Question

The values of x which satisfy the inequation $2^{\frac{1}{co{s}^{2}x}}\sqrt{y^2-y+\frac{1}{2}}\le 1$ is

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

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

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

Minimum value of $2^{\frac{1}{co{s}^{2}x}}=2$<\br> Minimum value of $\sqrt{(y-\frac{1}{2}{)}^2+(\frac{1}{2}{)}^2}=\frac{1}{2}$
$\Rightarrow$ minimum value of (1) is 1
$\Rightarrow$ (1) is possible when
$2^{\frac{1}{co{s}^{2}x}}\sqrt{(y-\frac{1}{2}{)}^2+(\frac{1}{2}{)}^2}=1$
$\Rightarrow co{s}^{2}x=1andy=\frac{1}{2}$
$y=\frac{1}{2},x=2n\pi$