JEE Main 2021 · 17 March · evening · Single correct · +4 / −1
Consider the function f : R → R defined by
f(x) = { ( 2 - sin ( 1/x ) )|x|,x ≠ 0; 0, x = 0 . Then f is :
- (a) not monotonic on (-∞, 0) and (0, ∞)
- (b) monotonic on (0, ∞) only
- (c) monotonic on (-∞, 0) only
- (d) monotonic on (-∞, 0) ∪ (0, ∞)
Answer and solution
Answer (a) not monotonic on (-∞, 0) and (0, ∞)
Solution
Maths sir · 4 steps
Differentiate: the 1/x cos 1/x term blows up near 0 and keeps flipping the sign of f'.
- Right side. For x > 0, |x| = x, so f(x) = 2x - x sin 1/x. Differentiating sin 1/x by the chain rule brings out a factor -1/x², which leaves a 1/x cos 1/x term.
f'(x) = 2 - sin 1/x + 1/x cos 1/x, x > 0
- Test two points. At x = 1/2nπ the sine is 0 and the cosine is 1, so f' = 2 + 2nπ > 0. At x = 1/(2n+1)π the sine is 0 and the cosine is -1, so f' = 2 - (2n+1)π < 0 because π > 2. Both kinds of points lie in (0, ∞), so f' takes both signs there.
- Left side. For x < 0, f(x) = -2x + x sin 1/x, so f'(x) = -2 + sin 1/x - 1/x cos 1/x. At x = -1/2nπ with n ≥ 1, f' = -2 + 2nπ > 0. At x = -1/(2n+1)π, f' = -2 - (2n+1)π < 0.
- So. On an interval where f is monotonic, f' cannot take both signs. Here f' is positive at some points and negative at others on both (-∞, 0) and (0, ∞). So f is monotonic on neither, which is option (a).
The idea to keep. To prove a function is not monotonic, find one point where f' > 0 and one where f' < 0 in the same interval. For sin 1/x-type terms, try 1/x = 2nπ and (2n+1)π.
Answer from the NTA final answer key. Solution by PrepWiser.