    Question

# A cubic polynomial f(x) vanishes at x=−2 and has a relative minimum/maximum at x=−1 and x=13. Then

A
If 11f(x) dx=133, then f(x)=14(5x3+5x25x+5)
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B
If 11f(x) dx=143, then f(x)=x3+x2x+2
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C
If 11f(x) dx=53, then f(x)=16(5x3+5x25x+15)
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D
If 11f(x) dx=103, then f(x)=17(5x3+5x25x+10)
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Solution

## The correct options are B If 1∫−1f(x) dx=143, then f(x)=x3+x2−x+2 D If ∫1−1f(x) dx=103, then f(x)=17(5x3+5x2−5x+10)As the function has relative minimum/maximum at x=−1 and x=13 So, f′(x)=a(x+1)(x−13) where a is constant. ⇒f′(x)=a(x2+2x3−13) ⇒f(x)=a(x33+x23−x3)+b where b is constant of integration. Now f(−2)=0 ⇒−8a3+4a3+2a3+b=0 ⇒b=2a3 ⇒f(x)=a3(x3+x2−x+2) When 1∫−1f(x) dx=143 ⇒a31∫−1(x3+x2−x+2) dx=143 ⇒a1∫−1(x2+2)=14 ⇒2a1∫0(x2+2)=14 ⇒2a(13+2)=14 ⇒143a=14⇒a=3 So the function will be f(x)=x3+x2−x+2 Similarly, when 1∫−1f(x) dx=103 ⇒143a=10 ⇒a=157 So the function will be f(x)=17(5x3+5x2−5x+10) Similarly, remaining two options can be found to be incorrect.  Suggest Corrections  1      Similar questions  Related Videos   Extrema
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