Let P(x) be a polynomial of degree 4, with $P (2) = - 1$ , $P^{'} (2) = 0$ , $P^{''} (2) = 2$ , $P^{'''} (2) = - 12$ and $P^{i v} (2) = 24$ , The value of $P^{''} (1)$ is

Open in App

Solution

Let us take $P (x) = a {(x - 2)}^{4} + b {(x - 2)}^{3} + c {(x - 2)}^{2} + d (x - 2) + e$
Thus $P (2) = e = - 1$
$\Rightarrow P^{'} (2) = d = 0$
$\Rightarrow P^{''} (2) = 2 c = 2 \Rightarrow c = 1$
$\Rightarrow P^{'''} (2) = 6 b = - 12 \Rightarrow b = - 2$
and $P^{i v} (2) = 24 a = 24 \Rightarrow a = 1$
$\Rightarrow P^{''} (x) = 12 {(x - 2)}^{2} - 12 (x - 2) + 2$
$∴$ $P^{''} (1) = 12 - 12 (- 1) + 2 = 26$

Suggest Corrections

Similar questions

Let $P (x)$ be a polynomial of degree 4, with $P (2) = 1,$ $P^{'} (2) = P^{''} (2) = 0$ , $P^{'''} (2) = 12$ and $P^{''''} (2) = 24$ . Then the value of $P^{''} (1)$ is

If P(x) be a polynomial of degree 4, with P(2)=-1, P'(2)=0, P”(2)=2, P”'(2)=-12 and P^ir(2) =24, then P”(1) is equal to

Q. Let p(x) be a fourth degree polynomial with coefficient of leading term 1and p(1)=p(2)=p(3)=0,then find the value ofp(0)+p(4).

Q. Let p(x) be a polynomial of degree 4 having extremum at x=1, 2 and

l i m x \to 0 (1 + \frac{p (x)}{x^{2}}) = 2

. Then the value of p(2) is ___

Join BYJU'S Learning Program

Let P(x) be a polynomial of degree 4, with P(2)=−1, P′(2)=0, P′′(2)=2, P′′′(2)=−12 and Piv(2)=24, The value of P′′(1) is

Let us take P(x)=a(x−2)4+b(x−2)3+c(x−2)2+d(x−2)+eThus P(2)=e=−1⇒P′(2)=d=0⇒P′′(2)=2c=2⇒c=1⇒P′′′(2)=6b=−12⇒b=−2and Piv(2)=24a=24⇒a=1⇒P′′(x)=12(x−2)2−12(x−2)+2∴ P′′(1)=12−12(−1)+2=26

Let P(x) be a polynomial of degree 4, with $P (2) = - 1$ , $P^{'} (2) = 0$ , $P^{''} (2) = 2$ , $P^{'''} (2) = - 12$ and $P^{i v} (2) = 24$ , The value of $P^{''} (1)$ is