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JEE Main 202529 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2025 · 29 January · Morning Shift · Mathematics · Definite Integration

Let (f:(0, ∞) arrow R be a twice differentiable function. If for...

Let (f:(0, ∞)→R\infty) \rightarrow \mathbf{R} be a twice differentiable function. If for some (a≠\mathrm{a} \neq0, \int_0^1 f(\lambda x) dλ=a\mathrm{d} \lambda=\mathrm{a}f(x), f(1)=1) and (f(16)=18\frac{1}{8}, then (16-f^{\prime}(116)\left(\frac{1}{16}\right) is equal to _______.

Correct answer

0

Step-by-step solution

Given, ∫01f(λx)dλ=af(x)\int_0^1 f(\lambda x) d \lambda=a f(x) Let λx=udλ=1xdu∴\lambda x=u d \lambda=\frac{1}{x} d u \therefore From (1) 1x∫0xf(u)du=af(x)⇒∫0xf(u)du=axf⁡(x)\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 f(x)=a(xf′(x)+f(x))⇒f(x)=axf′(x)+af(x)⇒(1−a)f(x)=axf′(x)⇒f′(x)f(x)=(1−a)a⋅1x\begin{aligned} & f(x)=a\left(x f^{\prime}(x)+f(x)\right) \\ & \Rightarrow \quad f(x)=a x f^{\prime}(x)+a f(x) \\ & \Rightarrow \quad(1-a) f(x)=a x f^{\prime}(x) \\ & \Rightarrow \quad \frac{f^{\prime}(x)}{f(x)}=\frac{(1-a)}{a} \cdot \frac{1}{x} \end{aligned} Integrate both side w.r.t. ( xx ) ⇒∫f′(x)f(x)dx=(1−a)a∫1xdx⇒ln⁡f(x)=(1−aa)ln⁡x+c\begin{aligned} & \Rightarrow \quad \int \frac{f^{\prime}(x)}{f(x)} d x=\frac{(1-a)}{a} \int \frac{1}{x} d x \\ & \Rightarrow \ln f(x)=\left(\frac{1-a}{a}\right) \ln x+c \end{aligned} Now at x=1f(1)=1⇒c=0x=1 f(1)=1 \Rightarrow c=0 Also given f(16)=18f(16)=\frac{1}{8}∴18=(16)1−aa⇒2−3=24−4aa⇒−3=4−4aa⇒−3a=4−4a⇒a=4∴f(x)=x−3/4f(x)=−34x−74\begin{aligned} & \therefore \quad \frac{1}{8}=(16)^{\frac{1-a}{a}} \\ & \Rightarrow \quad 2^{-3}=2^{\frac{4-4 a}{a}} \\ & \Rightarrow \quad-3=\frac{4-4 a}{a} \\ & \Rightarrow-3 a=4-4 a \\ & \Rightarrow a=4 \\ & \therefore \quad f(x)=x^{-3 / 4} \\ & f(x)=\frac{-3}{4} x^{-\frac{7}{4}} \end{aligned} Put x=116f′(116)=−34(116)−7/4=−34⋅2−4x(−74)=−96∴16−f′(116)⇒16−(−96)=112\begin{aligned} & \text{ Put } x=\frac{1}{16} \\ & f^{\prime}\left(\frac{1}{16}\right)=\frac{-3}{4}\left(\frac{1}{16}\right)^{-7 / 4}=\frac{-3}{4} \cdot 2^{-4 x\left(\frac{-7}{4}\right)}=-96 \\ & \therefore \quad 16-f^{\prime}\left(\frac{1}{16}\right) \Rightarrow 16-(-96)=112\end{aligned}$$
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