99 Percentile Qs Bank for JEE MainMathematicsTrigonometric Equations
The set of solutions of the equation ( 3 -1) +( 3 +1) =2 is
Options
- A2 n 4 + 12 : n Z
- B2 n 4 - 12 : n Z
- Cn +(-1)^n 4 + 12 : n Z
- Dn +(-1)^n 4 - 12 : n Z
Correct answer
A. 2 n 4 + 12 : n Z
Step-by-step solution
( 3 -1) +( 3 +1) =2 3 -1 2 + 3 +1 2 =1 ...(i) Comparing with a +b =1 . ie, a= 3 -1 2 , b= 3 +1 2 a^2+b^2 = ( 3 -1)^2 4 + ( 3 +1)^2 4 = 1 2 3+1-2 3 +3+1+2 3 = 1 2 8 = 1 2 2 2 = 2 Dividing on both sides by 2 in Eq. (i), we get ( 3 -1 2 2 ) + ( 3 +1 2 2 ) = 1 2 Let = 3 -1 2 2 Then, = 1- ( 3 -1 2 2 )^2 = 1- (4-2 3 ) 8 = 8-4+2 3 8 = 4+2 3 8 = 3 +1 4 = ( 3 +1)^2 8 = ( 3 +1 2 2 ) So, + = 1 2 ( - )= 1 2 = 4 - =2 n 4 , =2 n 4 + (ii) 15= 3 +1 2 2 ie, = 15^ = 12 = 12 From Eq. (ii), [ =2 n 4 + 12 ] . n Z