Question

I : lf $K={\sin }^{6}x+{\cos }^{6}x$, then $\frac{1}{4}\le K\le 1$
II: lf $P=\sin x+\cos x$, then $\frac{1}{2}\le P\le 1$
Which of the above is true ?

• A. only $I$
• B. only $II$
• C. Both $I$ and $II$
• D. Neither $I$ nor $II$

Answer 1

Answer: (A)

only $I$

I:

$K={\sin }^{6}x+{\cos }^{6}x=({\sin }^{2}x{)}^3+({\cos }^{2}x{)}^3=({\sin }^{2}x+{\cos }^{2}x)({\sin }^{4}x+{\cos }^{4}x-{\sin }^2{\cos }^{2}x)=({\sin }^{4}x+{\cos }^{4}x-{\sin }^2{\cos }^{2}x)=\left[\right({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\sin }^{2}x{\cos }^{2}x-{\sin }^2{\cos }^{2}x]=1-3{\sin }^{2}x{\cos }^{2}x$ ....... $\left(1\right)$

Now, we know that $-1\le \sin 2x\le 1$

$\Rightarrow -1\le 2\sin x\cos x\le 1$

$\Rightarrow -\frac{1}{2}\le \sin x\cos x\le \frac{1}{2}$

$\Rightarrow 0\le |\sin x\cos x|\le \frac{1}{2}$

$\Rightarrow 0\le |\sin x\cos x{|}^2\le \frac{1}{4}$ ........ $\left(2\right)$

Multiplying both sides by $-3$ in $\left(2\right)$

$\Rightarrow 0\ge -3|\sin x\cos x{|}^2\ge \frac{-3}{4}$ (as inequality changes when negative number is multiplied)

Now, adding $1$ we get,

$\Rightarrow 1+0\ge 1-3|\sin x\cos x{|}^2\ge 1+\frac{-3}{4}\Rightarrow 1\ge 1-3|\sin x\cos x{|}^2\ge \frac{1}{4}$

$\Rightarrow 1\ge K\ge \frac{1}{4}$ ..... (from $\left(1\right)$)

II:

$P=\sin x+\cos x$ ......... $\left(3\right)$

To find minimum and maximum of the function $P$, differentiate $P$ w.r.t. $x$ and equate it with 'zero'.

$\Rightarrow \cos x-\sin x=0$

$\Rightarrow \sin x=\cos x$

$\Rightarrow x=\frac{\pi }{4},\frac{5\pi }{4},\frac{9\pi }{4}$ and so on

Taking $x=\frac{5\pi }{4}$ for $P$ to be minimum, $P=-\frac{2}{\sqrt{2}}=-\sqrt{2}$

Taking $x=\frac{9\pi }{4}$ for $P$ to be maximum, $P=\frac{2}{\sqrt{2}}=\sqrt{2}$

$\Rightarrow -\sqrt{2}\le P\le \sqrt{2}$

Hence, only Statement I is true.