The minimum value of the function f(x) = |sin | vedclass

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

Question

(D) Let $p = \sin x + \cos x$. Then $p^2 = 1 + 2 \sin x \cos x$,which implies $\sin x \cos x = \frac{p^2 - 1}{2}$.We know that $	an x + \cot x = \fra

Vedclass · Accepted Answer

$2\sqrt{2} + 1$

Vedclass · Answer

(D) Let $p = \sin x + \cos x$. Then $p^2 = 1 + 2 \sin x \cos x$,which implies $\sin x \cos x = \frac{p^2 - 1}{2}$.We know that $	an x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x} = \frac{2}{p^2 - 1}$.Also,$\sec x + \csc x = \frac{1}{\cos x} + \frac{1}{\sin x} = \frac{\sin x + \cos x}{\sin x \cos x} = \frac{p}{\frac{p^2 - 1}{2}} = \frac{2p}{p^2 - 1}$.Substituting these into the function: $f(x) = |p + \frac{2}{p^2 - 1} + \frac{2p}{p^2 - 1}| = |p + \frac{2(1 + p)}{p^2 - 1}| = |p + \frac{2(1 + p)}{(p - 1)(p + 1)}| = |p + \frac{2}{p - 1}|$.We can rewrite this as $f(x) = |(p - 1) + \frac{2}{p - 1} + 1|$.For $p - 1 > 0$,by $AM$-$GM$ inequality,$(p - 1) + \frac{2}{p - 1} \geq 2\sqrt{(p - 1) \cdot \frac{2}{p - 1}} = 2\sqrt{2}$.Thus,$f(x) \geq 2\sqrt{2} + 1$.

The minimum value of the function $f(x) = |\sin x + \cos x + \tan x + \cot x + \sec x + \csc x|$ is equal to

Similar Questions

If $A + B + C = \pi ,$ then $\tan ^2 \frac{A}{2} + \tan ^2 \frac{B}{2} + \tan ^2 \frac{C}{2}$ is always

The minimum value of $27 \tan^2 \theta + 3 \cot^2 \theta$ is

If $5 \cos x + 12 \cos y = 13$,then the maximum value of $5 \sin x + 12 \sin y$ is:

Find the maximum and minimum values of the function given by $f(x) = |\sin 4x + 3|$.

If $A+B=\frac{\pi}{2}$,then the maximum value of $\cos A \cdot \cos B$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module