Let f : R - { 0 , 1 } be a function such that f ( x ) + f ( 1 1 - x ) = 1 + x . Then f ( 2 ) is equal to [2023]

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 $f : R - {0, 1} \to ℝ$ be a function such that $f (x) + f (\frac{1}{1 - x}) = 1 + x .$ Then $f (2)$ is equal to [2023]

1 $\frac{7}{3}$

2 $\frac{9}{2}$

3 $\frac{9}{4}$

4 $\frac{7}{4}$

Ans.
(3)

$f (x) + f (\frac{1}{1 - x}) = 1 + x$

$x = 2 \Rightarrow f (2) + f (- 1) = 3$ $\dots (i)$

$x = - 1 \Rightarrow f (- 1) + f (\frac{1}{2}) = 0$ $\dots (i i)$

$x = \frac{1}{2} \Rightarrow f (\frac{1}{2}) + f (2) = \frac{3}{2}$ $\dots (i i i)$

$(i) + (i i i) - (i i) \Rightarrow 2 f (2) = \frac{9}{2}$ $∴$ $f (2) = \frac{9}{4}$

Want Us to Email you About Special Offers & Updates?