cos2x +k(sinx) = 2k - 7 has solution for : (a) k < 3 (b) k < 2 (c) k > 3 (d) 2 < k < 6
Hello,
Given eqn is,
cos 2x + k ( sin x ) = 2k - 7
So, 1 - 2 sin ^ 2 x + k ( sin x ) = 2k - 7
So, 2 sin ^ 2 x - k ( sin x ) + 2k - 8 = 0
So, 2 sin ^ 2 x - k ( sin x ) + 2 ( k - 4 ) = 0............. ( 1 )
The eqn ( 1 ) is a quadratic eqn and after solving it, we get,
sin x = ( k - 4 ) / 2 and 2
SInce, value of sin x can never be greater than 1, sin x is not equal to 2
So, sin x = ( k - 4 ) / 2
Since the value of sin x ranges from -1 to 1,
-1 ≤ ( k - 4 ) / 2 ≤ 1
So, 2 ≤ k ≤ 6
So, the correct option is Option D.
Best Wishes.
Hello,
The correct answer is option d
When we solve the above equation in terms of quadratic form of sin x, then the value of k obtained solved within the range [-1,1] yields the result 2<k<6
Related Questions
Trending Articles/News
