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Question

A function y=f(x) has a second order derivative f"(x)=6(x1). If its graph passes through the point (2,1) and at that point the tangent to the graph is y=3x5, then the function is

A
(x+1)2
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B
(x1)3
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C
(x+1)3
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D
(x1)2
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Solution

The correct option is B (x1)3
f"(x)=6(x1). Integrating,we get
f(x)=3x26x+c
Slope at (2,1)=f(2)=c=3
[ slope of tangent at (2,1) is 3]
f(x)=3x26x+3=3(x1)2
Integrating again, we get f(x)=(x1)3+D
The curve passes through (2,1)
1=(21)3+DD=0
f(x)=(x1)3

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