All the pairs (x, y) that satisfy the inequal | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The given inequality is $2^{\sqrt{\sin^2 x - 2 \sin x + 5}} \cdot 2^{-2 \sin^2 y} \leq 1$. This simplifies to $2^{\sqrt{(\sin x - 1)^2 + 4}} \leq

Vedclass · Accepted Answer

$\sin x = |\sin y|$

Vedclass · Answer

(D) The given inequality is $2^{\sqrt{\sin^2 x - 2 \sin x + 5}} \cdot 2^{-2 \sin^2 y} \leq 1$. This simplifies to $2^{\sqrt{(\sin x - 1)^2 + 4}} \leq 2^{2 \sin^2 y}$. Since the base $2 > 1$,we have $\sqrt{(\sin x - 1)^2 + 4} \leq 2 \sin^2 y$. We know that $(\sin x - 1)^2 \geq 0$,so $\sqrt{(\sin x - 1)^2 + 4} \geq \sqrt{4} = 2$. Thus,$2 \sin^2 y \geq 2$,which implies $\sin^2 y \geq 1$. Since the maximum value of $\sin^2 y$ is $1$,we must have $\sin^2 y = 1$,which means $\sin y = \pm 1$. Substituting $\sin^2 y = 1$ into the inequality,we get $\sqrt{(\sin x - 1)^2 + 4} \leq 2(1) = 2$. This implies $(\sin x - 1)^2 + 4 \leq 4$,so $(\sin x - 1)^2 \leq 0$. Since a square cannot be negative,we must have $(\sin x - 1)^2 = 0$,which means $\sin x = 1$. Since $\sin x = 1$ and $|\sin y| = 1$,we conclude $\sin x = |\sin y|$.

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

Similar Questions

If $0 < \theta < \frac{\pi}{4}$ and $8 \cos \theta + 15 \sin \theta = 15$,then $15 \cos \theta - 8 \sin \theta = $

If $x = \sec \phi - \tan \phi$ and $y = \csc \phi + \cot \phi$,then:

If $2 \sinh^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)=\log \left(\frac{1+x}{1-x}\right)$,then $x$ is equal to

Which of the following statements is/are correct for $0 < \theta < \frac{\pi}{2}$?

The extreme values of $4 \cos \left(x^2\right) \cos \left(\frac{\pi}{3}+x^2\right) \cos \left(\frac{\pi}{3}-x^2\right)$ over $\mathbb{R}$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module