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, $\frac{2}{y-f \left(2\right)}+\frac{3}{y-f \left(3\right)}+\frac{4}{y-f\left(4\right)}=0$
$\Rightarrow \frac{2}{y-10}+\frac{3}{y-30}+\frac{4}{y-68}=0$
$\Rightarrow 2\left(y-30\right)\left(y-68\right)+3\left(y-10\right)$
$\left(y-68\right)+4\left(y-10\right)\left(y-30\right)=0$
$\Rightarrow 9y^{2}-590y+7320=0$
$y=\frac{590 \pm\sqrt{\left(590\right)^{2}-4\times9\times7320}}{2\times9}$
$y=\frac{590 \pm\sqrt{84580}}{18}$
$=\frac{590 \pm290.82}{18}$
$y=16.62$ or $48.93$
Clearly, exactly one root lie between $\left(f \left(3\right), f \left(4\right)\right)$