Question

Let $P\left(x\right)$ be a polynomial of degree $4$, with $P\left(2\right)=-1$, and . Then is equal to : $P\text{'}\left(2\right)=0,P\text{'}\text{'}\left(2\right)=2,P\text{'}\text{'}\text{'}\left(2\right)=–12,P\text{'}\text{'}\text{'}\text{'}\left(2\right)=24,P\text{'}\text{'}\left(1\right)$ is equal to :

• A. $20$
• B. $22$
• C. $24$
• D. $26$

Answer 1

The correct option is D

$26$

Explanation for the correct answer:

Finding the value of $P\text{'}\text{'}\left(1\right)$:

Step 1: Finding $\mathrm{P}\text{'}\left(2\right)$

Let us assume that $P\left(x\right)=a{(x-2)}^4+b{(x-2)}^3+c{(x-2)}^2+d(x-2)+e....\left(1\right)$

Differentiating both sides of $\left(1\right)$; w.r.t. $x$, we get

$P\text{'}\left(x\right)=4a{(x-2)}^3+3b{(x-2)}^2+2c(x-2)+d...\left(2\right)$

Putting$x=2$; we get

$$\begin{gathered} P\text{'}\left(2\right)=d \\ \Rightarrow d=0 \end{gathered}$$

Step 2: Finding $\mathrm{P}\text{'}\text{'}\left(2\right)$

Differentiating both sides of $\left(2\right)$ w.r.t. $x$ we get

$$\begin{gathered} P\text{'}\text{'}\left(x\right)=12a{(x-2)}^2 \\ =6b(x-2)+2c...\left(3\right) \end{gathered}$$

Putting$x=2,$ we get

$$\begin{gathered} P\text{'}\text{'}\left(2\right)=2c \\ \Rightarrow 2c=2 \\ \Rightarrow c=1 \end{gathered}$$

Step 3: Finding $\mathrm{P}\text{'}\text{'}\text{'}\left(2\right)$

Again differentiating $\left(3\right)$ w.r.t $x$, we get

$P\text{'}\text{'}\text{'}\left(x\right)=24a(x-2)+6b...\left(4\right)$

Substituting $x=2$, we get

$$\begin{gathered} \mathrm{P}\text{'}\text{'}\text{'}\left(2\right)=6\mathrm{b} \\ \Rightarrow -12=6\mathrm{b} \\ \Rightarrow \mathrm{b}=-2 \end{gathered}$$

Step 4: Finding $\mathrm{P}\text{'}\text{'}\text{'}\text{'}\left(2\right)$

Differentiating both sides of $\left(4\right)$ w.r.t $x$ we get

$P\text{'}\text{'}\text{'}\text{'}\left(x\right)=24a$

Substituting $x=2,$ we get

$$\begin{gathered} P\text{'}\text{'}\text{'}\text{'}\left(2\right)=24a \\ \Rightarrow 24=24a \\ \Rightarrow a=1 \end{gathered}$$

Step 5: Finding the value of $P\text{'}\text{'}\left(1\right)$

Now, substituting these values of $a,b,c,d,e,$ in$\left(3\right)$

$$\begin{gathered} P\text{'}\text{'}\left(x\right)=12\left(1\right){(x-2)}^2+6(-2)(x-2)+2\times \left(1\right) \\ \therefore P\text{'}\text{'}\left(1\right)=12{(1-2)}^2-12(1-2)+2 \\ \therefore P\text{'}\text{'}\left(1\right)=12+12+2=26 \end{gathered}$$

Hence, Option (D) is the correct answer.