangle. But the base doesn't necessarily have to be a right triangle. Wait, but usually, a right pyramid has the apex directly over the centroid or the center of the base. But if the base is a triangle, then the centroid is the intersection of medians. However, in this case, the apex is over vertex A, so it's not a right pyramid in that sense. Maybe the term "right" here is used differently.Alternatively, maybe "right triangular pyramid" refers to the fact that one of the lateral edges (AA₁) is perpendicular to the base, making it a right pyramid. But I think the definition can vary. Anyway, perhaps the key is that AA₁ is perpendicular to ABC, which is given.Given that, and AC = 2, BC perpendicular to A₁B. Let me try to find coordinates for B.We have A(0,0,0), A₁(0,0,2), C(2,0,0). Let B be (x, y, 0). Then, AC = 2, which is satisfied. BC is (2 - x, -y, 0). A₁B is (x, y, -2). Their dot product is x(2 - x) + y(-y) + 0*(-2) = 2x - x² - y² = 0. So, 2x = x² + y².Additionally, the problem doesn't specify any other conditions, so perhaps we need to find the circumradius in terms of x and y, but given that 2x = x² + y². However, since the answer is a specific number, the radius must be fixed. Therefore, there might be a unique position for B that satisfies all conditions, but I need to check.Wait, perhaps there is another condition we haven't considered. Let's think about the geometry. The pyramid is A₁-ABC, with AA₁ perpendicular to ABC. BC is perpendicular to A₁B. AC = 2, AA₁ = 2.Wait, maybe ABC is also a right triangle. If ABC is a right triangle, then the right angle must be at B or C, because we saw that if it's at A, it leads to a contradiction. Let's check.Suppose ABC is a right triangle at B. Then, AB is perpendicular to BC. Let's see. If AB is perpendicular to BC, then the vectors AB and BC are perpendicular. AB is (x, y, 0), BC is (2 - x, -y, 0). Their dot product is x(2 - x) + y(-y) = 2x - x² - y² = 0. But that's exactly the same equation as before. So, if ABC is a right triangle at B, then BC is perpendicular to AB, which gives 2x - x² - y² = 0. Which is already our condition. So, in this case, ABC is a right triangle at B. Therefore, the problem's pyramid is a right triangular pyramid with base ABC being a right triangle at B, and apex A₁ directly over A.Therefore, now we know ABC is a right triangle at B. So, AB is perpendicular to BC, and BC is perpendicular to A₁B. Wait, but BC is perpendicular to both AB and A₁B? That might mean that BC is perpendicular to the plane containing AB and A₁B. Wait, but BC is in the base plane z=0, and A₁B is a vector going up to z=2.But if BC is perpendicular to A₁B, which is a vector not in the base plane, then BC is perpendicular to A₁B in 3D space. Wait, but in 3D space, two vectors being perpendicular just means their dot product is zero, which we already have. So, given that, and ABC being a right triangle at B, we can proceed.So, with ABC being a right triangle at B, coordinates are A(0,0,0), B(x, y, 0), C(2,0,0), with AB perpendicular to BC. So AB · BC = 0, leading to 2x - x² - y² = 0, as before. So, point B lies on the circle (x - 1)^2 + y² = 1.But we need to find the circumradius of tetrahedron A₁-ABC. To find the circumradius, we can find the center of the sphere equidistant from all four points.Let me denote the center of the sphere as (p, q, r). The distance from (p, q, r) to A(0,0,0) is sqrt(p² + q² + r²).Distance to A₁(0,0,2) is sqrt(p² + q² + (r - 2)^2).Distance to B(x, y, 0) is sqrt((p - x)^2 + (q - y)^2 + r²).Distance to C(2,0,0) is sqrt((p - 2)^2 + q² + r²).Since all these distances are equal to the radius R, we can set up equations by squaring them:1. p² + q² + r² = R² (distance to A)2. p² + q² + (r - 2)^2 = R² (distance to A₁)3. (p - x)^2 + (q - y)^2 + r² = R² (distance to B)4. (p - 2)^2 + q² + r² = R² (distance to C)Subtract equation 1 from equation 2:[p² + q² + (r - 2)^2] - [p² + q² + r²] = 