For 0≤x≤2π, then 2cosec2⁡x12y2−y+1≤2

The given inequality can be written as2cosec2⁡x(y−1)2+1≤2---isince cosec2⁡x≥1 for all real x, we have 2cosec2⁡x≥2---ii Also (y−1)2+1≥1 ⇒ (y−1)2+1≥1---iiifrom Eqs. (i) and (ii), we get2cosec2⁡x(y+1)2+1≥2---ivTherefore, from Eqs. (i) and (iv), equality holds only when 2cosec2⁡x=2 and (y−1)2+1=1⇒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

For 0≤x≤2π, then 2cosec2⁡x12y2−y+1≤2

The given inequality can be written as2cosec2⁡x(y−1)2+1≤2---isince cosec2⁡x≥1 for all real x, we have 2cosec2⁡x≥2---ii Also (y−1)2+1≥1 ⇒ (y−1)2+1≥1---iiifrom Eqs. (i) and (ii), we get2cosec2⁡x(y+1)2+1≥2---ivTherefore, from Eqs. (i) and (iv), equality holds only when 2cosec2⁡x=2 and (y−1)2+1=1⇒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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For $0 \leq x \leq 2 π,$ 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 one 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
$2^{{cosec}^{2} x} \sqrt{(y - 1)^{2} + 1} \leq \sqrt{2} - - - (i)$
since ${cosec}^{2} x \geq 1 for all real x, we have$
$\begin{array}{l} 2^{{cosec}^{2} x} \geq 2 - - - (ii) \\ Also (y - 1)^{2} + 1 \geq 1 \Rightarrow \sqrt{(y - 1)^{2} + 1} \geq 1 - - - (iii) \end{array}$
from Eqs. (i) and (ii), we get
$2^{{cosec}^{2} x} \sqrt{(y + 1)^{2} + 1} \geq 2 - - - (iv)$
Therefore, from Eqs. (i) and (iv), equality holds only when $2^{{cosec}^{2} x} = 2 and \sqrt{(y - 1)^{2} + 1} = 1$
$\begin{array}{l} \Rightarrow {cosec}^{2} x = 1 and (y - 1)^{2} + 1 = 1 \Rightarrow \sin x = \pm 1 and y = 1 \\ \Rightarrow x = \frac{π}{2}, \frac{3 π}{2} and y = 1 \end{array}$
Hence, the solution of the given inequality is $x = \frac{π}{2}, \frac{3 π}{2} and y = 1$