Let y = log e ( 1 - x 2 1 + x 2 ) , - 1 < x < 1 . Then at x = 1 2 , the value of 225 ( y ' - y ' ' ) 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.
$Let y = \log_{e} (\frac{1 - x^{2}}{1 + x^{2}}), - 1 < x < 1 . Then at x = \frac{1}{2}, the value of 225 (y' - y'') is equal to$ [2024]

1 746

2 736

3 742

4 732

Ans.
(2)

We have, $y = \log_{e} (\frac{1 - x^{2}}{1 + x^{2}})$

$\Rightarrow y = \log_{e} (1 - x^{2}) - \log_{e} (1 + x^{2})$

$\Rightarrow y' = \frac{1}{1 - x^{2}} \times (- 2 x) - \frac{1}{1 + x^{2}} \times (2 x)$

$\Rightarrow y' = \frac{- 4 x}{1 - x^{4}} \Rightarrow y'' = \frac{(1 - x^{4}) (- 4) - (- 4 x) (- 4 x^{3})}{{(1 - x^{4})}^{2}}$

$\Rightarrow y'' = \frac{- 4 (1 + 3 x^{4})}{{(1 - x^{4})}^{2}}$

Now, $225 (y' - y'') = 225 [\frac{- 4 x}{1 - x^{4}} + \frac{4 (1 + 3 x^{4})}{{(1 - x^{4})}^{2}}]$

$\Rightarrow 225 (y' - y'') at x = \frac{1}{2} is equal to$ $225 [\frac{- 4 \times \frac{1}{2}}{1 - \frac{1}{16}} + \frac{4 (1 + \frac{3}{16})}{{(1 - \frac{1}{16})}^{2}}] = 736$

Want Us to Email you About Special Offers & Updates?