The maximum value of e^{(2 + sqrt{3} cos x + | vedclass

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

Question

(C) Let $f(x) = 2 + \sqrt{3} \cos x + \sin x$. To find the maximum value of $e^{f(x)}$,we need to find the maximum value of $f(x)$. We know that the e

Vedclass · Accepted Answer

$e^{4}$

Vedclass · Answer

(C) Let $f(x) = 2 + \sqrt{3} \cos x + \sin x$. To find the maximum value of $e^{f(x)}$,we need to find the maximum value of $f(x)$. We know that the expression $a \cos x + b \sin x$ lies in the interval $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$. Here,$a = \sqrt{3}$ and $b = 1$. So,$\sqrt{a^2 + b^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$. Thus,the maximum value of $\sqrt{3} \cos x + \sin x$ is $2$. Therefore,the maximum value of $f(x) = 2 + (\sqrt{3} \cos x + \sin x)$ is $2 + 2 = 4$. Hence,the maximum value of $e^{f(x)}$ is $e^{4}$.

The maximum value of $e^{(2 + \sqrt{3} \cos x + \sin x)}$ is

Similar Questions

Let $a = \max_{x \in R} \{8^{2 \sin 3x} \cdot 4^{4 \cos 3x}\}$ and $\beta = \min_{x \in R} \{8^{2 \sin 3x} \cdot 4^{4 \cos 3x}\}$. If $8x^2 + bx + c = 0$ is a quadratic equation whose roots are $\alpha^{1/5}$ and $\beta^{1/5}$,then the value of $c - b$ is equal to:

The equation $\sec^2 \theta = \frac{4xy}{(x + y)^2}$ is only possible when

Find the maximum and minimum values of the function given by $h(x) = \sin(2x) + 5$.

In triangle $ABC$,if $\cos A \cos B + \sin A \sin B \sin C = 1$,then $\sin A + \sin B + \sin C =$

If $p$ and $q$ are positive numbers such that $p^2 + q^2 = 1$,then the maximum value of $p + q$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module