Let f be a differentiable function in the interval (0, ∞) such that f(1)=1 and _ t arrow x (t^2 f(x)-x^2 f(t))/(t-x)=1 for each x>0 . Then 2 f(2)+3 f(3) is equal to _______

Correct answer: 0 aligned & _ t arrow x (t^2 f(x)-x^2 f(t))/(t-x)=1 & _ t arrow x 2 t . f(x)-x^2 f^ (x) 1 =1 & 2 x · f(x)-x 2 f^ (x)=1 & (d y)/(d x)-(2)/(x) · y=(-1)/(x^2) & I.f. =e^ ∫-(2)/(x) d x =(1)/(x^2) & (y)/(x^2)=∫-(1)/(x^4) d x+C & (y)/(x^2)=(1)/(3 x^3)+C aligned Put f(1)=1 aligned & C=(2)/(3) & y=(1)/(3 x)+(2 x^2)/(3) & y=(2 x^3+1)/(3 x) & f(2)=(17)/(6) & f(3)=(55)/(9) & 2 f(2)+3 f(3)=(17)/(3)+(55)/(3)=(72)/(3)=24 aligned

JEE Main 20245 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 5 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Limits

JEE Main 2024 · 5 April · Morning Shift · Mathematics · Limits

Let f be a differentiable function in the interval (0, ∞) such that...

Let

f

be a differentiable function in the interval

(0, \infty)

such that

f(1)=1

and

\lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1

for each

x>0

. Then

2 f(2)+3 f(3)

is equal to _______

Correct answer

Step-by-step solution

\begin{aligned} & \lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1 \\ & \lim _{t \rightarrow x} \frac{2 t . f(x)-x^2 f^{\prime}(x)}{1}=1 \\ & 2 x \cdot f(x)-x 2 f^{\prime}(x)=1 \\ & \frac{d y}{d x}-\frac{2}{x} \cdot y=\frac{-1}{x^2} \\ & \text{ I.f. }=e^{\int-\frac{2}{x} d x}=\frac{1}{x^2} \\ & \therefore \frac{y}{x^2}=\int-\frac{1}{x^4} d x+C \\ & \frac{y}{x^2}=\frac{1}{3 x^3}+C \end{aligned}

Put

f(1)=1

\begin{aligned} & C=\frac{2}{3} \\ & y=\frac{1}{3 x}+\frac{2 x^2}{3} \\ & y=\frac{2 x^3+1}{3 x} \\ & f(2)=\frac{17}{6} \\ & f(3)=\frac{55}{9} \\ & 2 f(2)+3 f(3)=\frac{17}{3}+\frac{55}{3}=\frac{72}{3}=24\end{aligned}

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