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Question

Let f:RR be given byf(x)=(x1)(x2)(x5). Define F(x)=x0f(t)dt, x>0.
Then which of the following options is/are correct?

A
F has a local minimum at x=1
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B
F has a local maximum at x=2
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C
F has two local maxima and one local minimum in (0,)
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D
F(x)0 for all x(0,5)
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Solution

The correct option is D F(x)0 for all x(0,5)
f(x)=(x1)(x2)(x5)
F(x)=x0f(t)dt,x>0
F(x)=f(x)=x38x2+17x10
F′′(x)=3x216x+17
For x=1, F′′(x)>0
Therefore F has a local minimum at x=1.
For x=2, F′′(x)<0
Therefore F has a local maximum at x=2.
For x=5, F′′(x)>0
Therefore F has a local minimum at x=5.

F(x)=x448x33+17x2210x
From the graph of y=F(x),
we can easily say F(x)0 x(0,5).

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