If ^ -1 (x - x^ 2 2 + x^ 3 4 -. ) + ^ -1 (x^ 2 - x^ 4 2 + x^ 6 4 -. ) = (π)/(2) for 0 < |x| < 2 , then x & is equal to

Correct answer: B. 1 only Since ^ -1 x + ^ -1 x = (π)/(2) for x ∈ ℝ . ∴ x - x^ 2 2 + x^ 3 4 -. .. = x^ 2 - x^ 4 2 + x^ 6 4 -. .. ⇒ (x)/(1 + (x)/(2)) = x^ 2 1 + x^ 2 2 (0 ⇒ (x)/(2 + x) = x^ 2 2 + x^ 2 ⇒ 2x + x^ 3 = 2x^ 2 + x^ 3 ⇒ x^ 2 = x ⇒ x = 0, 1 But x ≠ 0, ∴ x = 1 .

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

NTA Abhyas 2020 · Mathematics · Inverse Trigonometric Functions

If ^ -1 (x - x^ 2 2 + x^ 3 4 -. ) + ^ -1 (x^ 2 - x^ 4 2 + x^ 6 4 -....

\cot^{-1} \left(x - \dfrac{x^{2}}{2} + \dfrac{x^{3}}{4} -. \ldots \ldots\right) + \tan^{-1} \left(x^{2} - \dfrac{x^{4}}{2} + \dfrac{x^{6}}{4} -. \ldots \ldots\right) = \dfrac{\pi}{2}

for

0 < \left|x\right| < \sqrt{2},

then

x

& is equal to

Options

A $0$ only
B $1$ only
C $0, 1$ both
DNone of these

Correct answer

B. $1$ only

Step-by-step solution

Since

\tan^{-1} x + \cot^{-1} x = \dfrac{\pi}{2}

for

x \in \mathbb{R}

∴ x - \dfrac{x^{2}}{2} + \dfrac{x^{3}}{4} -. \ldots.. = x^{2} - \dfrac{x^{4}}{2} + \dfrac{x^{6}}{4} -. \ldots..

⇒ \dfrac{x}{1 + \dfrac{x}{2}} = \dfrac{x^{2}}{1 + \dfrac{x^{2}}{2}} \left(0 < \left| x\right| < \sqrt{2}\right)<br />

⇒

\dfrac{x}{2 + x} =\dfrac{x^{2}}{2 + x^{2}}

⇒ 2x + x^{3} = 2x^{2} + x^{3}

<br />

⇒ x^{2} = x ⇒ x = 0, 1

<br />But

x \ne 0, ∴ x = 1

.<br />

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