Question

Let P(x) be a fourth degree polynomial with derivative P'(x). Such that $P\left(1\right)=P\left(2\right)=P\left(3\right)=P^{\prime }\left(7\right)=0$. Let k is the real number such that $P\left(k\right)=0$, then k is equal to

• A. $\frac{317}{37}$
• B. $\frac{319}{37}$
• C. $\frac{321}{37}$
• D. $\frac{15}{37}$

Answer 1

The correct option is B $\frac{319}{37}$

Take $\alpha$, $\beta$, $\gamma$, k are roots of equation P(x) = 0
$P\left(x\right)=(x-\alpha )(x-\beta )(x-\gamma )(x-k)$
$lnP\left(x\right)=ln(x-\alpha )+ln(x-\beta )+ln(x-\gamma )+ln(x-k)$
take differentiation,
$\frac{p^{\prime }\left(x\right)}{p\left(x\right)}=\frac{1}{(x-\alpha )}+\frac{1}{(x-\beta )}+\frac{1}{(x-\gamma )}+\frac{1}{(x-k)}$
$x=-7,\alpha =1,\beta =2,\gamma =3$
$0=\frac{1}{7-1}+\frac{1}{7-2}+\frac{1}{7-3}+\frac{1}{7-k}\Rightarrow k=\frac{319}{37}$