Question

If $p\left(x\right)$ be a polynomial of degree three that has a local maximum value $8$ at $x=1$ and a local minimum value $4$ at $x=2$; then $p\left(0\right)$ is equal to

• A. $12$
• B. $-12$
• C. $-24$
• D. $6$

Answer 1

The correct option is B

$-12$

Explanation for the correct option.

Step 1: Find the factor of $p\left(x\right)$

Given that $x=1,x=2$ are points of local maximum & local minimum.

It means that $x=1,x=2$ must be critical point.

i.e; $p\text{'}\left(1\right)=0,p\text{'}\left(2\right)=0$

$\begin{array}{rcl}p\text{'}\left(x\right)& =& a(x-1)(x-2)\\ & =& a\left(x^2-3x+2\right)\end{array}$

Step 2: Finding the vlaue of $p\left(x\right)$

Integrating we get:

$p\left(x\right)=a\left[\frac{x^3}{3}-\frac{3x^2}{2}+2x\right]+b$

Also, $p\left(1\right)=8$.

$\begin{array}{rcl}& \Rightarrow & 8=a\left[\frac{1}{3}-\frac{3}{2}+2\right]+b\\ & \Rightarrow & 8=a\left[\frac{5}{6}\right]+b\\ & \Rightarrow & 48=5a+6b...\left(1\right)\end{array}$

For $p\left(2\right)=4$

$\begin{array}{rcl}& \Rightarrow & 4=a\left[\frac{8}{3}-\frac{12}{2}+4\right]+b\\ & \Rightarrow & 4=a\left[\frac{4}{6}\right]+b\\ & \Rightarrow & 24=4a+6b...\left(2\right)\end{array}$

Step 3: Finding the vlalue of $p\left(0\right)$

Solve equation (1) & (2) we get:

$a=24,b=-12$

Substitute value $a=24,b=-12$ in $p\left(x\right)=a\left[\frac{x^3}{3}-\frac{3x^2}{2}+2x\right]+b$we have:

$p\left(x\right)=24\left[\frac{x^3}{3}-\frac{3x^2}{2}+2x\right]-12$

$\Rightarrow p\left(0\right)=-12$

Hence, option B is correct.