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JEE Main 20199 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
9 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2019 · 9 April · Evening Shift · Mathematics · Functions

The domain of the definition of the function f (x) = 1 4 - x ^ 2 +...

The domain of the definition of the function ff\left(x\right) = \dfrac{1}{4 - {x}^{2}} + \log_{10}(x3−x)\left({x}^{3} - x\right) is:The domain of the definition of the function f (x) = 1 4 - x ^ 2 + _ 10 ( x ^ 3 - x) is:

Options

  1. A(−1,0)∪(1,2)∪(2,∞)\left(- 1,0\right) \cup \left(1, 2\right) \cup \left(2, \infty\right)
  2. B(1,2)∪(2,∞)( 1, 2 ) \cup \left(2, \infty\right)
  3. C(−2,−1)∪(−1,0)∪(2,∞)\left(- 2, - 1\right) \cup \left(- 1, 0\right) \cup \left(2, \infty\right)
  4. D(−1,0)∪(1,2)∪(3,∞)\left(- 1, 0\right) \cup \left(1, 2\right) \cup \left(3, \infty\right)

Correct answer

A. (−1,0)∪(1,2)∪(2,∞)\left(- 1,0\right) \cup \left(1, 2\right) \cup \left(2, \infty\right)

Step-by-step solution

We know that log⁡10(g(x))\log_{10}\left(g\left(x\right)\right) is defined when g(x)>0g\left(x\right) > 0 and for a fraction to be defined its denominator must be non-zero.

Thus, the given function is defined only when

x3−x>0{x}^{3} - x > 0 and 4−x2≠04 - {x}^{2} \ne 0

⇒x(x−1)(x+1)>0⇒ x\left(x - 1\right)\left(x + 1\right) > 0 and x≠2,−2x \ne 2, - 2

The wavy curve for the inequality is
The domain of the definition of the function f (x) = 1 4 - x ^ 2 + _ 10 ( x ^ 3 - x) is:
⇒x∈(−1,0)∪(1,∞)⇒ x \in \left(- 1, 0\right) \cup \left(1, \infty\right) and x≠2,−2.x \ne 2, - 2 .

Hence, domain is \left(- 1, 0\right) \cup \left(1, 2\right) \cup \left(2, \infty\right) .

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