If 0≤x≤2π, then 2cosec2⁡x12y2−y+1≤2

The given inequality can be written as 2cosec2⁡x(y−1)2+1≤2 (i)Since cosec2⁡x≥1 for all real x, we have2cosec2⁡x≥2 (ii)Also (y−1)2+1≥1⇒ (y−1)2+1≥1 (iii)From Eqs. (ii) and (iii), we get2cosec2⁡x(y−1)2+1≥2 (iv)Therefore, from Eqs. (i) and (iv), equality holds only when2cosec2⁡x=2 and (y−1)2+1=1,Thus,cosec2⁡x=1 and (y−1)2+1=1 ⇒ sin⁡x=±1 and y=1⇒ x=π2,3π2 and y=1Hence, the solution of the given inequality is x=π2,3π2 and y=1.

If 0≤x≤2π, then 2cosec2⁡x12y2−y+1≤2

The given inequality can be written as 2cosec2⁡x(y−1)2+1≤2 (i)Since cosec2⁡x≥1 for all real x, we have2cosec2⁡x≥2 (ii)Also (y−1)2+1≥1⇒ (y−1)2+1≥1 (iii)From Eqs. (ii) and (iii), we get2cosec2⁡x(y−1)2+1≥2 (iv)Therefore, from Eqs. (i) and (iv), equality holds only when2cosec2⁡x=2 and (y−1)2+1=1,Thus,cosec2⁡x=1 and (y−1)2+1=1 ⇒ sin⁡x=±1 and y=1⇒ x=π2,3π2 and y=1Hence, the solution of the given inequality is x=π2,3π2 and y=1.

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If $0 \leq x \leq 2 π, then 2^{{cosec}^{2} x} \sqrt{\frac{1}{2} y^{2} - y + 1} \leq \sqrt{2}$

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is satisfied by exactly one value of y

is satisfied by exactly two value of x

is satisfied by x for which cos x = 0

is satisfied by x for which sin x = 0

answer is A, B, C.

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Detailed Solution

The given inequality can be written as
$2^{{cosec}^{2} x} \sqrt{(y - 1)^{2} + 1} \leq 2 (i)$
Since ${cosec}^{2} x \geq 1$ for all real x, we have
$2^{{cosec}^{2} x} \geq 2 (ii)$
Also ${(y−1)}^{2} + 1 \geq 1$
$\Rightarrow \sqrt{(y - 1)^{2} + 1} \geq 1 (iii)$
From Eqs. (ii) and (iii), we get
$2^{{cosec}^{2} x} \sqrt{(y - 1)^{2} + 1} \geq 2 (iv)$
Therefore, from Eqs. (i) and (iv), equality holds only when
$2^{{cosec}^{2} x} = 2 and \sqrt{(y - 1)^{2} + 1} = 1$ ,Thus,
${cosec}^{2} x = 1 and (y - 1)^{2} + 1 = 1 \begin{aligned} \Rightarrow \sin x = \pm 1 and y = 1 \\ \Rightarrow x = \frac{π}{2}, \frac{3 π}{2} and y = 1 \end{aligned}$
Hence, the solution of the given inequality is $x = \frac{π}{2}, \frac{3 π}{2} and y = 1 .$