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Question

Let f:RR be a function such that f(x)=x3+x2f(1)+xf′′(2)+f′′′(3),xR. Then f(2) equals :

A
4
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B
30
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C
8
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D
2
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Solution

The correct option is D 2
f(x)=x3+x2f(1)+xf′′(2)+f′′′(3)
f(x)=3x2+2xf(1)+f′′(2) (i)
f′′(x)=6x+2f(1) (ii)
f′′′(x)=6 (iii)
​​​​​​f′′′(3)=6
Now from eqn (i)
f(1)=3+2f(1)+f′′(2)
f(1)+f′′(2)+3=0 (iv)
from eqn (ii)
f′′(2)=12+2f(1)
2f(1)f′′(2)+12=0 (v)
from eqn. (iv) and (v)
3f(1)+15=0
f(1)=5 and f′′(2)=2
So,
f(x)=x35x2+2x+6
f(2)=820+4+6=2

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