Let f(x) be a real differentiable function such that f(0) = 1 and for all x, y R. Then is equal to : [2025]

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(x) be a real differentiable function such that f(0) = 1 and $f (x + y) = f (x) f' (y) + f' (x) f (y)$ for all x, y $\in$ R. Then $\sum_{n = 1}^{100} \log_{e} f (n)$ is equal to : [2025]

1 2406

2 5220

3 2525

4 2384

Ans.
(3)

$f (x + y) = f (x) f' (y) + f' (x) f (y)$

When x = 0, y = 0, we have

$f (0) = f (0) f' (0) + f' (0) f (0)$

$\Rightarrow f (0) = 2 f' (0) f (0) \Rightarrow f' (0) = \frac{1}{2}$

When y = 0, $f (x) = f (x) f' (0) + f (0) f' (x)$

$\Rightarrow \frac{1}{2} f (x) = f' (x)$ [ $∵$ f(0) = 1]

Integrating both sides, we get $f (x) = e^{x / 2} \cdot C$

Now, $f (0) = 1 \Rightarrow f (x) = e^{x / 2} \Rightarrow \log_{e} f (x) = \frac{x}{2}$

$∴ \sum_{n = 1}^{100} \log_{e} f (n)$

$= \sum_{n = 1}^{100} \frac{n}{2} = \frac{1}{2} \frac{(100) (101)}{2} = 2525$ .

Want Us to Email you About Special Offers & Updates?