QJEEAdv16Mafb99: JEE Advanced 2016 Mathematics - Class 12 JEEAdv16Mafb99 Solution: Let , R be such that . Then 6( + ) equals _________.

Here, $\mathop {\lim }\limits_{x \to 0} {{{x^2}\sin (\beta x)} \over {\alpha x - \sin x}} = 1$

$\mathop {\lim }\limits_{x \to 0} {{{x^2}\left( {\beta x - {{{{(\beta x)}^3}} \over {3!}} + {{{{(\beta x)}^5}} \over {5!}} - ....} \right)} \over {\alpha x - \left( {x - {{{x^3}} \over {3!}} + {{{x^5}} \over {5!}} - ....} \right)}} = 1$

$\Rightarrow \mathop {\lim }\limits_{x \to 0} {{{x^3}\left( {\beta - {{{\beta ^3}{x^2}} \over {3!}} + {{{\beta ^5}{x^4}} \over {5!}} - ....} \right)} \over {(\alpha - 1)x + {{{x^3}} \over {3!}} + {{{x^5}} \over {5!}} - ...}} = 1$

Limit exists only, when $\alpha$ $-$ 1 = 0

$\Rightarrow$ $\alpha$ = 1 ...... (i)

$\therefore$ $\Rightarrow \mathop {\lim }\limits_{x \to 0} {{{x^3}\left( {\beta - {{{\beta ^3}{x^2}} \over {3!}} + {{{\beta ^5}{x^4}} \over {5!}} - ....} \right)} \over {{x^3}\left( {{1 \over {3!}} - {{{x^2}} \over {5!}} - ...} \right)}} = 1$

$\Rightarrow$ 6 $\beta$ = 1 ....... (ii)

From Eqs. (i) and (ii), we get

6( $\alpha$ + $\beta$ ) = 6 $\alpha$ + 6 $\beta$

= 6 + 1

= 7