Maths-
General
Easy

Question

Statement-I The equation  has exactly one solution in [0, 2].Statement-II For equations of type  to have real solutions in   should hold true.

Hint:

The correct answer is: Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for Statement-I.

Here , we have to find the which statement is correct and if its correct explanation or not.Firstly ,Statement-I: The equation √(3 ) cosx- sinx=2 has exactly one solution in [ 0 , 2 π ]We have, √(3 ) cosx- sinx=2⇒√(3 )/2 cosx-1/2 sinx=1⇒cos⁡〖π/6〗 cosx-sin⁡〖π/6〗 sinx=1 [ since we know that sinπ/3 = √(3 )/2 and cos⁡〖π/3〗 = 1/2 ]⇒cos⁡〖(π/6+ x )=1〗 [ since ,cos a cosb -sin a sinb = cos ( a + b ) ] ⇒cos⁡〖(x+ π/6)=1〗⇒ x + π/6 = 2nπ⇒ x = 2nπ - π/6For n = 0 , we have x = - π/6 , and ∉ ( 0,2π)For n = 1 , we have x = 11 π /6 , and ∈ ( 0, 2π)For n = 2, we have x = 23 π /6 and ∉ ( 0, 2π)it have only one solution in [ 0 , 2π]x = 11 π /6Therefore, the Statement-I is true.Now,Statement-II : For equations of type acos θ + bsin θ = c to have real solution in [ 0 , 2π], |c| ≤ √(a2+ b2 ) should hold true.We have ,acos θ + bsin θ = c⇒a /c cos θ + b/c sin θ = 1Let a/c = sinα and b/c = cosα We have,⇒ sinα cos θ + cosα sin θ = 1⇒ sin(θ + α ) = 1 [ since, sin a cosb + cosa sinb = sin( a+ b)]Therefore, (θ + α ) is definitely lies in [ 0 , 2π]. Hence statement – II is true. But it is not correct explanation of statement – I .The correct answer is , Statement-I is true, Statement-II is true ;Statement-II is NOT a correct explanation for Statement-I

In this question, we have to find the statements are the correct or not and statement 2 is correct explanation or not , is same like assertion and reason. Here, Start solving first Statement and try to prove it . Then solve the Statement-II . Remember cos a cosb -sin a sinb = cos ( a + b ) and sin a cosb + cosa sinb = sin( a+ b) .