Q

# How to solve this problem- - Complex numbers and quadratic equations - JEE Main-8

All the pairs ( x, y ) that satisfy the inequality

$2^{\sqrt{sin^{2}x-2sinx+5}\cdot \frac{1}{4^{sin^{2}y}}}\leq 1$  also satisfy the equation :

• Option 1)

$2|sinx|=3siny$

• Option 2)

$2 sinx=siny$

• Option 3)

$sinx=2siny$

• Option 4)

$sinx=|siny|$

Views

$2^{\sqrt{sin^{2}x-2sinx+5}\cdot \frac{1}{4^{sin^{2}y}}}\leq 1$

$2^{\sqrt{sin^{2}x-2sinx+5}}\leq 2^{2sin^{2}y}$

$\sqrt{{sin^{2}x-2sinx+5}}\leq 2 sin^{2}y$

$\sqrt{{(sinx-1)^{2}+4}}\leq 2 sin^{2}y$

=> $sinx=1$ & $|siny|=1$

So, correct option is (4).

Option 1)

$2|sinx|=3siny$

Option 2)

$2 sinx=siny$

Option 3)

$sinx=2siny$

Option 4)

$sinx=|siny|$

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