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Question

Let f(x) is a real valued function defined by f(x)=x2+x211tf(t)dt+x311f(t)dt, then


  1. 11tf(t)dt=1011

  2. f(1)+f(1)=3011

  3. 11tf(t)dt>11f(t)dt

  4. f(1)f(1)=2011


Solution

The correct options are
B

f(1)+f(1)=3011


D

f(1)f(1)=2011


f(x)=x2+ax2+bx3=(a+1)x2+bx3a=11tf(t)dt=11(a+1)t3+bt4dt=2b10t4dta=2b5....(i)b=11f(t)dt=11(a+1)t2+bt3dt=(a+1)t3311b=(a+1)23......(ii)
From (i)and (ii)
a=411 and b=1011
f(x)=15x211+10x311f(1)=2511f(1)=511

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