JEE Mains · Maths · STD 12 - 7.2 definite integral
Let \(f:(0, \infty) \rightarrow \mathbf{R}\) be a twice differentiable function. If for some \(\mathrm{a} \neq 0, \int_0^1 f(\lambda x) \mathrm{d} \lambda=\mathrm{a} f(x), f(1)=1\) and \(f(16)=\frac{1}{8}\), then \(16-f^{\prime}\left(\frac{1}{16}\right)\) is equal to _______.
- A 110
- B 111
- C 112
- D 113
Answer & Solution
Correct Answer
(C) 112
Step-by-step Solution
Detailed explanation
Given, \(\int_0^1 f(\lambda x) d \lambda=a f(x)\) Let \(\lambda x=u\) \(d \lambda=\frac{1}{x} d u\) \(\therefore\) From (1) \(\frac{1}{x} \int_0^x f(u) d u=a f(x)\) \(\Rightarrow \int_0^x f(u) d u=\operatorname{axf}(x)\) Differentiate both sides…
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