The set of all real values of λ for which the function
f(x) = ( 1 - cos ²x )( λ + sin x ),x ∈ ( - π/2,π/2 )
has exactly one maxima and exactly one minima, is :

Solution

Maths sir · 4 steps

Put t = sin x to get a cubic, then keep both of its turning points inside (-1, 1).

1. Change variable. Since 1 - cos² x = sin² x, put t = sin x. On (-π/2, π/2), t is strictly increasing and covers (-1, 1), so f = t²(λ + t) = t³ + λ t².

   f'(x) = cos x (3t² + 2λ t) = cos x · t (3t + 2λ)

2. Where f' changes sign. On this interval cos x > 0, so the sign of f' depends only on t(3t + 2λ). Its zeros are t = 0 and t = -2λ/3. One maximum and one minimum need two distinct simple zeros, both inside (-1, 1).
3. Range of lambda. -1 < -2λ/3 < 1 gives -3/2 < λ < 3/2. If λ = 0, the two zeros merge into t = 0. Then f = sin³ x, which is increasing and has no extremum, so λ = 0 must be removed.
4. So. λ ∈ (-3/2, 3/2) - {0}, which is option (a).

The idea to keep. If f depends only on sin x over (-π/2, π/2), put t = sin x and find the extrema in t. A repeated root of the derivative gives no extremum.

Answer from the NTA final answer key. Solution by PrepWiser.