Let ƒ(x) = xcos –1 (–sin|x|), x ∈ [ - π/2,π/2 ], then which of the following is true?

Solution

Maths sir · 4 steps

First tame the inverse cosine into π/2+|x|, and then f is just a polynomial with |x| in it.

1. Remove the minus sign. Use cos⁻¹(-y) = π - cos⁻¹ y. Then use cos⁻¹ y = π/2 - sin⁻¹ y, so cos⁻¹(-sin|x|) = π/2 + sin⁻¹(sin|x|).
2. Undo the sine. Here |x| lies in [0, π/2], which is inside the principal range of sin⁻¹. So sin⁻¹(sin|x|) = |x|, and f(x) = x(π/2 + |x|) = π/2x + x|x|.
3. Differentiate piecewise. For x > 0, f(x) = π/2x + x², so f'(x) = π/2 + 2x. For x < 0, f(x) = π/2x - x², so f'(x) = π/2 - 2x. Both sides give π/2 at x = 0, so f is differentiable there.

   f'(x) = π/2 + 2|x|, x ∈ (-π/2, π/2)

4. So. f' has a V shape with its lowest point at x = 0. It decreases on (-π/2, 0) and increases on (0, π/2), which is option (a).

The idea to keep. Inside an inverse trig function, first pull out the minus sign and change to the complementary function. Then check that the argument lies in the principal range before you cancel sin⁻¹(sin θ) to θ.

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