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 $\sin^2 2x = (2 \sin x \cos x)^2 = 4 \sin^2 x \cos^2 x = 4(1 - \cos^2 x) \cos^2 x$,we ge

Vedclass · Accepted Answer

$\frac{9}{4}$

Vedclass · Answer

(A) Let $f(x) = 4 + \frac{1}{2} \sin^2 2x - 2 \cos^4 x$.Using $\sin^2 2x = (2 \sin x \cos x)^2 = 4 \sin^2 x \cos^2 x = 4(1 - \cos^2 x) \cos^2 x$,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 $t \in [0, 1]$.Then $g(t) = -4t^2 + 2t + 4$.This is a downward 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 $t = \frac{1}{4}$:$M = g(\frac{1}{4}) = -4(\frac{1}{16}) + 2(\frac{1}{4}) + 4 = -\frac{1}{4} + \frac{1}{2} + 4 = \frac{1}{4} + 4 = \frac{17}{4}$.The minimum value $m$ occurs at the boundaries $t = 0$ or $t = 1$:$g(0) = 4$$g(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

If $\alpha+\beta=\frac{\pi}{2}$ and $\beta+\gamma=\alpha$,then $\tan \alpha$ equals

If $A + B + C = \pi$ $(A, B, C > 0)$ and the angle $C$ is obtuse,then:

If $A > 0, B > 0$ and $A + B = \frac{\pi}{6}$,then the minimum value of $\tan A + \tan B$ is

If $\alpha + \beta - \gamma = \pi ,$ then ${\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma = $

The maximum value of $3 \sin x + 4 \cos x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module