Q&A - Ask Doubts and Get Answers
Q
Engineering
1 week, 3 days ago

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|

 
Answers (1)
Views
V Vakul Arora
Answered 1 week, 3 days ago

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|

Knockout AEEE JEE Main May BUY NOW
Exams
Articles
Questions