0Expanding: r² -4r +4 + q² + p² - p² - q² - r² = 0 => -4r +4 =0 => r =1.So the z-coordinate of the center is 1.Now, substitute r=1 into equations 1, 3, 4:1. p² + q² + 1 = R²3. (p - x)^2 + (q - y)^2 + 1 = R²4. (p - 2)^2 + q² + 1 = R²Subtract equation 1 from equation 4:[(p - 2)^2 + q² +1] - [p² + q² +1] =0Expanding: p² -4p +4 + q² - p² - q² =0 => -4p +4=0 => p=1.So the x-coordinate of the center is 1.Now, substitute p=1 and r=1 into equation 1:1² + q² +1 = R² => q² +2 = R².Now, substitute p=1, r=1 into equation 3:(1 - x)^2 + (q - y)^2 +1 = R²But R² = q² +2, so:(1 - x)^2 + (q - y)^2 +1 = q² +2Simplify:(1 - x)^2 + (q - y)^2 = q² +1Expanding left side:1 - 2x + x² + q² - 2qy + y² = q² +1Cancel q² and 1 from both sides:-2x + x² - 2qy + y² =0But from earlier, we know that x² + y² = 2x (since BC · A₁B =0 gives 2x -x² - y²=0 => x² + y² =2x). So substitute x² + y² =2x into the above equation:-2x + (2x) -2qy =0 => (-2x +2x) -2qy=0 => 0 -2qy=0 => -2qy=0.So, either q=0 or y=0.But y=0 would mean point B is on the x-axis. However, in our setup, if y=0, then from x² + y²=2x, we get x²=2x => x=0 or x=2. But x=0 would be point A, and x=2 would be point C. But B cannot coincide with A or C, so y≠0. Therefore, we must have q=0.Therefore, q=0. Then, from equation 1, R²=0 +0 +1 +1=2? Wait, wait, equation 1 with p=1, q=0, r=1:R²=1² +0² +1²=1+0+1=2. Therefore, R=√2.Wait, but let me check. If q=0, then the center is at (1,0,1). Let's verify if this is equidistant to all four points.Distance to A(0,0,0): sqrt((1-0)^2 + (0-0)^2 + (1-0)^2)=sqrt(1+0+1)=sqrt(2).Distance to A₁(0,0,2): sqrt((1-0)^2 + (0-0)^2 + (1-2)^2)=sqrt(1+0+1)=sqrt(2).Distance to C(2,0,0): sqrt((1-2)^2 + (0-0)^2 + (1-0)^2)=sqrt(1+0+1)=sqrt(2).Distance to B(x, y, 0): sqrt((1 - x)^2 + (0 - y)^2 + (1 - 0)^2)=sqrt((1 -x)^2 + y² +1). But from x² + y² =2x, so (1 -x)^2 + y² =1 -2x +x² + y²=1 -2x +2x=1. Therefore, sqrt(1 +1)=sqrt(2). So yes, the distance from center (1,0,1) to B is also sqrt(2). Therefore, all points lie on a sphere with radius √2, centered at (1,0,1). Therefore, the surface area is 4π(√2)²=4π*2=8π.Wait, but hold on. This seems too straightforward. Let me verify again.We found that regardless of the position of B on the circle (x -1)^2 + y²=1, the center of the sphere is always (1,0,1), and radius √2. Therefore, the surface area is 8π. Is that possible?Wait, but how can the position of B not affect the circumradius? Because as B moves around the circle, the tetrahedron changes shape, but maybe due to the constraints given, the circumradius remains constant. Let me check with a specific example.Let me pick B at (1 +1, 0,0)=(2,0,0). But that's point C. So, invalid. Next, take θ=90 degrees. Then, B is (1 + cos90, sin90,0)=(1 +0,1,0)=(1,1,0). Let's check if this satisfies x² + y²=2x. 1² +1²=2≠2*1=2. Wait, 1 +1=2, 2*1=2. So, yes, 2=2. So, point B=(1,1,0) is valid.Then, check the distance from center (1,0,1) to B=(1,1,0):sqrt((1-1)^2 + (0-1)^2 + (1-0)^2)=sqrt(0 +1 +1)=sqrt(2). Correct.Another example: θ=180 degrees. Then, B=(1 + cos180, sin180,0)=(1 -1,0,0)=(0,0,0). But that's point A, which can't be. θ=60 degrees: B=(1 + cos60, sin60,0)=(1 +0.5, √3/2,0)=(1.5, √3/2,0). Check x² + y²: (1.5)^2 + (√3/2)^2=2.25 + 0.75=3. 2x=2*1.5=3. So, 3=3. Correct.Distance from center (1,0,1) to B=(1.5, √3/2,0):sqrt((1 -1.5)^2 + (0 - √3/2)^2 + (1 -0)^2)=sqrt(0.25 + 0.75 +1)=sqrt(2)=√2. Correct.So regardless of where B is on that circle, the distance remains √2. Therefore, the sphere centered at (1,0,1) with radius √2 always passes through all four points. Therefore, the surface area is 4π*(√2)^2=8π.Therefore, the answer should be 8π.But let me just confirm once again. The problem states that AA₁=AC=2. In our coordinate system, AA₁ is from (0,0,0) to (0,0,2), so length 2. AC is from (0,0,0) to (2,0,0), length 2. Correct. BC is perpendicular to A₁B, which we used to derive the condition x² + y²=2x, and the center of the sphere turned out