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

$\alpha, \beta, \gamma$ are real numbers satisfying $\alpha + \beta + \gamma = \pi$. The minimum value of the expression $\sin \alpha + \sin \beta + \sin \gamma$ is

For all values of $\theta$,the values of $3-\cos \theta+\cos \left(\theta+\frac{\pi}{3}\right)$ lie in the interval :

If $A+B+C=\pi$ and $\cos B=\cos A \cos C$,then $\tan A \tan C=$

If $A+B+C=\pi$,then $\sin A-\sin B+\sin C=$

If $A + B + C = \frac{3\pi}{2},$ then $\cos 2A + \cos 2B + \cos 2C = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module