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Question

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


A

22

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B

24

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C

26

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D

28

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Solution

The correct option is C

26


Let P(x)=ax4+bx3+cx2+dx+e

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

P'(x)=4ax3+3bx2+2cx+d

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

P''(x)=12ax2+6bx+2c

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

P'''(x)=24ax+6b

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

And piv (x) = 24 ⟹ piv (2) =24 a⟹24=24a

a=1,

From Eq. (iv), b=-10

From Eq. (iv), c=37

p' ' (x) =12x2 - 60x+74

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

=- 48+74

=26


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