JEE Main 2024 · 5 April · morning · Single correct · +4 / −1
For the function
f(x)=sin x+3 x-2/π(x²+x), where x ∈[0, π/2],
consider the following two statements :
(I) f is increasing in (0, π/2).
(II) f′ is decreasing in (0, π/2).
Between the above two statements,
- (a) only (I) is true.
- (b) both (I) and (II) are true.
- (c) only (II) is true.
- (d) neither (I) nor (II) is true.
Answer and solution
Answer (b) both (I) and (II) are true.
Solution
Maths sir · 4 steps
Use f'' twice: it shows f' is decreasing, so the smallest value of f' sits at x = π/2.
- Differentiate once. Expand 2/π(x² + x) and differentiate term by term. This gives f'(x) = cos x + 3 - 4x/π - 2/π.
- Differentiate again. f''(x) = -sin x - 4/π. On (0, π/2), sin x > 0, so both terms are negative and f''(x) < 0. So f' is decreasing, and (II) is true.
- Smallest value of f'. A decreasing f' is smallest at the right end, x = π/2. There, f'(π/2) = 0 + 3 - 2 - 2/π = 1 - 2/π ≈ 0.36, which is positive.
f'(x) > f'(π/2) = 1 - 2/π > 0 for 0 < x < π/2
- So. f' > 0 throughout the interval, so f is increasing and (I) is true as well. Both statements hold, so the answer is (b).
The idea to keep. To prove f' > 0 on an interval, first find whether f' rises or falls using f''. Then you only need to check f' at its lowest endpoint.
Answer from the NTA final answer key. Solution by PrepWiser.