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Question

Let f(x) is a real valued function defined by
f(x)=x2+x211t.f(t)dt+x311f(t)dt, then which of the following hold(s) good?

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

The correct options are
B f(1)+f(1)=3011
D f(1)f(1)=2011
Let
11tf(t)dt=a, and 11f(t)dt=b
f(x)=x2+ax2+bx3
f(t)=t2(1+a)+bt3
tf(t)=t3(1+a)+bt4
Now
a=11tf(t)dt=11(t3(1+a)+bt4)dt
=2[bt55]10=2b5 ...(1)
(using property of integral of odd and even function).

Similarly b=11f(t)dt=11(t2(1+a)+bt3)dt
=2[(1+a)t33]10=23(1+a) ....(2)
Solving the equations
a=2b5 and b=2(1+a)3 simultaneously, we get
a=411 and b=1011
Hence, options A and C are incorrect.
Substituting x=1 in the original equation we get,
f(1)=1+a+b ..(3)
Similarly, substituting x=1, we get
f(1)=1+ab ...(4)
Adding (3) and (4) we get ,
f(1)+f(1)=2(1+a)=2×1511=3011
Similarly subtracting (4) from (3) we get,
f(1)f(1)=2b=2011

Hence, options B and D are correct.

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