Theory Of Equations Ques 84
- All the pairs $(x, y)$ that satisfy the inequality $2^{\sqrt{\sin ^{2} x-2 \sin x+5}} \cdot \frac{1}{4^{\sin ^{2} y}} \leq 1$ also satisfy the equation
(2019 Main, 10 April I)
(a) $2|\sin x|=3 \sin y$
(b) $\sin x=|\sin y|$
(c) $\sin x=2 \sin y$
(d) $2 \sin x=\sin y$
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Answer:
Correct Answer: 84.(b)
Solution:
Formula:
Range of Quadratic Expression:
- Given, inequality is
$ \begin{aligned} & 2^{\sqrt{\sin ^{2} x-2 \sin x+5}} \cdot \frac{1}{4^{\sin ^{2} y}} \leq 1 \\ & \Rightarrow 2^{\sqrt{(\sin x-1)^{2}+4}} \cdot 2^{-2 \sin ^{2} y} \leq 1 \\ & \Rightarrow 2^{\sqrt{(\sin x-1)^{2}+4}} \leq 2^{2 \sin ^{2} y} \\ & \Rightarrow \sqrt{(\sin x-1)^{2}+4} \leq 2 \sin ^{2} y \\ & \quad\left[\text { if } a>1 \text { and } a^{m} \leq a^{n} \Rightarrow m \leq n\right] \\ & \because \text { Range of } \sqrt{(\sin x-1)^{2}+4} \text { is }[2,2 \sqrt{2}] \\ & \text { and range of } 2 \sin ^{2} y \text { is }[0,2] . \end{aligned} $
$\therefore$ The above inequality holds, iff
$ \begin{aligned} & \sqrt{(\sin x-1)^{2}+4}=2=2 \sin ^{2} y \\ & \Rightarrow \sin x=1 \text { and } \sin ^{2} y=1 \\ & \Rightarrow \sin x=|\sin y| \end{aligned} $
[from the options]
BETA
