\(3^{\mathrm{rd}}\) term in expansion of \(\left[(1 / \mathrm{x})+\mathrm{x}^{(\log )}{ }_{10}(\mathrm{x})\right]^{5}\) is 1000 then \(\mathrm{x}=\) (a) 10 (b) 100 (c) 1000 (d) None

Recall the binomial expansion formula: \((a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r\), where \(\binom{n}{r}\) is the binomial coefficient. We are given the expression \(\left[\frac{1}{x} + x^{\log _{10}(x)}\right]^5\), and we have to find the 3rd term in its expansion.

To find the 3rd term, we need to plug in r=2 (since the terms start from r=0) into the binomial expansion formula: 3rd term = \(\binom{5}{2} \left(\frac{1}{x}\right)^{5-2} \left(x^{\log _{10}(x)}\right)^2\)

Calculate the binomial coefficient \(\binom{5}{2}\): \(\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5\times 4}{2\times 1} = 10\)

Now, substitute the binomial coefficient back into the expression and simplify: 3rd term = \(10\left(\frac{1}{x}\right)^{3} \left(x^{\log _{10}(x)}\right)^2\) We are given that the 3rd term is 1000. So, we set the expression equal to 1000 and solve for x: \(10\left(\frac{1}{x}\right)^{3} \left(x^{\log _{10}(x)}\right)^2 = 1000\)

Divide both sides of the equation by 10 and simplify: \(\left(\frac{1}{x}\right)^{3} \left(x^{\log _{10}(x)}\right)^2 = 100\) Take the cube root of both sides: \(\frac{1}{x} \left(x^{\log _{10}(x)}\right) = 10\) Multiply both sides by x: \(x^{\log _{10}(x)} = 10x\) Since we want to find \(x^{\log _{10}(x)}\), let's take the logarithm base 10 on both sides: \(\log _{10}(x^{\log _{10}(x)}) = \log _{10}(10x)\) Use the property of logarithm \(\log _{10}(a^b)=b \log _{10}(a)\): \(\log _{10}(x) \cdot \log _{10}(x)= 1 + \log _{10}(x)\) Solve for \(x^{\log _{10}(x)}\): \(\log _{10}(x) \cdot \left(\log _{10}(x) - 1\right) = 0\) This equation has two possible solutions: \(\log _{10}(x)=0\) or \(\log _{10}(x)=1\). The first solution gives \(x=1\), which doesn't match any of the given options. The second solution gives \(x=10\), which matches option (a). Therefore, the correct answer is: \(\boxed{x=10}\) (option a)