Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 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)

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|

Posted by

Vakul

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’
anam-khan

NIT Puducherry Cut-off 2024: Computer Science and Engineering had highest closing rank last year
anam-khan

JEE Mains: Farmer's son from remote Maharashtra village aces exam

Latest Question