JEE Main 2022 · 24 June · morning · Single correct · +4 / −1
The sum of absolute maximum and absolute minimum values of the function f(x) = |2x² + 3x - 2| + sin x cos x in the interval [0, 1] is :
- (a) 3 + (sin (1)cos ²( 1/2 ))/2
- (b) 3 + 1/2(1 + 2 cos (1))sin (1)
- (c) 5 + 1/2(sin (1) + sin (2))
- (d) 2 + sin ( 1/2 )cos ( 1/2 )
Answer and solution
Answer (b) 3 + 1/2(1 + 2 cos (1))sin (1)
Solution
Maths sir · 3 steps
The modulus is zero at x = 1/2: f falls before it and rises after, so the minimum sits right there.
- Split the modulus. Factor 2x² + 3x - 2 = (2x - 1)(x + 2). On [0, 1] the factor x + 2 is positive, so the quadratic is negative for x < 1/2 and non-negative for x ≥ 1/2. Also write sin x cos x = 1/2 sin 2x.
- Check the slope. On [0, 1/2), f(x) = -2x² - 3x + 2 + 1/2 sin 2x, so f'(x) = -4x - 3 + cos 2x ≤ -3 + 1 < 0 and f is decreasing. On (1/2, 1], f(x) = 2x² + 3x - 2 + 1/2 sin 2x, so f'(x) = 4x + 3 + cos 2x > 0 and f is increasing.
Point f(x) Approx. value Role x = 0 2 + 0 = 2 2 left end, not the largest x = 1/2 0 + 1/2 sin 1 0.42 absolute minimum x = 1 3 + sin 1 cos 1 3.45 absolute maximum - So. The sum is 3 + sin 1 cos 1 + 1/2 sin 1 = 3 + 1/2(1 + 2 cos 1)sin 1, which is option (b).
The idea to keep. With a modulus, first find where the inside becomes zero and split there. Then check the sign of f' on each piece, and compare only the turning point and the two endpoints.
Answer from the NTA final answer key. Solution by PrepWiser.