to be (1,0,1) with radius √2. So surface area 8π. Seems correct.So, I think the answer is 8π, so the surface area is 8π, which would be boxed as boxed{8pi}.Final Answerboxed{8pi}</think>Given a right triangular pyramid (A_1-ABC) with (AA_1) perpendicular to the base (ABC), (BC) perpendicular to (A_1B), and (AA_1 = AC = 2), we need to find the surface area of the circumscribed sphere.1. Coordinates Setup: - Place (A) at ((0, 0, 0)), (A_1) at ((0, 0, 2)), and (C) at ((2, 0, 0)). - Let (B) be ((x, y, 0)). Given (AC = 2), (C) is correctly placed.2. Perpendicular Condition: - The vector (BC = (2 - x, -y, 0)) and vector (A_1B = (x, y, -2)) must be perpendicular. - Their dot product is zero: (x(2 - x) + y(-y) = 0 implies 2x - x^2 - y^2 = 0 implies x^2 + y^2 = 2x). This implies (B) lies on the circle ((x - 1)^2 + y^2 = 1).3. Circumscribed Sphere: - The center of the circumscribed sphere ((p, q, r)) must be equidistant from all vertices. - Solving the system of equations for the distances from the center to each vertex, we find: - (r = 1) (from (A) and (A_1)) - (p = 1) (from (A) and (C)) - (q = 0) (from (B)) - The center is ((1, 0, 1)) and radius (R = sqrt{2}).4. Surface Area Calculation: - The surface area of the sphere is (4pi R^2 = 4pi (sqrt{2})^2 = 8pi).Thus, the surface area of the circumscribed sphere is (boxed{8pi}).,<think>Okay, let me try to figure out this problem step by step. So, we have a right triangular pyramid A₁-ABC. AA₁ is perpendicular to the base ABC. Also, BC is perpendicular to A₁B, and we know that AA₁ and AC are both 2 units long. We need to find the surface area of the circumscribed sphere around this pyramid. First, let me visualize the pyramid. Since AA₁ is perpendicular to the base ABC, that makes A₁-ABC a right triangular pyramid, meaning the apex A₁ is directly above point A. The base is triangle ABC, and AA₁ is one of the edges perpendicular to the base.Given that AA₁ = AC = 2. So, the length from A to A₁ is 2, and from A to C is also 2. Then, BC is perpendicular to A₁B. Hmm, that's an important condition. So, BC and A₁B are perpendicular. I need to use that to find some relations between the sides.Since it's a triangular pyramid with a right triangle base maybe? Wait, but the base is ABC. Wait, let's clarify: in a right triangular pyramid, the apex is directly over one of the vertices of the base, which in this case is A. So, the base is triangle ABC, and AA₁ is the altitude of the pyramid.Given that BC is perpendicular to A₁B. So, BC ⊥ A₁B. That seems like a key geometric condition here. Let me try to sketch this mentally. So, points A, B, C, and A₁. AA₁ is vertical (perpendicular to ABC). AC is 2, so AC is a horizontal line (since AA₁ is vertical) of length 2.Now, BC is perpendicular to A₁B. Let me consider the coordinates. Maybe assigning coordinates would help. Let me place point A at the origin (0,0,0). Since AA₁ is perpendicular to the base ABC, which is in the xy-plane. Then, A₁ is at (0,0,2). Given AC = 2. Since AC is in the base ABC (the base is in the xy-plane), point C must be at (2,0,0), because AC is along the x-axis from A(0,0,0) to C(2,0,0). Now, point B is somewhere in the base ABC. Let's denote B as (x, y, 0). We need to determine x and y. Given that BC is perpendicular to A₁B. Let's express vectors BC and A₁B.First, vector BC is C - B = (2 - x, 0 - y, 0 - 0) = (2 - x, -y, 0).Vector A₁B is B - A₁ = (x - 0, y - 0, 0 - 2) = (x, y, -2).Since BC is perpendicular to A₁B, their dot product should be zero:(2 - x)x + (-y)y + 0*(-2) = 0So:(2 - x)x - y² = 0Simplify:2x - x² - y² = 0Equation (1): x² + y² = 2xThat's one equation relating x and y. But we need more information to find the coordinates of B.Since ABC is a triangle in the base (the xy-plane), and we have points A(0,0,0), C(2,0,0), and B(x, y, 0). We might need another condition. However, the problem doesn't specify any other lengths or angles. Wait, but the pyramid