JEE Main 2023 · 13 April · morning · Single correct · +4 / −1
max 0 ≤ x ≤ π{x-2 sin x cos x+1/3 sin 3 x}=
- (a) (5 π+2+3 √3)/6
- (b) 0
- (c) (π+2-3 √3)/6
- (d) π
Answer and solution
Answer (a) (5 π+2+3 √3)/6
Solution
Maths sir · 4 steps
Write f'(x) in terms of cos x alone. It factors neatly, and then you compare a few candidates.
- Tidy up f. Since 2 sin x cos x = sin 2x, the function is f(x) = x - sin 2x + 1/3 sin 3x. Differentiating gives f'(x) = 1 - 2 cos 2x + cos 3x.
f'(x) = 4 cos³ x - 4 cos² x - 3 cos x + 3 = (cos x - 1)(4 cos² x - 3)
- Critical points. Use cos 2x = 2 cos² x - 1 and cos 3x = 4 cos³ x - 3 cos x to get the factored form above. It is zero when cos x = 1 (x = 0) or cos x = ±√3/2 (x = π/6 or 5π/6).
- Read the sign. The factor cos x - 1 is never positive, so f'(x) > 0 exactly when 4 cos² x < 3, that is for π/6 < x < 5π/6. So f falls, then rises, then falls. That makes π/6 a local minimum and 5π/6 a local maximum.
Candidate x f(x) Approximate value 0 (endpoint) 0 0 π/6 (local min) π/6 - √3/2 + 1/3 = (π + 2 - 3√3)/6 -0.009 5π/6 (local max) 5π/6 + √3/2 + 1/3 = (5π + 2 + 3√3)/6 3.817 π (endpoint) π 3.142 - So. The largest value on [0, π] is at x = 5π/6, giving (5π + 2 + 3√3)/6. That is option (a).
The idea to keep. For the greatest value on a closed interval, list every critical point and both endpoints, then compare the actual values. A local max is not automatically the biggest value.
Answer from the NTA final answer key. Solution by PrepWiser.