Let f be a differentiable function in the interval (0, ) such that f(1) =1 and for each x > 0. Then 2f(2) + 3f(3) is equal to __________. [2024]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Let f be a differentiable function in the interval (0, $\infty$ ) such that f(1) =1 and $\lim_{t \to x} \frac{t^{2} f (x) - x^{2} f (t)}{t - x} = 1$ for each x > 0. Then 2f(2) + 3f(3) is equal to __________. [2024]

Ans.
(24)

We Have $\underset{t \to x}{l t} \frac{t^{2} f (x) - x^{2} f (t)}{t - x} = 1 (\frac{0}{0} form)$

$\Rightarrow \underset{t \to x}{l t} \frac{2 t f (x) - x^{2} f' (t)}{1} = 1 [L' Hospital Rule]$

$\Rightarrow 2 x f (x) - x^{2} f' (x) = 1 \Rightarrow f' (x) - \frac{2}{x} f (x) = \frac{- 1}{x^{2}}$

$I F = e^{\int \frac{- 2}{x} d x} = e^{- 2 \ln x} = \frac{1}{x^{2}}$

$∴ f (x) \frac{1}{x^{2}} = \int \frac{- 1}{x^{2}} \cdot \frac{1}{x^{2}} d x + C \Rightarrow f (x) \frac{1}{x^{2}} = \frac{1}{3 x^{3}} + C$

Now, f(1) = 1

$\Rightarrow \frac{2}{3} = C \Rightarrow f (x) = \frac{1}{3 x} + \frac{2 x^{2}}{3}$

Now, $2 f (2) + 3 f (3) = 2 [\frac{1}{6} + \frac{8}{3}] + 3 [\frac{1}{9} + \frac{18}{3}]$

$= \frac{1}{3} + \frac{16}{3} + \frac{1}{3} + 18 = 6 + 18 = 24$ .

Want Us to Email you About Special Offers & Updates?