is a right triangular pyramid. Maybe the base ABC is a right triangle? If the pyramid is a right triangular pyramid, does that mean the base is a right triangle? Hmm, sometimes "right pyramid" refers to the apex being over the centroid or a vertex, but here it's specified as a "right triangular pyramid," which probably means that the apex is directly over one of the vertices, which is already given as A. So the base ABC can be any triangle, but in this case, we might need to use other given conditions.Wait, the only given lengths are AA₁=2 and AC=2. So AC is 2, which we've placed from (0,0,0) to (2,0,0). Then, point B is (x, y, 0). We need to find x and y such that BC is perpendicular to A₁B, leading to equation (1). But with only that equation, we can't find x and y. So maybe there are other constraints?Wait, the pyramid is right triangular, but perhaps the base ABC is a right triangle? If so, then ABC must have a right angle. Since AA₁ is perpendicular to the base, and the apex is over A, so if the base is a right triangle, the right angle would be at A. But in that case, AB and AC would be perpendicular. But AC is given as 2, but AB is not given. However, if ABC is a right triangle at A, then AB and AC are perpendicular. Then, we can have coordinates: A(0,0,0), C(2,0,0), B(0, b, 0). Then, point B would be (0, b, 0). Then, let's check if BC is perpendicular to A₁B in that case.Wait, let's suppose ABC is a right triangle at A. Then, B is (0, b, 0), C is (2,0,0). Then vector BC is (2, -b, 0), and vector A₁B is (0, b, -2). Their dot product is (2)(0) + (-b)(b) + (0)(-2) = -b². For them to be perpendicular, the dot product should be zero. Therefore, -b² = 0 → b = 0. But then point B would be (0,0,0), which coincides with A. That can't be. Therefore, ABC cannot be a right triangle at A. So maybe the right angle is at B or C? Let's check.If ABC is a right triangle at B, then AB and BC are perpendicular. So, vector AB = (x, y, 0) - (0,0,0) = (x, y, 0), and vector BC = (2 - x, -y, 0). Their dot product should be zero:x(2 - x) + y(-y) = 0Which is 2x - x² - y² = 0. Wait, that's the same as equation (1). So if ABC is a right triangle at B, then equation (1) is satisfied. But we also have the condition that BC is perpendicular to A₁B, which also gives equation (1). So, in this case, ABC is a right triangle at B, and also BC is perpendicular to A₁B. So, maybe these conditions are the same? Wait, but if ABC is a right triangle at B, then BC is perpendicular to AB, but here we have BC perpendicular to A₁B, which is different. So perhaps ABC is a right triangle at B, but also BC is perpendicular to A₁B, which is a different vector. So maybe these two conditions can coexist?Wait, let me check again. If ABC is a right triangle at B, then AB is perpendicular to BC. But we also have BC perpendicular to A₁B. So, BC is perpendicular to both AB and A₁B. If BC is perpendicular to both AB and A₁B, then BC is perpendicular to the plane containing AB and A₁B. But AB and A₁B are two vectors emanating from B. Wait, AB is from B to A, and A₁B is from B to A₁. So the plane containing AB and A₁B is the plane that contains points A, B, and A₁. Since AA₁ is vertical, this plane would be the plane containing the vertical edge AA₁ and the point B.But if BC is perpendicular to this plane, then BC must be perpendicular to both AB and A₁B. But in our case, we are given that BC is perpendicular to A₁B, which gives equation (1). But if ABC is a right triangle at B, then BC is also perpendicular to AB, which gives the same equation (1). Therefore, in this problem, the two conditions are equivalent? So, if ABC is a right triangle at B, then BC is perpendicular to both AB and A₁B. Therefore, in this case, ABC is a right triangle at B, and BC is also perpendicular to A₁B. Thus, equation (1) is satisfied.Therefore, ABC is a right triangle at B. So, point B is such that AB is perpendicular to