Let S be the set of all solutions of the equation ^-1(2x) - 2 ^-1( 1 - x^2) =, x [ - 12, 12 ]. Then x S 2 ^-1(x^2 - 1) is equal to:

Given: ^-1(2x) - 2 ^-1( 1-x^2) =, x [- 12, 12 ] Find: x S 2 ^-1(x^2-1) the solution concludes that the correct option is B. Using the identity shown in the solution, 2 ^-1(t) = ^-1(2t^2-1) with t = 1-x^2, the equation becomes ^-1(2x) - ^-1(2(1-x^2)-1) = that is, ^-1(2x) - ^-1(1-2x^2) = Rearranging as in the extracted working, - ^-1(1-2x^2) = - ^-1(2x) Taking cosine on both sides, (- ^-1(1-2x^2)) = ( - ^-1(2x)) So, 1-2x^2 = -2x which gives 2x^2 - 2x - 1 = 0 Solving the quadratic gives x = 1- 32, 1+ 32 From the interval [- 12, 12 ], the value 1+ 32 is rejected. Hence x = 1- 32 Then, as stated in the solution, x^2 - 1 = - 32 Therefore, 2 ^-1(x^2-1) = 2 ^-1 (- 32) = - 2 3 So the correct option is B.

Let S be the set of all solutions of the equation ^-1(2x) - 2 ^-1… — Solution | JEE Mathematics

MCQMediumJEE 2023Inverse Trigonometric Functions

Let $S$ be the set of all solutions of the equation $\cos^{-1}(2x) - 2\cos^{-1}(\sqrt{1 - x^2}) = \pi, \quad x \in \left[ -\frac{1}{2}, \frac{1}{2} \right] .$ Then $\sum_{x \in S} 2 \sin^{-1}(x^2 - 1)$ is equal to:

A
$0$
B
$\frac{-2\pi}{3}$
C
$\pi - \sin^{-1} \left( \frac{\sqrt{3}}{4} \right)$
D
$\pi - 2\sin^{-1} \left( \frac{\sqrt{3}}{4} \right)$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given:

\cos^{-1}(2x) - 2\cos^{-1}(\sqrt{1-x^2}) = \pi, \quad x \in \left[-\frac{1}{2}, \frac{1}{2}\right]

Find:

\sum_{x \in S} 2\sin^{-1}(x^2-1)

the solution concludes that the correct option is B.

Using the identity shown in the solution,

2\cos^{-1}(t) = \cos^{-1}(2t^2-1)

with $t = \sqrt{1-x^2}$ , the equation becomes

\cos^{-1}(2x) - \cos^{-1}(2(1-x^2)-1) = \pi

that is,

\cos^{-1}(2x) - \cos^{-1}(1-2x^2) = \pi

Detailed Working

Rearranging as in the extracted working,

-\cos^{-1}(1-2x^2) = \pi - \cos^{-1}(2x)

Taking cosine on both sides,

\cos\left(-\cos^{-1}(1-2x^2)\right) = \cos\left(\pi - \cos^{-1}(2x)\right)

So,

1-2x^2 = -2x

which gives

2x^2 - 2x - 1 = 0

Final Evaluation

Solving the quadratic gives

x = \frac{1-\sqrt{3}}{2}, \; \frac{1+\sqrt{3}}{2}

From the interval $\left[-\frac{1}{2}, \frac{1}{2}\right]$ , the value $\frac{1+\sqrt{3}}{2}$ is rejected. Hence

x = \frac{1-\sqrt{3}}{2}

Then, as stated in the solution,

x^2 - 1 = -\frac{\sqrt{3}}{2}

Therefore,

2\sin^{-1}(x^2-1) = 2\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{2\pi}{3}

So the correct option is B.

Common mistakes

Using the answer key instead of the solution working. The solution explicitly concludes option B, so answer resolution must follow the solution as primary source.
Ignoring the domain restriction $x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$ . One root of the quadratic is outside the interval and must be rejected.
Applying inverse-trigonometric identities without checking principal values. For expressions involving $\cos^{-1}$ and $\sin^{-1}$ , principal-value ranges matter and can change the valid transformation.

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Answer

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