Question

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

• A. 22
• B. 24
• C. 26
• D. 28

Answer 1

The correct option is C

26

Let P(x)=$a{x}^4+b{x}^3+c{x}^2$+dx+e

P(2)=16a +8b+4c+2d+e=-1 ...(i)

P'(x)=$4a{x}^3+3b{x}^2+2cx+d$

P'(2)=32a+12b+4c+d=0 ...(ii)

P''(x)=12a$x^2$+6bx+2c

P''(2)=48a+12b+2c=2 ...(iii)

$\therefore$ P'''(x)=24ax+6b

$\therefore$ P'''(2)=48a+6b=-12 ...(iv)

And $p^{iv}$ (x) = 24 ⟹ $p^{iv}$ (2) =24 a⟹24=24a

$\therefore$ a=1,

From Eq. (iv), b=-10

From Eq. (iv), c=37

$\therefore$ p' ' (x) =12$x^2$ - 60x+74

$\therefore$ p' ' (1) =12x - 60+74

=- 48+74

=26