JEE Main 2023 · 24 January · morning · Single correct · +4 / −1
Let f(x) = { x²sin ( 1/x ), , x ≠ 0; 0, , x = 0
Then at x=0
- (a) f is continuous but f' is not continuous
- (b) f and f' both are continuous
- (c) f is continuous but not differentiable
- (d) f' is continuous but not differentiable
Answer and solution
Answer (a) f is continuous but f' is not continuous
Solution
Maths sir · 5 steps
Get f'(0) from the limit definition, then check whether f'(x) actually approaches that value.
- Continuity of f. Since |sin 1/x| ≤ 1, we have |x²sin 1/x| ≤ x², which tends to 0. So lim(x→ 0) f(x) = 0 = f(0), and f is continuous at 0.
- Derivative at zero. By definition, f'(0) = lim(h→ 0)(h²sin(1/h) - 0)/h = lim(h→ 0) h sin 1/h = 0, by the same squeeze. So f is differentiable at 0, with f'(0) = 0.
- Derivative elsewhere. For x ≠ 0, use the product rule and the chain rule:
f'(x) = 2x sin 1/x - cos 1/x, x ≠ 0
- Limit of f'. As x → 0, the term 2x sin 1/x goes to 0. But cos 1/x keeps swinging between -1 and 1, so lim(x→ 0) f'(x) does not exist and cannot equal f'(0) = 0.
- So. f is continuous (and even differentiable) at 0, but f' is not continuous at 0. The answer is (a).
The idea to keep. Never find f'(0) by plugging 0 into the formula for f'(x). Use the limit definition at the point, and test continuity of f' separately by taking the limit of f'(x).
Answer from the NTA final answer key. Solution by PrepWiser.