ExamHabit
Practise free

JEE Advanced 2013

JEE Main 2013 Mathematics question on Definite Integrals and Areas

Let f:[1/2,1]→Rf : [1/2, 1] \to R (the set of all real numbers) be a positive, non-constant and differentiable function such that f′(x)<2f(x)f'(x) < 2f(x) and f(1/2)=1f(1/2) = 1. Then, the value of ∫1/21f(x) dx\int_{1/2}^{1} f(x)\, dx lies in the interval
  1. A.
    (2e−1,2e)(2e-1, 2e)
  2. B.
    (e−1,2e−1)(e-1, 2e-1)
  3. C.
    (e−12,e−1)\left(\dfrac{e-1}{2}, e-1\right)
  4. D.
    (0,e−12)\left(0, \dfrac{e-1}{2}\right)
Show answer and solution

Correct option: D

Solution

From f'(x) < 2f(x), divide by f(x) > 0: f'(x)/f(x) < 2, so d/dx[ln f(x)] < 2.

Integrating from 1/2 to x: ln f(x) - ln f(1/2) < 2(x - 1/2), giving ln f(x) < 2x - 1, so f(x) < e^(2x-1).

Since f is positive and non-constant with f'(x) < 2f(x), and f(1/2) = 1, we have 0 < f(x) < e^(2x-1) on (1/2, 1].

Therefore: 0 < ∫(1/2 to 1) f(x)dx < ∫(1/2 to 1) e^(2x-1)dx = [e^(2x-1)/2](1/2 to 1) = (e¹ - e⁰)/2 = (e-1)/2.

So the integral lies in (0, (e-1)/2).

[JEE Advanced 2013]

Practise Definite Integrals and Areas the way it is asked

A timed set of real JEE Main Definite Integrals and Areas questions, graded as the exam grades. No sign-up.

Start a Definite Integrals and Areas test

More: all JEE Main Definite Integrals and Areas questions · JEE Main 2013 · Mathematics chapter list