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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Mathematics
Chapter
Functions

BITSAT 2023 · Paper 2 · Mathematics · Functions

The domain of the function f(x)= x- 1-x^2 is

The domain of the function f(x)=x−1−x2f(x)=\sqrt{x-\sqrt{1-x^2}} is

Options

  1. A[−1,−12]∪[12,1]\left[-1,-\frac{1}{\sqrt{2}}\right]\cup \left[\frac{1}{\sqrt{2}},1\right]
  2. B[−1,1][-1,1]
  3. C\left(-\infty,−-\frac{1}{2}\right] \cup \left[\frac{1}{\sqrt{2}},+\infty\right)
  4. D[12,1]\left[\frac{1}{\sqrt{2}}, 1\right]

Correct answer

D. [12,1]\left[\frac{1}{\sqrt{2}}, 1\right]

Step-by-step solution

For f(x)f(x) to be defined, we must havex−1−x2≥0 or x≥1−x2>0 or x2≥1−x2x-\sqrt{1-x^2} \geq 0 \text{ or } x \geq \sqrt{1-x^2} > 0 \text{ or } x^2 \geq 1-x^2or x2≥12x^2 \geq \frac{1}{2}.Now, x2≥12⇒(x−12)(x+12)≥0⇒x≤−12 or x≥12x^2 \geq \frac{1}{2} \Rightarrow\left(x-\frac{1}{\sqrt{2}}\right)\left(x+\frac{1}{\sqrt{2}}\right) \geq 0 \Rightarrow x \leq-\frac{1}{\sqrt{2}} \quad \text{ or } \quad x \geq \frac{1}{\sqrt{2}}Also, 1−x2≥0or1-x^2 \geq 0 orx^2 \leq 1.Also,.Also,x^2 \leq 1 \Rightarrow(x-1)(x+1) \leq 0 \Rightarrow-1 \leq x \leq 1Thus,Thus,x > 0, x^2 \geq \frac{1}{2}andandx^2 \leq 1 \Rightarrow x \in \left[\frac{1}{\sqrt{2}},1\right]
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