JEE Main 2024 · 8 April · morning · Single correct · +4 / −1
The number of critical points of the function f(x)=(x-2)^(2 / 3)(2 x+1) is
- (a) 2
- (b) 1
- (c) 0
- (d) 3
Answer and solution
Answer (a) 2
Solution
Maths sir · 4 steps
Critical points come from two places: where f'(x)=0 and where f' fails to exist inside the domain.
- Differentiate. Use the product rule on (x-2)^(2/3)(2x+1). This gives f'(x) = 2/3(x-2)^(-1/3)(2x+1) + 2(x-2)^(2/3).
- Factor it. Take (x-2)^(-1/3) out of both terms. The bracket becomes 2(2x+1)/3 + 2(x-2) = (4x + 2 + 6x - 12)/3 = 10(x-1)/3.
f'(x) = 10(x-1)/3(x-2)^(1/3)
- Find both kinds. The numerator is zero at x = 1, so f'(1) = 0. The denominator is zero at x = 2, so f'(2) does not exist. But f(2) = 0 is defined, so x = 2 is in the domain and still counts.
- So. The critical points are x = 1 and x = 2, which makes 2 critical points.
The idea to keep. A critical point is any point in the domain where f' = 0 or f' does not exist. With fractional powers, always check where the derivative blows up.
Answer from the NTA final answer key. Solution by PrepWiser.