Minimum value of the function f(x) = left| {s | vedclass

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

Question

Put $=\sin x+\cos x=p \Rightarrow 1+2 \sin x \cos x=p^{2}$

$\sin x \cos x=\left(\frac{p^{2}-1}{2}ight)$

$	herefore f(x) = \left| {p + \frac{2

Vedclass · Accepted Answer

$2\sqrt 2  - 1$

Vedclass · Answer

Put $=\sin x+\cos x=p \Rightarrow 1+2 \sin x \cos x=p^{2}$

$\sin x \cos x=\left(\frac{p^{2}-1}{2}ight)$

$	herefore f(x) = \left| {p + \frac{2}{{\left( {{p^2} - 1} ight)}} + \frac{{2p}}{{{p^2} - 1}}} ight|$

${=\left|p+\frac{2}{(p-1)}ight|} $

${=\left|(p-1)+\frac{2}{(p-1)}+1ight|}$

$	herefore \frac{(p-1)+\left(\frac{2}{p-1}ight)}{2} \geq\left((p-1) \cdot \frac{2}{p-1}ight)^{1 / 2}$

$	herefore(p-1)+\left(\frac{2}{p-1}ight) \geq 2 \sqrt{2}$

$	herefore \quad f(x)_{\min } \geq 2 \sqrt{2}+1$

Minimum value of the function $f(x) = \left| {\sin \,x + \cos \,x + \tan \,x + \cot \,x + \sec \,x + \ cosec\ x} \right|$ is equal to

Similar Questions

If $\alpha,-\frac{\pi}{2}<\alpha<\frac{\pi}{2}$ is the solution of $4 \cos \theta+5 \sin \theta=1$, then the value of $\tan \alpha$ is

If $sin\, \theta = sin\, \alpha$ then $sin\, \frac{\theta }{3}$ =

Find the general solution of the equation $\cos 3 x+\cos x-\cos 2 x=0$

If $2\,cos\,\theta + sin\, \theta \, = 1$ $\left( {\theta \ne \frac{\pi }{2}} \right)$ , then $7\, cos\,\theta + 6\, sin\, \theta $ is equal to

The number of solutions of the equation $\cos \left(x+\frac{\pi}{3}\right) \cos \left(\frac{\pi}{3}-x\right)=\frac{1}{4} \cos ^{2} 2 x, x \in[-3 \pi$ $3 \pi]$ is