Question

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

• 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 $\cos x=\frac{1}{2}$
• D. is satisfied by $x$ for which $\sin x=0$

Answer 1

Answer: (A), (B)

The correct options are
A is satisfied by exactly one value of $y$
B is satisfied by exactly two values of $x$

$2^{{\text{cosec}}^{2}x}\sqrt{\frac{1}{2}{y}^2-y+1}\le \sqrt{2}$
$\Rightarrow 2^{{\text{cosec}}^{2}x}\sqrt{\frac{y^2-2y+2}{2}}\le \sqrt{2}$
$\Rightarrow 2^{{\text{cosec}}^{2}x}\sqrt{(y-1{)}^2+1}\le 2\cdots \left(1\right)$

Since, ${\text{cosec}}^{2}x\ge 1,\forall x\in R$, we have
$2^{{\text{cosec}}^{2}x}\ge 2$
Also, $\sqrt{(y-1{)}^2+1}\ge 1$
$\Rightarrow 2^{{\text{cosec}}^{2}x}\sqrt{(y-1{)}^2+1}\ge 2\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$, equality holds only when
$2^{{\text{cosec}}^{2}x}=2$ and $\sqrt{(y-1{)}^2+1}=1$

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