The function
f(x) = (4x³ - 3x²)/6 - 2 sin x + ( 2x - 1 )cos x :

Solution

Maths sir · 3 steps

Differentiate, let the cos x terms cancel, and f'(x) factors as (2x-1)(x-sin x).

1. Differentiate. The polynomial part gives (12x²-6x)/6 = 2x² - x. By the product rule, (2x-1)cos x gives 2 cos x - (2x-1)sin x. Its 2 cos x cancels the -2 cos x that comes from -2 sin x.

   f'(x) = x(2x-1) - (2x-1)sin x = (2x-1)(x-sin x)

2. Second factor. Since |sin x| < |x| for x ≠ 0, the factor x - sin x is positive for x > 0, negative for x < 0, and zero at x = 0.

   Interval	2x-1	x-sin x	f'(x)
   x < 0	negative	negative	positive, so f increases
   0 < x < 1/2	negative	positive	negative, so f decreases
   x > 1/2	positive	positive	positive, so f increases

3. So. On [1/2, ∞), f'(x) ≥ 0 and it is zero only at x = 1/2, so f increases there. On (-∞, 1/2] it first rises and then falls, so it is neither increasing nor decreasing there. The answer is (c).

The idea to keep. When f'(x) comes out as a product, check the sign of each factor on every interval between their zeros. Also remember that x - sin x always has the same sign as x.

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