BC. Therefore, we can use that to find coordinates of B. Wait, but how?Wait, since ABC is a right triangle at B, then AB and BC are perpendicular. Also, BC is perpendicular to A₁B. So, as we saw, these two conditions lead to the same equation. Therefore, perhaps we can use that ABC is a right triangle at B, and then use equation (1) to find coordinates of B.Wait, but if ABC is a right triangle at B, then coordinates of B must satisfy that AB · BC = 0. Let's see:Vector AB = (x, y, 0)Vector BC = (2 - x, -y, 0)Their dot product is x(2 - x) + y(-y) = 2x - x² - y² = 0, which is the same as equation (1). So, the coordinates of B must satisfy x² + y² = 2x.Additionally, since ABC is a triangle in the base, with AC = 2. So, the distance from A to C is 2, which is already satisfied as we placed C at (2,0,0). The other sides AB and BC can be found once we have coordinates of B.But we need another condition to determine x and y. Wait, the problem doesn't mention any other lengths. Hmm, maybe we can use the fact that the pyramid is right, so AA₁ is perpendicular to the base, and thus the apex A₁ is directly above A. Then, perhaps the lateral edges are related? But not necessarily. Wait, unless there are other perpendicular conditions.Alternatively, maybe we need to use the fact that the circumscribed sphere exists, and all vertices lie on the sphere. But since we need to find the surface area of the sphere, we need to find the radius. The surface area is 4πr², so if we can find the radius of the circumscribed sphere, then we can compute the surface area.To find the radius, we need to find the center of the sphere, which is equidistant from all five vertices: A, B, C, A₁, and possibly other edges, but in a triangular pyramid, there are four vertices: A, B, C, A₁. Wait, no, in a pyramid, the base is a triangle, and the apex. So, four vertices total: A, B, C, A₁. So, the circumscribed sphere must pass through these four points.Therefore, the center of the sphere (let's call it O) must be equidistant from A, B, C, and A₁. Let me denote O as (h, k, l). Then, the distance from O to A (0,0,0) is √(h² + k² + l²). The distance from O to C (2,0,0) is √((h - 2)² + k² + l²). The distance from O to A₁ (0,0,2) is √(h² + k² + (l - 2)²). The distance from O to B (x, y, 0) is √((h - x)² + (k - y)² + l²).Since all these distances must be equal, we can set up equations:1. OA = OC: √(h² + k² + l²) = √((h - 2)² + k² + l²)Squaring both sides: h² = (h - 2)² → h² = h² -4h +4 → 0 = -4h +4 → 4h =4 → h =1.2. OA = OA₁: √(h² + k² + l²) = √(h² + k² + (l - 2)²)Squaring: h² + k² + l² = h² + k² + (l - 2)² → l² = l² -4l +4 → 0 = -4l +4 → 4l=4 → l=1.So, from OA = OC and OA = OA₁, we have h=1 and l=1. So the center O is at (1, k, 1). Now, we need to find k such that OA = OB and OA = OC (already satisfied OA=OC since h=1 and l=1). Let's write OA = OB.OA is √(1² + k² +1²) = √(2 + k²)OB is √((1 - x)² + (k - y)² +1²)So, set √(2 + k²) = √((1 - x)² + (k - y)² +1)Square both sides:2 + k² = (1 - x)² + (k - y)² +1Simplify:2 + k² = (1 - 2x + x²) + (k² - 2ky + y²) +1Simplify the right side:1 - 2x + x² + k² - 2ky + y² +1 = x² + y² -2x -2ky + k² +2Set equal to left side:2 + k² = x² + y² -2x -2ky + k² +2Subtract 2 + k² from both sides:0 = x² + y² -2x -2kyBut from equation (1), we have x² + y² = 2x. Substitute into above:0 = 2x -2x -2ky → 0 = -2ky → -2ky =0Thus, either k=0 or y=0.If y=0, then from equation (1): x² +0=2x → x²=2x →x=0 or x=2. But x=0 would mean point B is at (0,0,0), same as A, which can't be. x=2 would mean point B is at (2,0,0), same as C, which also can't be. Therefore, y cannot be 0, so k must be 0.Therefore, k=0.So the center of the sphere is at (1,0,1). Now, let's verify if OA = OB. The center is (1,0,1). Let's compute OA: √(1² +0² +1²)=√2. OB is √((1 - x)² + (0 - y)² +1²) = √((1 -x)^2 + y² +1). We need this to equal √2. So:√((1 -x)^2 + y² +1) = √2 → (1 -x)^2 + y² +1 =2 → (1 -x)^2 + y² =1But from