Question

A non zero polynomial with real coefficients has the property that $g\left(x\right)=g^{\prime }\left(x\right).g^{\prime \prime }\left(x\right)$. Let the leading coefficient of $g\left(x\right)$ be $a$. Then $36a=$

• A. $6$
• B. $4$
• C. $3$
• D. $2$

Answer 1

The correct option is D $2$

Let $g\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+...a_{1}x+a_0$
On differentiating we get,
$g^{\prime }\left(x\right)=n\times \left(a_n{x}^{n-1}\right)+(n-1)\left(a_{n-1}{x}^{n-2}\right)+...a_1$
Differentiating again,
$g^{\prime \prime }\left(x\right)=\left(n\right)(n-1)\left(a_n{x}^{n-2}\right)+(n-1)(n-2)\left(a_{n-1}{x}^{n-2}\right)+...2a_2$
Given,
$g\left(x\right)=g^{\prime }\left(x\right)\times g^{\prime \prime }\left(x\right)$
Comparing the coefficient of the highest power of $x$ we get,
$a_n{x}^n=\left(n\right)\left(n\right)(n-1)\left(a_n{x}^{n-1}\right)\left(a_n{x}^{n-2}\right)$
$\Rightarrow n=2n-3$ and $a_n=n^2(n-1){a_n}^2$
$\Rightarrow n=3$
$\Rightarrow a_n=9\left(2\right){a_n}^2$
$\Rightarrow a_n=a=\frac{1}{18}$
$\Rightarrow 36a=2$