The ratio of the maximum and minimum values a | vedclass

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

Question

(B) Given,$f(x) = 1 + 2 \sin x + 3 \cos^2 x$. Substituting $\cos^2 x = 1 - \sin^2 x$,we get: $f(x) = 1 + 2 \sin x + 3(1 - \sin^2 x) = 4 + 2 \sin x - 3

Vedclass · Accepted Answer

$13 : 9$

Vedclass · Answer

(B) Given,$f(x) = 1 + 2 \sin x + 3 \cos^2 x$. Substituting $\cos^2 x = 1 - \sin^2 x$,we get: $f(x) = 1 + 2 \sin x + 3(1 - \sin^2 x) = 4 + 2 \sin x - 3 \sin^2 x$. Let $t = \sin x$. Since $0 \leq x \leq \frac{2\pi}{3}$,the range of $t$ is $[0, 1]$. $f(t) = -3t^2 + 2t + 4$. This is a downward parabola with vertex at $t = -\frac{b}{2a} = -\frac{2}{2(-3)} = \frac{1}{3}$. Since $\frac{1}{3} \in [0, 1]$,the maximum value is $f(\frac{1}{3}) = -3(\frac{1}{9}) + 2(\frac{1}{3}) + 4 = -\frac{1}{3} + \frac{2}{3} + 4 = \frac{13}{3}$. The minimum value in the interval $[0, 1]$ occurs at the endpoints $t = 0$ or $t = 1$. $f(0) = 4$ and $f(1) = -3(1)^2 + 2(1) + 4 = 3$. Thus,the minimum value is $3$. The ratio of maximum to minimum is $\frac{13/3}{3} = \frac{13}{9}$ or $13 : 9$.

The ratio of the maximum and minimum values attained by the function $f(x) = 1 + 2 \sin x + 3 \cos^2 x$ for $0 \leq x \leq \frac{2\pi}{3}$ is

Similar Questions

If $A = \sin^2 x + \cos^4 x$,then for all real $x :$

If $\left| \cos \theta \left\{ \sin \theta + \sqrt{\sin^2 \theta + \sin^2 \alpha} \right\} \right| \le k$,then the value of $k$ is

Given that $\cos \left( \frac{\alpha - \beta}{2} \right) = 2\cos \left( \frac{\alpha + \beta}{2} \right)$,then $\tan \frac{\alpha}{2} \tan \frac{\beta}{2}$ is equal to

If $A+B+C=2S$,then $\sin(S-A)+\sin(S-B)-\sin C=$

The maximum value of the function $f(x) = a \sin x + b \cos x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module