QJEEAdv18Mad11d: JEE Advanced 2018 Mathematics - Class 12 JEEAdv18Mad11d Solution: The number of real solutions of the equation sin − 1 ( ∑ i = 1…

Question

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The number of real solutions of the equation

\begin{aligned} \sin^{- 1} (\sum_{i = 1}^{\infty} x^{i + 1} - x \sum_{i = 1}^{\infty} {(\frac{x}{2})}^{i}) \\ = \frac{π}{2} - \cos^{1} (\sum_{i = 1}^{\infty} {(\frac{- x}{2})}^{i} - \sum_{i = 1}^{\infty} {(- x)}^{i}) \end{aligned}

lying in the interval

\left( { - {1 \over 2},{1 \over 2}} \right)

is ........... .

(Here, the inverse trigonometric functions sin^{$-$ 1} x and cos^{$-$ 1} x assume values in

{\left[ { - {\pi \over 2},{\pi \over 2}} \right]}

and

{\left[ {0,\pi } \right]}

, respectively.)

Accepted Answer

We have,

{\sin ^{ - 1}}\left( {\sum\limits_{i = 1}^\infty {} {x^{i + 1}} - x\sum\limits_{i = 1}^\infty {} {{\left( {{x \over 2}} \right)}^i}} \right)

= {\pi \over 2} - {\cos ^1}\left( {\sum\limits_{i = 1}^\infty {} {{\left( {{{ - x} \over 2}} \right)}^i} - \sum\limits_{i = 1}^\infty {} {{\left( { - x} \right)}^i}} \right)

\Rightarrow {\sin ^{ - 1}}\left[ {{{{x^2}} \over {1 - x}} - {{x.{x \over 2}} \over {1 - {x \over 2}}}} \right]

= {\pi \over 2} - {\cos ^{ - 1}}\left[ {{{{{ - x} \over 2}} \over {1 + {x \over 2}}} - {{( - x)} \over {1 + x}}} \right]

\because

\left[ {\sum\limits_{i = 1}^\infty {} {x^{i + 1}} = {x^2} + {x^3} + {x^4} + .... = {{{x^2}} \over {1 - x}}} \right]

using sum of infinite terms of GP

\Rightarrow {\sin ^{ - 1}}\left[ {{{{x^2}} \over {1 - x}} - {{{x^2}} \over {2 - x}}} \right] = {\pi \over 2} - {\cos ^{ - 1}}\left[ {{x \over {1 + x}} - {x \over {2 + x}}} \right]

\Rightarrow {\sin ^{ - 1}}\left[ {{{{x^2}} \over {1 - x}} - {{{x^2}} \over {2 - x}}} \right] = {\sin ^{ - 1}}\left[ {{x \over {1 + x}} - {x \over {2 + x}}} \right]

\because

\left[ {{{\sin }^{ - 1}}x = {\pi \over 2} - {{\cos }^{ - 1}}x} \right]

\Rightarrow {{{x^2}} \over {1 - x}} - {{{x^2}} \over {2 - x}} = {x \over {1 + x}} - {x \over {2 + x}}

\Rightarrow {x^2}\left( {{{2 - x - 1 + x} \over {(1 - x)(2 - x)}}} \right) = x{{(2 + x - 1 - x)} \over {(1 + x)(2 + x)}}

\Rightarrow {x \over {2 - 3x + {x^2}}} = {1 \over {2 + 3x + {x^2}}}

or x = 0

\Rightarrow {x^3} + 3{x^2} + 2x = {x^2} - 3x + 2

\Rightarrow {x^3} + 2{x^2} + 5x - 2 = 0

or x = 0

Let

f(x) = {x^3} + 2{x^2} + 5x - 2

f'(x) = 3{x^2} + 4x + 5

f'(x) &gt; 0,\,\forall x \in R

\therefore

{x^3} + 2{x^2} + 5x - 2

has only one real roots

Therefore, total number of real solution is 2.

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