equation (1): x² + y² =2x. Let's write (1 -x)^2 + y² =1 → 1 -2x +x² + y² =1 → (x² + y²) -2x +1 =1. Substitute x² + y² =2x:2x -2x +1 =1 →1=1. Which holds true. So, as long as x and y satisfy equation (1), then OB=√2. Therefore, the center is indeed (1,0,1), and all points A, B, C, A₁ lie on the sphere with radius √2.Wait, but hold on. Let me confirm this. If the radius is √2, then the surface area is 4π*(√2)^2 =4π*2=8π. But let's verify with point A₁. The distance from center (1,0,1) to A₁(0,0,2) is √((1-0)^2 + (0-0)^2 + (1 -2)^2) = √(1 +0 +1)=√2. Similarly, distance to C(2,0,0): √((1-2)^2 +0 + (1 -0)^2)=√(1 +1)=√2. Distance to A(0,0,0): √(1 +0 +1)=√2. Distance to B(x,y,0): √((1 -x)^2 + y² +1). As we saw earlier, since (1 -x)^2 + y² =1, then the distance is √(1 +1)=√2. So yes, all points are at distance √2 from center (1,0,1). Therefore, the radius is √2, so surface area is 4π*(√2)^2=8π.But wait, that seems too straightforward. Let me check if there's something missing. Because in this case, the coordinates of B are any point on the circle x² + y² =2x, which is a circle with center at (1,0) and radius 1. But the sphere's center is at (1,0,1) and radius √2. However, the problem doesn't specify the position of B beyond the given conditions. But since all such B would lie on the sphere, maybe the sphere is uniquely determined regardless of B's position? Wait, but in 3D space, four non-coplanar points determine a sphere. Here, points A, B, C, A₁. If B is moving on a circle, does the sphere remain the same? Wait, in our calculation, the center (1,0,1) and radius √2 works for any B on the circle x² + y²=2x. Because when we derived the center, we found that the only possible center is (1,0,1) with radius √2, regardless of the position of B as long as it's on that circle. So, even though B can move around that circle in the base ABC, the circumscribed sphere remains the same. Therefore, the surface area is 8π.But let me double-check. Let's take a specific example. Suppose B is at (1,1,0). Let's check if this satisfies x² + y²=2x. 1 +1=2*1 → 2=2. Yes. Then, the distance from B to center (1,0,1) is √((1-1)^2 + (1-0)^2 + (0 -1)^2)=√(0 +1 +1)=√2. Correct. So, B is on the sphere. If B is at (1,-1,0), same thing. Distance is √(0 +1 +1)=√2. If B is at (2,0,0), which is point C, distance is √((1-2)^2 +0 +1)=√(1 +1)=√2. So, yes, all points are on the sphere. Therefore, regardless of where B is on the circle x² + y²=2x, the sphere is fixed with center (1,0,1) and radius √2. Hence, the surface area is 8π.Therefore, the answer should be 8π, so surface area is 8π, which in boxed form is boxed{8pi}.Final Answerboxed{8pi}</think>Given a right triangular pyramid ( A_1-ABC ) where ( AA_1 ) is perpendicular to the base ( ABC ), ( BC ) is perpendicular to ( A_1B ), and ( AA_1 = AC = 2 ). We need to find the surface area of the circumscribed sphere of this pyramid.1. Coordinates Assignment: - Place ( A ) at the origin ((0, 0, 0)). - ( A_1 ) is at ((0, 0, 2)) since ( AA_1 ) is vertical. - ( C ) is at ((2, 0, 0)) because ( AC = 2 ).2. Vectors and Perpendicular Condition: - Let ( B ) be ((x, y, 0)). - Vector ( BC = (2 - x, -y, 0) ). - Vector ( A_1B = (x, y, -2) ). - Dot product ( BC cdot A_1B = 0 ) gives the equation ( x(2 - x) - y^2 = 0 ) leading to ( x^2 + y^2 = 2x ).3. Sphere Center and Radius: - The center of the circumscribed sphere must be equidistant from all vertices ( A, B, C, A_1 ). - Solving for the center coordinates, we find ( h = 1 ), ( l = 1 ), and ( k = 0 ). Thus, the center is ((1, 0, 1)). - The radius is calculated as ( sqrt{2} ).4. Surface Area Calculation: - The radius of the sphere is ( sqrt{2} ). - The surface area of the sphere is ( 4pi r^2 = 4pi (sqrt{2})^2 = 8pi ).Thus, the surface area of the circumscribed sphere is (boxed{8pi}).]

