JEE Advance - Mathematics (2018 - Paper 1 Offline - No. 10)
The number of real solutions of the equation $$\eqalign{
& {\sin ^{ - 1}}\left( {\sum\limits_{i = 1}^\infty {} {x^{i + 1}} - x\sum\limits_{i = 1}^\infty {} {{\left( {{x \over 2}} \right)}^i}} \right) \cr
& = {\pi \over 2} - {\cos ^1}\left( {\sum\limits_{i = 1}^\infty {} {{\left( {{{ - x} \over 2}} \right)}^i} - \sum\limits_{i = 1}^\infty {} {{\left( { - x} \right)}^i}} \right) \cr} $$ 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.)
(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.)
Answer
2
Explanation
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 $$f(x) = {x^3} + 2{x^2} + 5x - 2$$
$$f'(x) = 3{x^2} + 4x + 5$$
$$f'(x) > 0,\,\forall x \in R$$
$$ \therefore $$ $${x^3} + 2{x^2} + 5x - 2$$ has only one real roots
Therefore, total number of real solution is 2.
$${\sin ^{ - 1}}\left( {\sum\limits_{i = 1}^\infty {} {x^{i + 1}} - x\sum\limits_{i = 1}^\infty {} {{\left( {{x \over 2}} \right)}^i}} \right)$$
$$ = {\pi \over 2} - {\cos ^1}\left( {\sum\limits_{i = 1}^\infty {} {{\left( {{{ - x} \over 2}} \right)}^i} - \sum\limits_{i = 1}^\infty {} {{\left( { - x} \right)}^i}} \right)$$
$$ \Rightarrow {\sin ^{ - 1}}\left[ {{{{x^2}} \over {1 - x}} - {{x.{x \over 2}} \over {1 - {x \over 2}}}} \right]$$
$$ = {\pi \over 2} - {\cos ^{ - 1}}\left[ {{{{{ - x} \over 2}} \over {1 + {x \over 2}}} - {{( - x)} \over {1 + x}}} \right]$$
$$ \because $$ $$\left[ {\sum\limits_{i = 1}^\infty {} {x^{i + 1}} = {x^2} + {x^3} + {x^4} + .... = {{{x^2}} \over {1 - x}}} \right]$$ 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 $$f(x) = {x^3} + 2{x^2} + 5x - 2$$
$$f'(x) = 3{x^2} + 4x + 5$$
$$f'(x) > 0,\,\forall x \in R$$
$$ \therefore $$ $${x^3} + 2{x^2} + 5x - 2$$ has only one real roots
Therefore, total number of real solution is 2.
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