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Question

The third derivative of a function f(x) vanishes for all x. If f(0)=1, f'1(1)=2 and f”(1) = – 1, then f(x) is equal to

A
(32)x2+3x+9
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B
(12)x23x+1
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C
(12)x2+3x+1
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D
(32)x27x+2
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Solution

The correct option is C (12)x2+3x+1
Given, f’’’(x) = 0
On integrating
f′′(x)=cf′′(1)=c1=c (given)
f’’(x) = – 1
f’(x) = – x + d
f’(1) = – 1 + d = 2
d = 3
f’(x) = 3 – x
or f(x)=x22+3x+ef(0)=0+0+e=1e=1
Hence, f(x)=x22+3x+1

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