Question

The equation $2{{\cos }^{-1}}x={{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right)$ is satisfied by:
(a). $-1\le x\le 1$
(b). $0\le x\le 1$
(c). $x\ge 1$
(d). $x\le 1$

Answer 1

Hint: In the equation given above we have to find the solutions in x which we are going to find by using the relation that in ${{\cos }^{-1}}x$, the values which $x$ can take lies from -1 to 1. Then whatever is written inside ${{\cos }^{-1}}\left( x \right)$, make that function lie from -1 to 1 and then find the interval of x by taking the intersection of two solutions.

Complete step-by-step solution:
The inverse trigonometric equation given in the above problem of which we have to find the solutions are as follows:
$2{{\cos }^{-1}}x={{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right)$
Now, to find the solutions in x for the above equation, we know that for ${{\cos }^{-1}}x$, the values of x lies in the range of -1 to 1 so using this interval of x and putting the expression written inside the ${{\cos }^{-1}}$ from -1 to 1, we are going to find the intervals of x.
First of all, the ${{\cos }^{-1}}$ written on the L.H.S of the above equation we get,
${{\cos }^{-1}}x$
The range of x which can be possible for the above inverse trigonometric function is as follows:
$-1\le x\le 1$ ………….(1)
Now, writing the interval of expression ${{\cos }^{-1}}$ written on the R.H.S of the above equation we get,
${{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right)$
The expression written inside the ${{\cos }^{-1}}$ is equal to:
$2{{x}^{2}}-1$
Now, writing the above expression in between -1 to 1 we get,
$-1\le 2{{x}^{2}}-1\le 1$
Adding 1 on all the three sides of the above inequality we get,
$\begin{align}
  & -1+1\le 2{{x}^{2}}\le 1+1 \\
 & \Rightarrow 0\le 2{{x}^{2}}\le 2 \\
\end{align}$
Dividing 2 on all the three sides of the above inequality we get,
$\begin{align}
  & 0\le \dfrac{2{{x}^{2}}}{2}\le \dfrac{2}{2} \\
 & \Rightarrow 0\le {{x}^{2}}\le 1 \\
\end{align}$
Now, ${{x}^{2}}$ is always non negative so we are not considering the inequality which is showing that ${{x}^{2}}\ge 0$ and only taking the other inequality and then the above inequality will look like:
${{x}^{2}}\le 1$
Subtracting 1 on both the sides of the above inequality and we get,
${{x}^{2}}-1\le 0$
We know the algebraic identity which states that:
${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$
Substituting $a$ as x and b as 1 in the above equation we get,
${{x}^{2}}-{{1}^{2}}=\left( x-1 \right)\left( x+1 \right)$
So, using the above relation in ${{x}^{2}}-1\le 0$ we get,
$\left( x-1 \right)\left( x+1 \right)\le 0$
The solution interval of the above inequality is as follows:
$-1\le x\le 1$ ………… (2)
Finding the intersection of two inequalities (1 and 2) as:
$-1\le x\le 1$
$-1\le x\le 1$
As both the inequalities are same so the intersection of the above two inequalities are as follows:
$-1\le x\le 1$
Now, substituting any negative value of x in the given equation and see whether the negative values hold true or not. Let us take the value of x as $-\dfrac{1}{2}$ in the given equation and we get,
$\begin{align}
  & 2{{\cos }^{-1}}x={{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right) \\
 & \Rightarrow 2{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)={{\cos }^{-1}}\left( 2{{\left( -\dfrac{1}{2} \right)}^{2}}-1 \right) \\
 & \Rightarrow 2\left( \pi -\dfrac{\pi }{3} \right)={{\cos }^{-1}}\left( 2\left( \dfrac{1}{4} \right)-1 \right) \\
 & \Rightarrow 2\left( \dfrac{2\pi }{3} \right)={{\cos }^{-1}}\left( \dfrac{1}{2}-1 \right) \\
 & \Rightarrow \dfrac{4\pi }{3}={{\cos }^{-1}}\left( -\dfrac{1}{2} \right) \\
 & \Rightarrow \dfrac{4\pi }{3}=\dfrac{2\pi }{3} \\
\end{align}$
As you can see that L.H.S is not equal to R.H.S so for negative values of x. So the equation won’t hold true.
Hence, the solutions of x for the above equation are $0\le x \le 1$ and the correct option is (b).

Note: The mistake that could be possible in the above solution is that you might forget to check the negative values of x and there are higher possibilities for committing the mistake because you can see one of the options has $-1\le x\le 1$ so make sure you won’t make this mistake.