JEE Main 2021 · 24 February · evening · Single correct · +4 / −1
Let f:R → R be defined as
f(x) = { - 55x,, if x < - 5; 2x³ - 3x² - 120x,, if - 5 ≤ x ≤ 4; 2x³ - 3x² - 36x - 336,, if x > 4,
Let A = x ∈ R : f is increasing. Then A is equal to :
- (a) ( - 5,∞ )
- (b) ( - ∞ , - 5) ∪ (4,∞ )
- (c) ( - 5, - 4) ∪ (4,∞ )
- (d) ( - ∞ , - 5) ∪ ( - 4,∞ )
Answer and solution
Answer (c) ( - 5, - 4) ∪ (4,∞ )
Solution
Maths sir · 3 steps
Differentiate each piece on its own interval, factorise, and read the sign of f' there.
- Derivative of each piece. For x < -5, f'(x) = -55. For -5 < x < 4, f'(x) = 6x² - 6x - 120 = 6(x-5)(x+4). For x > 4, f'(x) = 6x² - 6x - 36 = 6(x-3)(x+2).
Interval f'(x) Sign of factors Behaviour x < -5 -55 always negative decreasing -5 < x < -4 6(x-5)(x+4) (-)(-), positive increasing -4 < x < 4 6(x-5)(x+4) (-)(+), negative decreasing x > 4 6(x-3)(x+2) (+)(+), positive increasing - Check the joins. At x = -5 both formulas give 275, and at x = 4 both give -400. So f is continuous and has no jumps that could change the answer.
- So. f' > 0 only on (-5, -4) and (4, ∞), so A = (-5, -4) ∪ (4, ∞), which is option (c).
The idea to keep. For a piecewise function, differentiate each piece only on its own interval, factorise f', and check the sign there. Also check the joins for continuity.
Answer from the NTA final answer key. Solution by PrepWiser.