If f(x) = left| begin{array}{ccc} cos(2x) & c | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Expanding the determinant $f(x)$:$f(x) = \cos(2x)(\cos x \cos x - (-\sin x \sin x)) - \cos(2x)(-\cos x \cos x - (-\sin x \sin x)) + \sin(2x)(-\cos

Vedclass · Accepted Answer

$B, C$

Vedclass · Answer

(D) Expanding the determinant $f(x)$:$f(x) = \cos(2x)(\cos x \cos x - (-\sin x \sin x)) - \cos(2x)(-\cos x \cos x - (-\sin x \sin x)) + \sin(2x)(-\cos x \sin x - \cos x \sin x)$$f(x) = \cos(2x)(\cos^2 x + \sin^2 x) - \cos(2x)(-\cos^2 x + \sin^2 x) + \sin(2x)(-2 \sin x \cos x)$$f(x) = \cos(2x)(1) - \cos(2x)(-\cos 2x) + \sin(2x)(-\sin 2x)$$f(x) = \cos 2x + \cos^2 2x - \sin^2 2x = \cos 2x + \cos 4x$.Now,$f'(x) = -2 \sin 2x - 4 \sin 4x = -2 \sin 2x - 8 \sin 2x \cos 2x = -2 \sin 2x (1 + 4 \cos 2x)$.Setting $f'(x) = 0$,we get $\sin 2x = 0$ or $\cos 2x = -1/4$.In $(-\pi, \pi)$,$\sin 2x = 0$ at $x = 0, \pm \pi/2$. ($3$ points)$\cos 2x = -1/4$ has $4$ solutions in $(-\pi, \pi)$.Thus,$f'(x) = 0$ at $3 + 4 = 7$ points,which is more than $3$. So $B$ is correct.For $f(x) = \cos 2x + \cos 4x$,at $x = 0$,$f(0) = \cos 0 + \cos 0 = 2$.Since $\cos 	heta \le 1$,the maximum value of $f(x)$ is $2$,which occurs at $x = 0$. So $C$ is correct.Therefore,the correct options are $B$ and $C$.

Similar Questions

If $f(x) = \left| \begin{array}{ccc} \sin(x + \alpha) & \sin(x + \beta) & \sin(x + \gamma) \\ \cos(x + \alpha) & \cos(x + \beta) & \cos(x + \gamma) \\ \sin(\alpha + \beta) & \sin(\beta + \gamma) & \sin(\gamma + \alpha) \end{array} \right|$ and $f(10) = 10$,then $f(\pi)$ is equal to

Let $f(x) = \left| \begin{array}{ccc} 2\cos^2 x & \sin(2x) & -\sin x \\ \sin(2x) & 2\sin^2 x & \cos x \\ \sin x & -\cos x & 0 \end{array} \right|$. Then,evaluate $\int_{0}^{\frac{\pi}{2}} [f(x) + f'(x)] dx$.

The rank of the matrix $A = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 0 & 1 \\ 4 & 1 & 4 \end{bmatrix}$ is

Let $f(x) = \left|\begin{array}{ccc} a & -1 & 0 \\ ax & a & -1 \\ ax^2 & ax & a \end{array}\right|$,where $a \in R$. Then the sum of the squares of all the values of $a$ for which $2f'(10) - f'(5) + 100 = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module