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Question

Let f be a real valued function defined as f(x)=x2+x211tf(t) dt+x311f(t) dt. Then which of the following hold(s) good ?

A
11tf(t) dt=1011
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B
f(1)+f(1)=3011
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C
11tf(t) dt>11f(t) dt
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D
f(1)f(1)=2011
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Solution

The correct option is D f(1)f(1)=2011
We have f(x)=x2+ax2+bx3
where a=11tf(t) dt and b=11f(t) dt

Now, a=11t[(a+1)t2+bt3]dt
a=(a+1)11t3 dt+b11t4 dt
a=0+2b10t4dt
a=2b5 (1)

Again b=11((a+1)t2+bt3)dt
b=210(a+1)t2 dt
b=2(a+1)3 (2)

From (1) and (2),
5a2=2(a+1)3
a=411 and b=1011

11tf(t)dt=411 and 11f(t)dt=1011

f(x)=(a+1)x2+bx3
Now, f(1)+f(1)=2(a+1)=3011
And f(1)f(1)=2b=2011

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