The number of real solutions of the equation ^ - 1 ( ∑ _ i = 1 ^ ∞ x ^ i + 1 - x ∑ _ i = 1 ^ ∞ ((x)/(2)) ^ i ) = (π)/(2) - ^ - 1 ( ∑ _ i = 1 ^ ∞ (- (x)/(2)) ^ i - ∑ _ i = 1 ^ ∞ (- x) ^ i ) lying in the interval (

Correct answer: 0 ∑ _ i = 1 ^ ∞ x ^ i + 1 = x ^ 2 1 - x ∑ _ i = 1 ^ ∞ ((x)/(2)) ^ i = (x)/(2 - x) ∑ _ i = 1 ^ ∞ (- (x)/(2)) ^ i = (- x)/(2 + x) ∑ _ i = 1 ^ ∞ (- x) ^ i = (- x)/(1 + x) For the given equation to have real solutions ∑ _ i = 1 ^ ∞ x ^ i + 1 - x ∑ _ i = 1 ^ ∞ ((x)/(2)) ^ i = ∑ _ i = 1 ^ ∞ (- (x)/(2)) ^ i - ∑ _ i = 1 ^ ∞ (- x) ^ i x ^ 2 1 - x - x ^ 2 2 - x = (- x)/(2 + x) + (x)/(1 + x) x( x ^ 3 + 2 x ^ 2 + 5x - 2) = 0 ⇒ x = 0isasolution , also f(x) = x ^ 3 + 2 x ^ 2 + 5x - 2 ⇒ f ^ ' ( x ) = 3 x ^ 2 + 4x + 5D = 16 - 60 = - 44 and ∵ f((1)/(2)) . f(- (1)/(2)) Hencetwosolutionexist .

JEE Advanced 2018Paper 1

Exam: JEE Advanced
Year: 2018
Paper: Paper 1
Subject: Mathematics
Chapter: Application of Derivatives

JEE Advanced 2018 · Paper 1 · Mathematics · Application of Derivatives

The number of real solutions of the equation ^ - 1 ( ∑ _ i = 1 ^ ∞...

The number of real solutions of the equation

\sin^{- 1} \left(\mathop{∑}_{i = 1}^{\infty}{x}^{i + 1} - x\mathop{∑}_{i = 1}^{\infty}{\left(\dfrac{x}{2}\right)}

^{i}\right) = \dfrac{\pi}{2} - \cos^{- 1} \left(\mathop{∑}_{i = 1}^{\infty}{\left(- \dfrac{x}{2}\right)}^{i} - \mathop{∑}_{i = 1}^{\infty}{\left(- x\right)}^{i}\right)

lying in the interval \left(- \dfrac{1}{2}, \dfrac{1}{2}\right)

is____.
(Here, the inverse trigonometric functions

\sin^{- 1} xand\cos^{- 1} x

assume values in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]and\left[0, \pi\right]$
respectively.)

Correct answer

Step-by-step solution

\mathop{∑}_{i = 1}^{\infty}{x}^{i + 1} = \dfrac{{x}^{2}}{1 - x} $ $ \mathop{∑}_{i = 1}^{\infty}{\left(\dfrac{x}{2}\right)}^{i} = \dfrac{x}{2 - x} $ $ \mathop{∑}_{i = 1}^{\infty}{\left(- \dfrac{x}{2}\right)}^{i} = \dfrac{- x}{2 + x} $ $ \mathop{∑}_{i = 1}^{\infty}{\left(- x\right)}^{i} = \dfrac{- x}{1 + x} $ For the given equation to have real solutions $ \mathop{∑}_{i = 1}^{\infty}{x}^{i + 1} - x\mathop{∑}_{i = 1}^{\infty}{\left(\dfrac{x}{2}\right)}^{i} = \mathop{∑}_{i = 1}^{\infty}{\left(- \dfrac{x}{2}\right)}^{i} - \mathop{∑}_{i = 1}^{\infty}{\left(- x\right)}^{i} $ $ \dfrac{{x}^{2}}{1 - x} - \dfrac{{x}^{2}}{2 - x} = \dfrac{- x}{2 + x} + \dfrac{x}{1 + x} $ $ x\left({x}^{3} + 2{x}^{2} + 5x - 2\right) = 0 $$ ⇒ x = 0isasolution$ $, also$

$f\left(x\right) = {x}^{3} + 2{x}^{2} + 5x - 2 ⇒ {f}^{'} ( x ) = 3{x}^{2} + 4x + 5D = 16 - 60 = - 44 < 0 ⇒ fisstrictlyincreasing$
$and ∵ f\left(\dfrac{1}{2}\right) . f\left(- \dfrac{1}{2}\right) < 0$
Hencetwosolutionexist .$

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