Mathematics

Let $$\displaystyle f(x)=ax^{3}+bx^{2}+cx$$ have relative extrema x=1 and at $$\displaystyle x=5$$.If $$\displaystyle \int_{-1}^{1}f(x)dx=6$$ then


ANSWER

$$a=-1$$

$$b=9$$


SOLUTION
$$\displaystyle \int _{ -1 }^{ 1 }{ f\left( x \right)  } dx=6\Rightarrow \int _{ -1 }^{ 1 }{ \left( { ax }^{ 2 }+{ bx }^{ 2 }+cx \right)  } dx=6$$
$$\displaystyle \Rightarrow \left[ \frac { { ax }^{ 4 } }{ 4 } +\frac { { bx }^{ 3 } }{ 3 } +\frac { { cx }^{ 2 } }{ 2 }  \right] _{ -1 }^{ 1 }{ =6 }$$
$$\displaystyle \Rightarrow \left[ \frac { a }{ 4 } +\frac { b }{ 3 } +\frac { c }{ 2 } -\frac { a }{ 4 } +\frac { b }{ 3 } -\frac { c }{ 2 }  \right] =16$$
$$\displaystyle \Rightarrow \frac { 2b }{ 3 } =6\Rightarrow b=9$$  ...(1)
$$f\left( x \right) ={ ax }^{ 3 }+{ bx }^{ 2 }+cx\\ f'\left( x \right) ={ 3ax }^{ 2 }+bx+c$$
$$\therefore f'\left( 1 \right) =0\Rightarrow 3a+bc+c=0$$   ...(2)
$$f'\left( 5 \right) =0\Rightarrow 15a+5b+c=0$$   ...(3)
From (1),(2) and (3)
$$a=-1$$
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Multiple Correct Medium Published on 17th 09, 2020
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