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Name: Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.1 Question 3 Subquestion (iii) Maths Textbook Solution
Uploaded: Mon, 2021-07-05 23:48
Description: Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.1 Question 3 Subquestion (iii) Maths Textbook Solution

Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.1 Question 3 Subquestion (iii) Maths Textbook Solution

Answers (1)

Answer:
$\left [ -\sqrt{2},1 \right ]\cup \left [ 1,\sqrt{2} \right ]$

Hint:
$The\: domain \: o\! f \sin^{-1}x \: is \left [ -1,1 \right ].$
Given:
$f\! \left ( x \right )= \sin^{-1}\sqrt{x^{2}-1}$
Explanation:
$-1\leq \sqrt{x^{2}-1}\leq 1$
$x^{2}-1\leq 1 \: [Squaring\: both \: sides]$
$x^{2}-1+1\leq 1+1$
$x^{2}\leq 2$
$-\sqrt{2}\leq x\leq \sqrt{2} \cdot \cdot \cdot \cdot (i)$
$Also, x^{2}-1\geq 0$
$x^{2}\geq 1$
$\Rightarrow x\leq -1 \: and \: x\geq 1 \cdot \cdot \cdot \cdot (ii)$
$From (i) and (ii), the\: value\: o\! f \: x \: lies\: between$
$1\leq x\leq\sqrt{2} \: and \: -\sqrt{2}\leq x\leq -1$
$The \: domain \: o\! f \sin^{-1}\sqrt{x^{2}-1} \: is \left [ -\sqrt{2,}-1 \right ]\cup \left [ 1,\sqrt{2} \right ]$