ExamHabit
Practise free

JEE Main 2023 · 29 Jan · Shift 1

JEE Main 2023 Mathematics question on Sets Relations and Functions

The domain of f(x)=log⁡(x+1)(x−2)e2log⁡ex−(2x+3)f(x) = \dfrac{\log_{(x+1)}(x-2)}{e^{2\log_e x} - (2x+3)}, x∈Rx \in \mathbb{R} is
  1. A.
    R−{−1,3}\mathbb{R} - \{-1, 3\}
  2. B.
    (2,∞)−{3}(2, \infty) - \{3\}
  3. C.
    R−{3}\mathbb{R} - \{3\}
  4. D.
    (−1,∞)−{3}(-1, \infty) - \{3\}
Show answer and solution

Correct option: B

Solution

For f(x) to be defined, three conditions must hold simultaneously.

1) Logarithm log(x+1)(x-2) requires: argument x-2 > 0 → x > 2, and base x+1 > 0 and x+1 ≠ 1 → x > -1 and x ≠ 0. Combined with x > 2, this gives x > 2.

2) The denominator: e^(2·lnx) = x², so denominator = x² - (2x+3) = x² - 2x - 3 = (x-3)(x+1) ≠ 0 → x ≠ 3 and x ≠ -1. Since we already need x > 2, only x ≠ 3 is relevant.

3) Also, lnx in e^(2lnx) requires x > 0, which is already satisfied by x > 2.

Combining everything: x > 2 and x ≠ 3, giving domain (2, ∞) − {3}.

[JEE Main 2023 · 29 Jan · Shift 1]

Practise Sets Relations and Functions the way it is asked

A timed set of real JEE Main Sets Relations and Functions questions, graded as the exam grades. No sign-up.

Start a Sets Relations and Functions test

More: all JEE Main Sets Relations and Functions questions · JEE Main 2023 · Mathematics chapter list