🔑:Given that AA_1 is perpendicular to the base ABC and BC is perpendicular to A_1B in the right triangular pyramid A_1-ABC, we can extend the pyramid to form a rectangular box. The diagonal of this box will be the diameter of the circumscribed sphere.Since AA_1=AC=2, we can find the length of the diagonal by using the Pythagorean theorem in the right triangle AA_1C:(AA_1C)^2 = (AA_1)^2 + (AC)^2 = 2^2 + 2^2 = 8So, AA_1C = sqrt{8} = 2sqrt{2}. This is the diameter of the circumscribed sphere, so the radius is half of that, which is sqrt{2}.Now, we can find the surface area of the sphere using the formula S = 4pi R^2, where R is the radius:S = 4pi (sqrt{2})^2 = 4pi (2) = 8piTherefore, the surface area of the circumscribed sphere is boxed{8pi}.This problem tests your understanding of the surface area calculation of a sphere and your ability to analyze and solve problems by recognizing that extending the right triangular pyramid to form a rectangular box allows us to find the diameter of the circumscribed sphere.

🧠:[<think>Okay, let's see. So, the problem is about Mother Hubbard and her chocolate bar. She gave each child one-twelfth of the chocolate bar, and after that, one-third of the bar was left. The question is asking how many children she had, with the options given. Alright, let's break this down step by step.First, let's visualize the chocolate bar. It's a Giant Bear chocolate bar, which I assume is just a standard bar, so the total amount we can consider as 1 whole bar. She gave each child one-twelfth of the bar. So each child gets 1/12 of the chocolate. Then, after giving away these portions, there's still one-third of the bar left. We need to find out how many children received the chocolate, which would be the number of portions she gave out.So, the total chocolate given to the children plus the leftover should equal the whole bar. Let's express that in an equation. Let the number of children be x. Each child gets 1/12 of the bar, so the total given out is x*(1/12). The leftover is 1/3. Therefore:Total given to children + Leftover = Total barx*(1/12) + 1/3 = 1Now, let's solve for x. Let me write that down:x/12 + 1/3 = 1To solve for x, first, subtract 1/3 from both sides:x/12 = 1 - 1/31 minus 1/3 is 2/3, so:x/12 = 2/3Now, to isolate x, multiply both sides by 12:x = (2/3)*12Calculating that, 12 divided by 3 is 4, so 2*4 is 8. So x = 8.Hmm, so the answer should be 8? Let me check again. If she has 8 children, each gets 1/12. So 8*(1/12) is 8/12, which simplifies to 2/3. Then, adding the leftover 1/3, 2/3 + 1/3 = 1. That checks out. So yes, 8 children. So the answer is B, which is 8.Wait, but let me make sure I didn't make any miscalculations. Let's go through it again. The total chocolate given to children is number of children times 1/12. Leftover is 1/3. So total chocolate is 1. Therefore:Number of children * (1/12) + 1/3 = 1If we rearrange:Number of children * (1/12) = 1 - 1/3 = 2/3So, Number of children = (2/3) / (1/12) = (2/3) * (12/1) = 24/3 = 8Yes, that's correct. So 8 is indeed the answer. Let me see if any of the other options could make sense. Suppose if there were 6 children: 6*(1/12)=6/12=1/2. Then leftover would be 1/2, but the problem says leftover is 1/3. So that's not. If it was 12 children: 12*(1/12)=1. Then there would be 0 left, which isn't the case. 15 children: 15/12=1.25, which is more than the bar, which doesn't make sense. Similarly, 18 would be 1.5, which is impossible. So yes, 8 is the correct answer.Alternatively, maybe think of it in terms of fractions. If each child gets 1/12, how many portions of 1/12 are in the amount given to the children. The amount given to children is total minus leftover: 1 - 1/3 = 2/3. So how many 1/12s are in 2/3?Divide 2/3 by 1/12:(2/3) ÷ (1/12) = (2/3)*(12/1) = 24/3 = 8. Yep, same result.So whether you set up the equation or think of it as dividing the portion given by the portion per child, you get 8. So the answer is B, 8. That must be the correct choice. Just to confirm once more. 8 children each get 1/12. 8*1/12=2/3. 