Solveeit Logo
Login

Question

Question: The equation \(2{{\cos }^{-1}}x={{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right)\) is satisfied by: (a). ...

The equation 2cos⁡−1x=cos⁡−1(2x2−1)2{{\cos }^{-1}}x={{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right) is satisfied by:
(a). −1≤x≤1-1\le x\le 1
(b). 0≤x≤10\le x\le 1
(c). x≥1x\ge 1
(d). x≤1x\le 1

Explanation

Solution

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⁡−1x{{\cos }^{-1}}x, the values which xx can take lies from -1 to 1. Then whatever is written inside cos⁡−1(x){{\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:
2cos⁡−1x=cos⁡−1(2x2−1)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⁡−1x{{\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{{\cos }^{-1}} from -1 to 1, we are going to find the intervals of x.
First of all, the cos⁡−1{{\cos }^{-1}} written on the L.H.S of the above equation we get,
cos⁡−1x{{\cos }^{-1}}x
The range of x which can be possible for the above inverse trigonometric function is as follows:
−1≤x≤1-1\le x\le 1 ………….(1)
Now, writing the interval of expression cos⁡−1{{\cos }^{-1}} written on the R.H.S of the above equation we get,
cos⁡−1(2x2−1){{\cos }^{-1}}\left( 2{{x}^{2}}-1 \right)
The expression written inside the cos⁡−1{{\cos }^{-1}} is equal to:
2x2−12{{x}^{2}}-1
Now, writing the above expression in between -1 to 1 we get,
−1≤2x2−1≤1-1\le 2{{x}^{2}}-1\le 1
Adding 1 on all the three sides of the above inequality we get,
−1+1≤2x2≤1+1 ⇒0≤2x2≤2 \begin{aligned} & -1+1\le 2{{x}^{2}}\le 1+1 \\\ & \Rightarrow 0\le 2{{x}^{2}}\le 2 \\\ \end{aligned}
Dividing 2 on all the three sides of the above inequality we get,
0≤2x22≤22 ⇒0≤x2≤1 \begin{aligned} & 0\le \dfrac{2{{x}^{2}}}{2}\le \dfrac{2}{2} \\\ & \Rightarrow 0\le {{x}^{2}}\le 1 \\\ \end{aligned}
Now, x2{{x}^{2}} is always non negative so we are not considering the inequality which is showing that x2≥0{{x}^{2}}\ge 0 and only taking the other inequality and then the above inequality will look like:
x2≤1{{x}^{2}}\le 1
Subtracting 1 on both the sides of the above inequality and we get,
x2−1≤0{{x}^{2}}-1\le 0
We know the algebraic identity which states that:
a2−b2=(a−b)(a+b){{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)
Substituting aa as x and b as 1 in the above equation we get,
x2−12=(x−1)(x+1){{x}^{2}}-{{1}^{2}}=\left( x-1 \right)\left( x+1 \right)
So, using the above relation in x2−1≤0{{x}^{2}}-1\le 0 we get,
(x−1)(x+1)≤0\left( x-1 \right)\left( x+1 \right)\le 0
The solution interval of the above inequality is as follows:
−1≤x≤1-1\le x\le 1 ………… (2)
Finding the intersection of two inequalities (1 and 2) as:
−1≤x≤1-1\le x\le 1
−1≤x≤1-1\le x\le 1
As both the inequalities are same so the intersection of the above two inequalities are as follows:
−1≤x≤1-1\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 −12-\dfrac{1}{2} in the given equation and we get,
2cos⁡−1x=cos⁡−1(2x2−1) ⇒2cos⁡−1(−12)=cos⁡−1(2(−12)2−1) ⇒2(π−π3)=cos⁡−1(2(14)−1) ⇒2(2π3)=cos⁡−1(12−1) ⇒4π3=cos⁡−1(−12) ⇒4π3=2π3 \begin{aligned} & 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{aligned}
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≤x≤10\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≤x≤1-1\le x\le 1 so make sure you won’t make this mistake.