Let f(x) = 2cos ⁻¹x + 4cot ⁻¹x - 3x² - 2x + 10, x ∈ [ - 1,1]. If [a, b] is the range of the function f, then 4a - b is equal to :

Solution

Maths sir · 4 steps

Show f is decreasing on [-1, 1]; then the range runs from f(1) up to f(-1).

1. Differentiate. Both cos⁻¹ x and cot⁻¹ x are decreasing, but -3x²-2x is not monotonic on [-1,1]. So check the sign of f'(x) directly.

   f'(x) = -2/√(1-x²) - 4/(1+x²) - (6x+2)

2. Sign of f'. For |x|<1, 2/√(1-x²) ≥ 2 and 4/(1+x²) > 2, so the first two terms together are less than -4. Also -(6x+2) < 4 for x > -1. Hence f'(x) < 0, and f is decreasing on [-1,1].
3. End values. Least value: a = f(1) = 0 + 4 × π/4 - 3 - 2 + 10 = π + 5. Greatest value: b = f(-1) = 2π + 4 × 3π/4 - 3 + 2 + 10 = 5π + 9.
4. So. 4a - b = 4π + 20 - 5π - 9 = 11 - π, which is option (b).

The idea to keep. Prove the function is monotonic with the derivative; then the range is just its two end values. And cot⁻¹(-1) = 3π/4, since cot⁻¹ lives in (0, π).

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