The domain of the function f(x) = sin^{-1}lef | vedclass

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(C) For the function $f(x) = \sin^{-1}\left(\frac{3x^2+x-1}{(x-1)^2}\right) + \cos^{-1}\left(\frac{x-1}{x+1}\right)$ to be defined,both parts must be

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$\left[\frac{1}{4}, \frac{1}{2}\right] \cup \{0\}$

Vedclass · Answer

(C) For the function $f(x) = \sin^{-1}\left(\frac{3x^2+x-1}{(x-1)^2}\right) + \cos^{-1}\left(\frac{x-1}{x+1}\right)$ to be defined,both parts must be defined.$1$. For $\cos^{-1}\left(\frac{x-1}{x+1}\right)$,we require $-1 \leq \frac{x-1}{x+1} \leq 1$.Solving $\frac{x-1}{x+1} \leq 1 \Rightarrow \frac{x-1-x-1}{x+1} \leq 0 \Rightarrow \frac{-2}{x+1} \leq 0 \Rightarrow x+1 > 0 \Rightarrow x > -1$.Solving $\frac{x-1}{x+1} \geq -1 \Rightarrow \frac{x-1+x+1}{x+1} \geq 0 \Rightarrow \frac{2x}{x+1} \geq 0 \Rightarrow x \in (-\infty, -1) \cup [0, \infty)$.Combining these,we get $x \in [0, \infty)$.$2$. For $\sin^{-1}\left(\frac{3x^2+x-1}{(x-1)^2}\right)$,we require $-1 \leq \frac{3x^2+x-1}{(x-1)^2} \leq 1$.Solving $\frac{3x^2+x-1}{(x-1)^2} \leq 1 \Rightarrow 3x^2+x-1 \leq x^2-2x+1 \Rightarrow 2x^2+3x-2 \leq 0 \Rightarrow (2x-1)(x+2) \leq 0 \Rightarrow x \in [-2, 1/2]$.Solving $\frac{3x^2+x-1}{(x-1)^2} \geq -1 \Rightarrow 3x^2+x-1 \geq -x^2+2x-1 \Rightarrow 4x^2-x \geq 0 \Rightarrow x(4x-1) \geq 0 \Rightarrow x \in (-\infty, 0] \cup [1/4, \infty)$.Combining these,we get $x \in [-2, 0] \cup [1/4, 1/2]$.Taking the intersection of the domains from $(1)$ and $(2)$,we get $x \in [0, \infty) \cap ([-2, 0] \cup [1/4, 1/2]) = \{0\} \cup [1/4, 1/2]$.

The domain of the function $f(x) = \sin^{-1}\left(\frac{3x^2+x-1}{(x-1)^2}\right) + \cos^{-1}\left(\frac{x-1}{x+1}\right)$ is:

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