If K = sin^6x + cos^6x, then K belongs to the | vedclass

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

Question

$\mathrm{K}=\left(\sin ^{2} \mathrm{x}+\cos ^{2} \mathrm{x}\right)^{3}-3 \sin ^{2} \mathrm{x} \cos ^{2} \mathrm{x}\left(\sin ^{2} \mathrm{x}+\cos ^{2}

Vedclass · Accepted Answer

$\left[ {\frac{1}{4},1} \right]$

Vedclass · Answer

$\mathrm{K}=\left(\sin ^{2} \mathrm{x}+\cos ^{2} \mathrm{x}\right)^{3}-3 \sin ^{2} \mathrm{x} \cos ^{2} \mathrm{x}\left(\sin ^{2} \mathrm{x}+\cos ^{2} \mathrm{x}\right)$

$=(1)^{3}-3 \sin ^{2} x \cos ^{2} x(1)$

$=1-\frac{3}{4} \sin ^{2} 2 x$

Now, $0 \leq \sin ^{2} 2 x \leq 1 \quad \Rightarrow \quad 0 \leq \frac{3}{4} \sin ^{2} 2 x \leq \frac{3}{4}$

$\Rightarrow \quad-\frac{3}{4} \leq-\frac{3}{4} \sin ^{2} 2 x \leq 0$

$\Rightarrow \quad 1-\frac{3}{4} \leq 1-\frac{3}{4} \sin ^{2} 2 x \leq 1$

$\Rightarrow \quad \frac{1}{4} \leq 1-\frac{3}{4} \sin ^{2} 2 x \leq 1$

$\Rightarrow \quad \frac{1}{4} \leq \mathrm{K} \leq 1 \Rightarrow \mathrm{K} \in\left[\frac{1}{4}, 1\right]$

If $K = sin^6x + cos^6x$, then $K$ belongs to the interval

Similar Questions

If $\sec 4\theta - \sec 2\theta = 2$, then the general value of $\theta $ is

Number of solution$(s)$ of the equation $\sin 2\theta + \cos 2\theta = - \frac{1}{2},\theta \in \left( {0,\frac{\pi }{2}} \right)$ is-

If $\cos \,\alpha + \cos \,\beta = \frac{3}{2}$ and $\sin \,\alpha + \sin \,\beta = \frac{1}{2}$ and $\theta $ is the the arithmetic mean of $\alpha $ and $\beta $ , then $\sin \,2\theta + \cos \,2\theta $ is equal to

If $\cos A\sin \left( {A - \frac{\pi }{6}} \right)$ is maximum, then the value of $A$ is equal to

If $\sin 2\theta = \cos \theta ,\,\,0 < \theta < \pi $, then the possible values of $\theta $ are