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Differential Calculus> Q. THE RANGE OF THE FUNCTION sin(inverse)...

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Q. THE RANGE OF THE FUNCTION sin(inverse)( x(square) + 2x)

Bharat Makkar , 10 Years ago

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1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello Student,
Please find answer to your question below

$f(x) = sin^{-1}(x^2+2x)$
First we need to check the domain of the function
$-1\leq x^2+2x\leq 1$
$x^2+2x\geq -1$
$x^2+2x+1\geq 0$
$(x+1)^2\geq 0$
$\Rightarrow x\in R$
$x^2+2x\leq 1$
$x^2+2x-1\leq 0$
$(x-(-1-\sqrt{2}))(x-(\sqrt{2}-1))\leq 0$
$\Rightarrow x\in [-1-\sqrt{2},\sqrt{2}-1]$ ….......(1)
So the domain of f(x) is (1).
Since it covers all the principle domain of sin^-1(x).
So the range of f(x)
$\frac{-\pi }{2}\leq f(x)\leq \frac{\pi }{2}$

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