Question

If $\alpha =\underset{n\to \infty }{lim}(\frac{1}{n^3+1}+\frac{4}{n^3+1}+\frac{9}{n^3+1}+\cdots +\frac{n^2}{n^3+1})$ and $\beta =\underset{x\to 0}{lim}\frac{\sin 2x}{\sin 8x},$ then a quadratic equation whose roots are $\alpha$ and $\beta$ is

• A. $12x^2-7x+1=0$
• B. $x^2+19x-120=0$
• C. $x^2-17x+66=0$
• D. $x^2-7x+12=0$

Answer 1

The correct option is A $12x^2-7x+1=0$

$\alpha =\underset{n\to \infty }{lim}\left(\frac{1^2+2^2+3^2+\cdots +n^2}{n^3+1}\right)\Rightarrow \alpha =\underset{n\to \infty }{lim}\frac{n(n+1)(2n+1)}{6(n^3+1)}\Rightarrow \alpha =\frac{2}{6}=\frac{1}{3}$

Also,
$\beta =\underset{x\to 0}{lim}\frac{2x\cdot \left(\frac{\sin 2x}{2x}\right)}{8x\cdot \left(\frac{\sin 8x}{8x}\right)}=\frac{1}{4}$

Now, sum of roots $=\alpha +\beta =\frac{1}{3}+\frac{1}{4}=\frac{7}{12}$
and product of roots $=\alpha \beta =\left(\frac{1}{3}\right)\left(\frac{1}{4}\right)=\frac{1}{12}$

So, required quadratic equation is $x^2-(\alpha +\beta )x+\alpha \beta =0$
$\Rightarrow 12x^2-7x+1=0$