Question

Let $f(x)=x^3+x$, then the equation
$\frac{2}{y-f(2)}+\frac{3}{y-f(3)}+\frac{4}{y-f(4)}=0$, has

• A. both roots lying in $(f(2),f(3))$
• B. exactly one root lying in $(f(3),f(4))$
• C. exactly one root lying in $(-\infty ,f(2))$
• D. exactly one root lying in $(f(4),\infty )$

Answer 1

Answer: (B)

We have, $f\left(x\right)=x^3+x$
$f\left(2\right)=10$,
$f\left(3\right)=30$,
$f\left(4\right)=68$
Given, $\frac{2}{y-f\left(2\right)}+\frac{3}{y-f\left(3\right)}+\frac{4}{y-f\left(4\right)}=0$
$\Rightarrow \frac{2}{y-10}+\frac{3}{y-30}+\frac{4}{y-68}=0$
$\Rightarrow 2\left(y-30\right)\left(y-68\right)+3\left(y-10\right)$
$\left(y-68\right)+4\left(y-10\right)\left(y-30\right)=0$
$\Rightarrow 9y^2-590y+7320=0$
$y=\frac{590\pm \sqrt{{\left(590\right)}^2-4\times 9\times 7320}}{2\times 9}$
$y=\frac{590\pm \sqrt{84580}}{18}$
$=\frac{590\pm 290.82}{18}$
$y=16.62$ or $48.93$
Clearly, exactly one root lie between $\left(f\left(3\right),f\left(4\right)\right)$