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

Answers (1)

Answer:
\left [ -1,1 \right ]
Hint:
The\: domain\: o\! f \sin^{-1}x \: is \left [ -1,1 \right ]
Given:
\sin^{-1}x^{2}
Explanation:
Let \: f\! \left ( x \right )= \left ( x^{2} \right )
\! \! \! \! \! \! \! \! \! \! \! \! The\, domain \, o\! f \sin^{-1} is \left [ -1,1 \right ] which\, implies\, that \, the\, value\, o\! f \, x\, lies\, between -1 and\, 1
i.e; -1\leq x\leq 1
but\: here\: we\: have\: the\: f\! unction \: x^{2}
so,-1\leq x^{2}\leq 1\; \; \; \; \; \; \; [As\, x^{2}\: is\: always\: positive]
\Rightarrow 0\leq x^{2}\leq 1
\Rightarrow\left | x \right | \leq 1    
\Rightarrow x\leq 1\: and x\geq -1\: \: \: \: \: \: \: \: \: \: \: \left [ \because a^{2}-b^{2}=\left ( a+b \right )\left ( a-b \right ) \right ]
Hence, the\: domain\: o\! f \sin^{-1}x^{2} \: is \left [ -1,1 \right ].

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