Question

if $x_1,x_2,x_3...x_n$ denote the values of $x$ where $f\left(x\right)$ vanishes such that $x_1>x_2>x_3...x_n$,then $\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{r}{x_r}$is equal to

• A. $-8$
• B. $-4$
• C. $-2$
• D. $-1$

Answer 1

Answer: (B)

$-4$

$f\left(\frac{x+y}{3}\right)=\frac{2+f\left(x\right)+f\left(y\right)}{3}$

$f\left(x\right)=kx+\frac{2}{4}$

$f^{\prime }\left(x\right)=k$

Given, $f^{\prime }\left(2\right)=2$

Therefore, $k=2$

$f\left(x\right)=2x+\frac{2}{4}$

For finding vanishing points

$2x+\frac{2}{4}=0$

$x=-4$

Now, ${lim}_{n\to \infty }{\sum }_{r=1}^n\frac{r}{x_r}\Rightarrow {lim}_{n\to \infty }\frac{1}{1/-4}\Rightarrow -4$ (Answer)