If m and M are the minimum and the maximum va | vedclass

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

Question

(A) Let $f(x) = 4 + \frac{1}{2} \sin^2 2x - 2 \cos^4 x$.Using $\sin^2 2x = 4 \sin^2 x \cos^2 x = 4(1 - \cos^2 x) \cos^2 x$,we get:$f(x) = 4 + \frac{1}

Vedclass · Accepted Answer

$\frac{9}{4}$

Vedclass · Answer

(A) Let $f(x) = 4 + \frac{1}{2} \sin^2 2x - 2 \cos^4 x$.Using $\sin^2 2x = 4 \sin^2 x \cos^2 x = 4(1 - \cos^2 x) \cos^2 x$,we get:$f(x) = 4 + \frac{1}{2} [4(1 - \cos^2 x) \cos^2 x] - 2 \cos^4 x$$f(x) = 4 + 2 \cos^2 x - 2 \cos^4 x - 2 \cos^4 x$$f(x) = 4 + 2 \cos^2 x - 4 \cos^4 x$.Let $t = \cos^2 x$,where $0 \le t \le 1$.$f(t) = -4t^2 + 2t + 4$.This is a downward-opening parabola with vertex at $t = -\frac{b}{2a} = -\frac{2}{2(-4)} = \frac{1}{4}$.Since $\frac{1}{4} \in [0, 1]$,the maximum value $M$ occurs at $t = \frac{1}{4}$:$M = f(\frac{1}{4}) = -4(\frac{1}{16}) + 2(\frac{1}{4}) + 4 = -\frac{1}{4} + \frac{1}{2} + 4 = \frac{1}{4} + 4 = \frac{17}{4}$.The minimum value $m$ occurs at the boundaries $t=0$ or $t=1$:$f(0) = 4$.$f(1) = -4(1)^2 + 2(1) + 4 = 2$.Thus,$m = 2$.$M - m = \frac{17}{4} - 2 = \frac{17 - 8}{4} = \frac{9}{4}$.

If $m$ and $M$ are the minimum and the maximum values of $4 + \frac{1}{2} \sin^2 2x - 2 \cos^4 x$ for $x \in R$,then $M - m$ is equal to

Similar Questions

$\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = $ (in $/16$)

If $\left| a\sin^2 \theta + b\sin \theta \cos \theta + c\cos^2 \theta - \frac{1}{2}(a + c) \right| \le \frac{1}{2}k,$ then $k^2$ is equal to

$\sqrt{3} \csc 20^\circ - \sec 20^\circ = $

If $\sin \beta$ is the geometric mean between $\sin \alpha$ and $\cos \alpha,$ then $\cos 2\beta$ is equal to

$\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}+\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module