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

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

Grade:Select Grade

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
8 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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