Question

For $0<x<2\pi ,$ $2^{{\text{cosec}}^{2}x}\sqrt{\frac{1}{2}{y}^2-y+1}\le \sqrt{2}$

• A. is satisfied by exactly one value of $y$
• B. is satisfied by exactly two values of $x$
• C. is satisfied by $x$ for which $\cos x=0$
• D. is satisfied by $x$ for which $\sin x=0$

Answer 1

Answer: (A), (B), (C)

is satisfied by $x$ for which $\cos x=0$

The given inequality can be written as
$2^{{\text{cosec}}^{2}x}\sqrt{{(y-1)}^2+1}\le 2$
Since, ${\text{cosec}}^{2}x\ge 1$ for all $x\in (0,2\pi ),$ we have
$2^{{\text{cosec}}^{2}x}\ge 2\cdots \left(1\right)$

Also, ${(y-1)}^2+1\ge 1$
$\Rightarrow \sqrt{{(y-1)}^2+1}\ge 1...\left(2\right)$

From $\left(1\right)$ and $\left(2\right),$ we get
$2^{{\text{cosec}}^{2}x}\sqrt{{(y-1)}^2+1}\ge 2$

Equality holds only when
$2^{{\text{cosec}}^{2}x}=2$ and $\sqrt{{(y-1)}^2+1}=1$
$\Rightarrow {\text{cosec}}^{2}x=1$ and ${(y-1)}^2+1=1$
$\Rightarrow \sin x=\pm 1$ and $y=1$
$\Rightarrow x=\frac{\pi }{2},\frac{3\pi }{2}$ and $y=1$

Hence, the solution of the given inequality is $x=\frac{\pi }{2},\frac{3\pi }{2}$ and $y=1$