Question

If $0\le x\le 2\pi$ then $2^{co{\sec }^{2}x}\sqrt{\frac{1}{2}{y}^2-y+1}\le \sqrt{2}$

• A. is satisfied by exactly one value of $y$
• B. is satisfied by exactly two value of $x$
• C. is satisfied by $x$ for which $cosx=0$
• D. is satisfied by $x$ for which $sinx=0$

Answer 1

Answer: (C)

is satisfied by $x$ for which $cosx=0$

Given,

$2^{co{\sec }^{2}x}\sqrt{\frac{1}{2}{y}^2-y+1}\le \sqrt{2}$

Simplifying, we get
$2^{cose{c}^{2}x}.\sqrt{y^2-2y+2}.\frac{1}{\sqrt{2}}\le \sqrt{2}$.

Or
$2^{cose{c}^{2}x}.\sqrt{y^2-2y+2}\le 2$

Or
$2^{cose{c}^{2}x-1}.\sqrt{(y-1{)}^2+1}\le 1$

$2^{co{t}^{2}x}.\sqrt{(y-1{)}^2+1}\le 1$

Let
$y=1$ we get

$2^{co{t}^{2}x}\le 1$

Now $co{t}^2\left(x\right)>0$

Hence
The above inequality is true for $cotx=0$
Or

$cos\left(x\right)=0$ ........ $\because \frac{\cos A}{\sin A}=\cot A$
Or
$x=\frac{\pi }{2}$.