Question

The range of the function ${\sin }^{2n}x+{\cos }^{2n}x;x\in R,n\in N$ is

• A. $(0,2)$
• B. $[0,1]$
• C. $(0,1]$
• D. $(0,\frac{1}{2})$

Answer 1

The correct option is C $(0,1]$

We know that, if $-1\le x\le 1,$ then
$0\le x^{2n}\le x^2,n\in N$
So,
$0\le {\sin }^{2n}x\le {\sin }^{2}x\Rightarrow 0\le {\cos }^{2n}x\le {\cos }^{2}x$
$\Rightarrow 0\le {\sin }^{2n}x+{\cos }^{2n}x\le {\sin }^{2}x+{\cos }^{2}x\Rightarrow 0\le {\sin }^{2n}x+{\cos }^{2n}x\le 1\Rightarrow 0\le ({\sin }^{n}x{)}^2+({\cos }^{n}x{)}^2\le 1$

For the expression to be minimum,
$({\sin }^{n}x{)}^2+({\cos }^{n}x{)}^2=0$
which is possible only when,
$\sin x=0,\cos x=0$
Which is not possible.

Hence, $({\sin }^{n}x{)}^2+({\cos }^{n}x{)}^2\in (0,1]$