Question

Let $P\left(x\right)$ be a polynomial of degree $4$, with $P\left(2\right)=-1,P^{\prime }\left(2\right)=0,P^{\prime \prime }\left(2\right)=2,P^{\prime \prime \prime }\left(2\right)=-12,P^{\prime \prime \prime \prime }\left(2\right)=24$ then the value of $P^{{}^{\prime \prime }}\left(1\right)=$

• A. $-12$
• B. $2$
• C. $24$
• D. $26$

Answer 1

Answer: (D)

$26$

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

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

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

Putting $x=2;$ we get

$P^{\prime }\left(2\right)=d\Rightarrow d=0$

Differentiating both sides of (2) w.r.t. $x,$ we get

$P^{\prime \prime }\left(x\right)=12a{(x-2)}^2=6b(x-2)+2c$ ...(3)

Putting $x=2,$ we get

$P^{\prime \prime }\left(2\right)=2c\Rightarrow 2c=2\Rightarrow c=1$

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

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

Putting $x=2,$ we get

$P^{\prime \prime \prime }\left(2\right)=6b\Rightarrow -12=6b\Rightarrow b=-2$

Differentiating both sides of (4) w.r.t $x,$ we get

$P^{\prime \prime \prime \prime }\left(x\right)=24a$

Putting $x=2,$ we get

$P^{\prime \prime \prime \prime }\left(2\right)=24a\Rightarrow 24=24a\Rightarrow a=1$

Now, substituting this values of $a,b,c,d,e,$ in (3)

$P^{\prime \prime }\left(x\right)=12\left(1\right){(x-2)}^2+6(-2)(x-2)+2\times \left(1\right)$

$\therefore P^{\prime \prime }\left(1\right)=12{(1-2)}^2-12(1-2)+2$

$=12+12+2=26$