QJEEM22Mac044: JEE Main 2022 Mathematics - Class 12 JEEM22Mac044 Solution: If , then is equal to :

$\mathop {\lim }\limits_{n \to \alpha } \left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right) = 0$

[ This limit will be zero when $\alpha$ < 0 as when $\alpha$ > 0 then overall limit will be $\infty$ . ]

$\Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{\left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right)\left( {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)} \right)} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = 0$

$\Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{\left( {{n^2} - n - 1} \right) - {{\left( {n\alpha + \beta } \right)}^2}} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = 0$

$\Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{{n^2} - n - 1 - {n^2}{\alpha ^2} - 2n\alpha\beta - {\beta ^2}} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = 0$

$\Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{{n^2}\left( {1 - {\alpha ^2}} \right) - n\left( {1 + 2\alpha \beta } \right) - \left( {1 + {\beta ^2}} \right)} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}}$

Here power of "n" in the numerator is 2 and power of "n" in the denominator is 1.

To get the value of limit equal to zero power of "n" should be equal in both numerator and denominator, otherwise value of limit will be infinite ( $\infty$ ).

$\therefore$ Coefficient of n² should be 0 in this case.

$\therefore$ $1 - {\alpha ^2} = 0$

$\Rightarrow \alpha = \, \pm \,1$

But $\alpha$ should be < 0

$\therefore$ $\alpha = \, + \,1$ not possible

$\therefore$ $\alpha = - 1$ .

$\Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{0 - n\left( {1 + 2\alpha \beta } \right) - \left( {1 + \beta } \right)} \over {n\left[ {\sqrt {1 - {1 \over n} - {1 \over {{n^2}}}} - \alpha - {\beta \over n}} \right]}} = 0$

Divide numerator and denominator by n then we get,

$\Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{ - \left( {1 + 2\alpha \beta } \right) - {{(1 + \beta )} \over n}} \over {\sqrt {1 - {1 \over n} - {1 \over {{n^2}}}} - \alpha - {\beta \over n}}} = 0$

$\Rightarrow {{ - \left( {1 + 2\alpha \beta } \right) - 0} \over {\sqrt {1 - 0 - 0} - \alpha - 0}} = 0$

$\Rightarrow {{ - \left( {1 + 2\alpha \beta } \right)} \over {1 - \alpha }} = 0$

$\Rightarrow - \left( {1 + 2\alpha \beta } \right) = 0$

$\Rightarrow 1 + 2\alpha \beta = 0$

$\Rightarrow 2\alpha \beta = - 1$

$\Rightarrow \beta = - {1 \over {2\alpha }} = - {1 \over {2( - 1)}} = {1 \over 2}$

$\therefore$ $8\left( {\alpha + \beta } \right)$

$= 8\left( { - 1 + {1 \over 2}} \right)$

$= 8 \times - {1 \over 2}$

$= - 4$