Let f:[1, ∞) arrow[2, ∞) be a differentiable function, If 10 ∫_1^ x f(t) dt=5 x f(x)-x^5-9 for all x ≥ 1 , then the value of f(3) is :

Correct answer: B. 32 aligned & 10 ~d dx ∫_1^ x f(t) dt= d dx (5 xf(x)-x^5-9) & 10 f(x)=5 f(x)+5 xf^ (x)-5 x^4 & f(x)+x^4=xf^ (x) & y+x^4=x dy dx & (d y)/(d x)+y(-(1)/(x))=x^3 & y e^ ∫-(1)/(x) d x =∫ x^3 e^ ∫-(1)/(x) d x +c & ye^ -tn|x| =∫ x^3 e^ -tr|x| +c & y |x| =∫ x^3 |x| +c & y x =∫ x^2+c & y x = x^3 3 +c aligned Put x=1 in given equation aligned & 0=5 f(1)-1-9 f(1)=2 & (2)/(1)=(1)/(3)+c c=(5)/(3) & f(3) 3 =(27)/(3)+(5)/(3) & f(3)=32 aligned

JEE Main 20252 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 2 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2025 · 2 April · Evening Shift · Mathematics · Differential Equations

Let f:[1, ∞) arrow[2, ∞) be a differentiable function, If 10 ∫_1^ x...

Let

f:[1, \infty) \rightarrow[2, \infty)

be a differentiable function, If

10 \int_1^{\mathrm{x}} f(\mathrm{t}) \mathrm{dt}=5 \mathrm{x} f(\mathrm{x})-\mathrm{x}^5-9

for all

\mathrm{x} \geq 1

, then the value of

f(3)

is :

Options

A $18$
B $32$
C $22$
D $26$

Correct answer

B. $32$

Step-by-step solution

\begin{aligned}<br/>& 10 \frac{\mathrm{~d}}{\mathrm{dx}} \int_1^{\mathrm{x}} \mathrm{f}(\mathrm{t}) \mathrm{dt}=\frac{\mathrm{d}}{\mathrm{dx}}\left(5 \mathrm{xf}(\mathrm{x})-\mathrm{x}^5-9\right) \\ & \Rightarrow 10 f(x)=5 f(x)+5 \mathrm{xf}^{\prime}(\mathrm{x})-5 \mathrm{x}^4 \\ & \Rightarrow \mathrm{f}(\mathrm{x})+\mathrm{x}^4=\mathrm{xf}^{\prime}(\mathrm{x}) \\ & \Rightarrow \mathrm{y}+\mathrm{x}^4=\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}} \\ & \Rightarrow \frac{d y}{d x}+y\left(-\frac{1}{x}\right)=x^3 \\ & \Rightarrow y e^{\int-\frac{1}{x} d x}=\int x^3 e^{\int-\frac{1}{x} d x}+c \\ & \Rightarrow \mathrm{ye}^{-\operatorname{tn}|x|}=\int \mathrm{x}^3 \mathrm{e}^{-\operatorname{tr}|x|}+\mathrm{c} \\ & \Rightarrow \frac{\mathrm{y}}{|\mathrm{x}|}=\int \frac{\mathrm{x}^3}{|\mathrm{x}|}+\mathrm{c} \\ & \Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\int \mathrm{x}^2+\mathrm{c} \\ & \Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\frac{\mathrm{x}^3}{3}+\mathrm{c}<br/>\end{aligned}

Put

x=1

in given equation

\begin{aligned}<br/>& \Rightarrow 0=5 \mathrm{f}(1)-1-9 \Rightarrow \mathrm{f}(1)=2 \\ & \Rightarrow \frac{2}{1}=\frac{1}{3}+\mathrm{c} \Rightarrow \mathrm{c}=\frac{5}{3} \\ & \Rightarrow \frac{\mathrm{f}(3)}{3}=\frac{27}{3}+\frac{5}{3} \\ & \Rightarrow \mathrm{f}(3)=32<br/>\end{aligned}

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