Let f : R R be a function defined by f ( x ) = log m { 2 ( sin x - cos x ) + m - 2 } , for some m , such that the range of f is [0, 2]. Then the value of m is [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 $f : R \to R$ be a function defined by $f (x) = \log_{\sqrt{m}} {\sqrt{2} (\sin x - \cos x) + m - 2},$ for some $m$ , such that the range of $f$ is [0, 2]. Then the value of $m$ is [2023]

1 3

2 5

3 4

4 2

Ans.
(2)

Since, $- \sqrt{2} \leq \sin x - \cos x \leq \sqrt{2}$

$∴$ $- 2 \leq \sqrt{2} (\sin x - \cos x) \leq 2$

Let $\sqrt{2} (\sin x - \cos x) = t$

$- 2 \leq t \leq 2$ ...(i)

$f (x) = \log_{\sqrt{m}} (t + m - 2)$

We have, $0 \leq f (x) \leq 2$

$\Rightarrow 0 \leq \log_{\sqrt{m}} (t + m - 2) \leq 2$

$1 \leq t + m - 2 \leq m$

$- m + 3 \leq t \leq 2$ ...(ii)

From (i) and (ii), we get $- m + 3 = - 2 \Rightarrow m = 5$

Want Us to Email you About Special Offers & Updates?