Determinants Matrices Question 4
Question: If $ f(x)= \begin{vmatrix} 1+{{\sin }^{2}}x & {{\cos }^{2}}x & 4\sin 2x \\ {{\sin }^{2}}x & 1+{{\cos }^{2}}x & 4\sin 2x \\ {{\sin }^{2}}x & {{\cos }^{2}}x & 1+4\sin 2x \\ \end{vmatrix} $ What is the maximum value of $ f(x) $ -
Options:
A) 2
B) 4
C) 6
D) 8
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] $ f(x)= \begin{vmatrix} 1+{{\sin }^{2}}x & {{\cos }^{2}}x & 4\sin 2x \\ {{\sin }^{2}}x & 1+{{\cos }^{2}}x & 4\sin 2x \\ {{\sin }^{2}}x & {{\cos }^{2}}x & 1+4\sin 2x \\ \end{vmatrix} $
Applying $ C_1\to C_1+C_2 $
$ = \begin{vmatrix} 2 & {{\cos }^{2}}\theta & 4\sin 2x \\ 2 & 1+{{\cos }^{2}}\theta & 4\sin 2x \\ 1 & {{\cos }^{2}}\theta & 1+4\sin 2x \\ \end{vmatrix} $ (Applying $ R_2\to R_2-R_1 $ and $ R_3\to R_3-R_1 $ ) $ = \begin{vmatrix} 2 & {{\cos }^{2}}\theta & 4\sin 2x \\ 0 & 1 & 0 \\ -1 & 0 & 1 \\ \end{vmatrix} $
$ f(x)=2+4\sin 2x $
$ \therefore -1\le \sin 2x\le 1, $ maximum value of $ sin2x=1 $ Thus, maximum value of $ f(x)=2+4=6 $
BETA
