Question

If $f\left(x\right)={\cos }^{-1}\left(x-x^2\right)+\sqrt{1-\frac{1}{\left|x\right|}}+\frac{1}{\left[x^2-1\right]}$, then the domain of $f\left(x\right)$ is

• A. $\left[-\sqrt{2},\frac{1-\sqrt{5}}{2}\right]$
• B. $\left[\sqrt{2},\frac{1+\sqrt{5}}{2}\right]$
• C. $(\sqrt{2},\frac{1+\sqrt{5}}{2}]$
• D. None of these

Answer 1

The correct option is C

$(\sqrt{2},\frac{1+\sqrt{5}}{2}]$

Explanation for the correct option.

Step 1. Find the domain of $f_1\left(x\right)={\cos }^{-1}\left(x-x^2\right)$.

For inverse cosine function the domain is $\left[-1,1\right]$, so for the function $f_1\left(x\right)={\cos }^{-1}\left(x-x^2\right)$ to be defined: $-1\le x-x^2\le 1$. Now,

$$\begin{gathered} x-x^2\ge -1 \\ \Rightarrow x^2-x\le 1 \\ \Rightarrow x^2-x-1\le 0 \end{gathered}$$

Now, the roots of the equation $x^2-x-1=0$ is given as:

$\begin{array}{rcl}x& =& \frac{-\left(-1\right)\pm \sqrt{{\left(-1\right)}^2-4\times 1\times \left(-1\right)}}{2\times 1}\\ & =& \frac{1\pm \sqrt{5}}{2}\end{array}$

So the solution of the inequality $x^2-x-1\le 0$ is $\frac{1-\sqrt{5}}{2}\le x\le \frac{1+\sqrt{5}}{2}$.

Again for $x-x^2\le 1$ it can be seen that

$$\begin{gathered} x^2-x\ge -1 \\ \Rightarrow x^2-x+1\ge 0 \end{gathered}$$

This inequality is true for all real numbers.

Thus the domain of the function $f_1\left(x\right)={\cos }^{-1}\left(x-x^2\right)$ is $\frac{1-\sqrt{5}}{2}\le x\le \frac{1+\sqrt{5}}{2}$.

Step 2. Find the domain of $f_2\left(x\right)=\sqrt{1-\frac{1}{\left|x\right|}}$.

The expression inside a square root should be non-negative, so the function $f_2\left(x\right)=\sqrt{1-\frac{1}{\left|x\right|}}$ is defined when $1-\frac{1}{\left|x\right|}\ge 0$. So

$$\begin{gathered} 1-\frac{1}{\left|x\right|}\ge 0 \\ \Rightarrow -\frac{1}{\left|x\right|}\ge -1 \\ \Rightarrow \frac{1}{\left|x\right|}\le 1 \\ \Rightarrow \left|x\right|\ge 1 \end{gathered}$$

So the solution of the inequality is $x\in (-\infty ,-1]\cup [1,\infty )$.

So the domain of the function $f_2\left(x\right)=\sqrt{1-\frac{1}{\left|x\right|}}$ is $x\in (-\infty ,-1]\cup [1,\infty )$.

Step 3. Find the domain of $f_3\left(x\right)=\frac{1}{\left[x^2-1\right]}$.

The function $f_3\left(x\right)=\frac{1}{\left[x^2-1\right]}$ is not defined when the denominator is equal to $0$. So $\left[x^2-1\right]\ne 0$. As $\left[\right]$ is the greatest integer function.

So the given function is not defined on $0\le x^2-1<1$. On solving

$$\begin{gathered} 1\le x^2<2 \\ \Rightarrow 1\le x<\sqrt{2} \end{gathered}$$

Thus, the function is not defined on $1\le x<\sqrt{2}$.

So the domain of the function $f_3\left(x\right)=\frac{1}{\left[x^2-1\right]}$ is $x\in (-\infty ,1]\cup [\sqrt{2},\infty )$.

Step 4. Find the domain of $f\left(x\right)$.

The function $f\left(x\right)$ is defined as $f\left(x\right)={\cos }^{-1}\left(x-x^2\right)+\sqrt{1-\frac{1}{\left|x\right|}}+\frac{1}{\left[x^2-1\right]}$.

Now, the domain of the function $f_1\left(x\right)={\cos }^{-1}\left(x-x^2\right)$ is $\frac{1-\sqrt{5}}{2}\le x\le \frac{1+\sqrt{5}}{2}$, the domain of the function $f_2\left(x\right)=\sqrt{1-\frac{1}{\left|x\right|}}$ is $x\in (-\infty ,-1]\cup [1,\infty )$, and the domain of the function $f_3\left(x\right)=\frac{1}{\left[x^2-1\right]}$ is $x\in (-\infty ,1]\cup [\sqrt{2},\infty )$.

So the domain of $f\left(x\right)$ is the intersection of the domain of all three functions and so its domain is $\sqrt{2}<x\le \frac{1+\sqrt{5}}{2}$.

Thus the domain of $f\left(x\right)$ is $(\sqrt{2},\frac{1+\sqrt{5}}{2}]$.

Hence, the correct option is C.