JEE Main 2020 · 4 September · evening · Single correct · +4 / −1
Let f:( 0,∞ ) → ( 0,∞ ) be a differentiable function such that f(1) = e and
lim _(t → x) (t²f²(x) - x²f²(t))/(t - x) = 0. If f(x) = 1, then x is equal to :
- (a) 1/e
- (b) e
- (c) 1/2e
- (d) 2e
Answer and solution
Answer (a) 1/e
Solution
Maths sir · 4 steps
The limit is a derivative with respect to t in disguise. Differentiate in t, then put t = x.
- Spot the 0/0 form. At t = x the numerator t² f²(x) - x² f²(t) becomes x² f²(x) - x² f²(x) = 0. So the limit is the derivative of the numerator with respect to t, taken at t = x. This is the same as using L'Hôpital's rule.
- Differentiate in t. Treat x as a constant. The derivative is 2t f²(x) - 2x² f(t)f'(t). Put t = x and set the result equal to 0.
2x f²(x) - 2x² f(x)f'(x) = 0 ⇒ f(x) = x f'(x)
- Solve the equation. Dividing by f(x) is allowed because f > 0. Since x > 0 too, f'(x)/f(x) = 1/x. Integrating gives ln f = ln x + C, so f(x) = cx. Then f(1) = e gives c = e, so f(x) = ex.
- So. f(x) = 1 means ex = 1, so x = 1/e. That is option (a).
The idea to keep. A limit as t → x of an expression that becomes 0/0 is a derivative in disguise. Differentiate with respect to the moving variable, put t = x, and you usually get a separable differential equation.
Answer from the NTA final answer key. Solution by PrepWiser.