2/3 given away, 1/3 left. Perfect. So yes, B.Final Answerboxed{B}</think>Mother Hubbard gave each of her children one-twelfth of the chocolate bar, and one-third of the bar was left. We need to determine how many children she had.1. Let ( x ) be the number of children.2. Each child received ( frac{1}{12} ) of the chocolate bar.3. The total amount given to the children is ( frac{x}{12} ).4. The leftover amount is ( frac{1}{3} ).The equation representing the total chocolate bar is:[frac{x}{12} + frac{1}{3} = 1]Subtract ( frac{1}{3} ) from both sides:[frac{x}{12} = 1 - frac{1}{3}][frac{x}{12} = frac{2}{3}]Multiply both sides by 12:[x = frac{2}{3} times 12][x = 8]Thus, the number of children Mother Hubbard had is (boxed{B}).,<think>Okay, let's see. The problem says that Old Mother Hubbard had a Giant Bear chocolate bar in her cupboard. She gave each of her children one-twelfth of the chocolate bar. After that, one third of the bar was left. The question is asking how many children she had, and the options are A through E: 6, 8, 12, 15, 18.Hmm, so first, I need to figure out how much of the chocolate bar she gave out in total, and then divide that by the amount each child received to find the number of children. Let me break it down step by step.The entire chocolate bar is considered as 1 whole. She gave away some portions to her children, each child getting 1/12 of the bar. After giving those portions away, 1/3 of the bar was left. So, the amount she gave to the children must be the whole bar minus the leftover. That is, 1 - 1/3. Let me compute that.1 minus 1/3 is 2/3. So, she gave away 2/3 of the chocolate bar to her children. Each child received 1/12 of the bar. Therefore, the number of children would be the total amount given away divided by the amount per child. So, 2/3 divided by 1/12. Let me do that division.Dividing fractions: 2/3 ÷ 1/12 is the same as multiplying by the reciprocal, so 2/3 * 12/1. Let's calculate that. 2 times 12 is 24, and 3 times 1 is 3. So, 24/3 equals 8. So, 8 children. Let me check the answer options. Option B is 8. So, that's probably the correct answer.Wait, let me verify once again to make sure I didn't make a mistake. If there are 8 children, each getting 1/12, then total given is 8 * 1/12 = 8/12, which simplifies to 2/3. Subtract that from the whole bar, 1 - 2/3 = 1/3, which matches the problem statement. So, that checks out.Alternatively, maybe I can think of it in terms of fractions. The whole bar is 12/12. If each child gets 1/12, then each child takes one part out of twelve. If she gave away some parts and has 1/3 left, which is 4/12 (since 1/3 is equivalent to 4/12). So, original 12/12 minus 4/12 equals 8/12 given to the children. Since each child gets 1/12, 8/12 divided by 1/12 per child is 8 children. Yep, that also gives 8. So, definitely, the answer is B) 8.Final Answerboxed{B}</think>Old Mother Hubbard had a Giant Bear chocolate bar, which is considered as 1 whole. She gave each child one-twelfth of the bar, and one-third of the bar was left. 1. The total amount of chocolate left is ( frac{1}{3} ).2. The amount given to the children is ( 1 - frac{1}{3} = frac{2}{3} ).3. Each child received ( frac{1}{12} ) of the bar.4. To find the number of children, divide the total amount given away by the amount each child received: [ frac{2}{3} div frac{1}{12} = frac{2}{3} times frac{12}{1} = frac{24}{3} = 8 ]Verification: If 8 children each receive ( frac{1}{12} ) of the bar, the total given is ( 8 times frac{1}{12} = frac{8}{12} = frac{2}{3} ). Subtracting this from the whole bar leaves ( 1 - frac{2}{3} = frac{1}{3} ), which matches the problem statement.Thus, the number of children is (boxed{B}).]

🔑:Solution: BOne third of the bar was left, so Mother Hubbard's children ate two-thirds of the bar. Since they ate one-twelfth of the bar each, Mother Hubbard hadfrac{2 / 3}{1 / 12}=frac{2}{3} times frac{12}{1}=frac{24}{3}=8 children.