JEE Main 2022 — Mathematics Calculus

Given, $\underset{n\rightarrow \infty }{\mathrm{lim}}\sqrt{{n}^{2}-n-1}+\alpha n+\beta =0$n→∞lim​n2−n−1​+αn+β=0

Now rewriting the expression we get,

$$\underset{n\rightarrow \infty }{\mathrm{lim}}{({n}^{2}-n-1)}^{\frac{1}{2}}+\alpha n=-\beta$$n→∞lim​(n2−n−1)21​+αn=−β

$\Rightarrow$⇒$\underset{n\rightarrow \infty }{\mathrm{lim}}n{(1-\frac{1}{n}-\frac{1}{{n}^{2}})}^{\frac{1}{2}}+\alpha n=-\beta$n→∞lim​n(1−n1​−n21​)21​+αn=−β

Now on using binomial approximation ${(1+x)}^{n}=1+nx$(1+x)n=1+nx we get,

$$\underset{n\rightarrow \infty }{\mathrm{lim}}n(1-\frac{1}{2n}-\frac{1}{2{n}^{2}})+\alpha n=-\beta$$n→∞lim​n(1−2n1​−2n21​)+αn=−β

$$\Rightarrow \underset{n\rightarrow \infty }{\mathrm{lim}}n-\frac{1}{2}-\frac{1}{2n}+\alpha n=-\beta$$⇒n→∞lim​n−21​−2n1​+αn=−β

$$\Rightarrow \underset{n\rightarrow \infty }{\mathrm{lim}}n(1+\alpha )-\frac{1}{2}-\frac{1}{2n}=-\beta$$⇒n→∞lim​n(1+α)−21​−2n1​=−β

Now for limit to exist $(1+\alpha )=0\Rightarrow \alpha =-1$(1+α)=0⇒α=−1

Now on putting the value of $\alpha$α we get,

$$\underset{n\rightarrow \infty }{\mathrm{lim}}n(1+\alpha )-\frac{1}{2}-\frac{1}{2n}=-\beta$$n→∞lim​n(1+α)−21​−2n1​=−β

$$\Rightarrow \underset{n\rightarrow \infty }{\mathrm{lim}}-\frac{1}{2}-\frac{1}{2n}=-\beta$$⇒n→∞lim​−21​−2n1​=−β

$$\Rightarrow -\frac{1}{2}-0=-\beta$$⇒−21​−0=−β

$$\Rightarrow \beta =\frac{1}{2}$$⇒β=21​

So, $8(\alpha +\beta )=8(-1+\frac{1}{2})=8\times \frac{-1}{2}=-4$8(α+β)=8(−1+21​)=8×2−1​=−4