Let M and m respectively be the maximum and the minimum values of . 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 M and m respectively be the maximum and the minimum values of $f (x) =$ $| \begin{matrix} 1 + \sin^{2} x & \cos^{2} x & 4 \sin 4 x \\ \sin^{2} x & 1 + \cos^{2} x & 4 \sin 4 x \\ \sin^{2} x & \cos^{2} x & 1 + 4 \sin 4 x \end{matrix} |$ $, x \in R$ . Then $M^{4} - m^{4}$ is equal to : [2025]

1 1295

2 1040

3 1215

4 1280

Ans.
(4)

$f (x) =$ $| \begin{matrix} 1 + \sin^{2} x & \cos^{2} x & 4 \sin 4 x \\ \sin^{2} x & 1 + \cos^{2} x & 4 \sin 4 x \\ \sin^{2} x & \cos^{2} x & 1 + 4 \sin 4 x \end{matrix} |$

$=$ $| \begin{matrix} 2 & \cos^{2} x & 4 \sin 4 x \\ 2 & 1 + \cos^{2} x & 4 \sin 4 x \\ 1 & \cos^{2} x & 1 + 4 \sin 4 x \end{matrix} |$ $(Applying C_{1} \to C_{1} + C_{2})$

$=$ $| \begin{matrix} 2 & \cos^{2} x & 4 \sin 4 x \\ 0 & 1 & 0 \\ 1 & \cos^{2} x & 1 + 4 \sin 4 x \end{matrix} |$ $(Applying R_{2} \to R_{2} - R_{1})$

Expanding along $R_{2}$ , we get f(x) = 2(1 + 4 sin 4x) – 4 sin 4x

$\Rightarrow f (x) = 2 + 4 \sin 4 x$

Maximum value of f(x), M = 6

Minimum value of f(x), m = –2

$∴ M^{4} - m^{4} = 1280$ .

Want Us to Email you About Special Offers & Updates?