Question

Let $f\left(x\right)=a{x}^3+b{x}^2+cx$ have relative extrema x=1 and at $x=5$.If ${\int }_{-1}^{1}f\left(x\right)dx=6$ then

• A. $a=-1$
• B. $b=9$
• C. $c=15$
• D. $a=1$

Answer 1

Answer: (A), (B)

The correct options are
A $a=-1$
B $b=9$

${\int }_{-1}^{1}f\left(x\right)dx=6\Rightarrow {\int }_{-1}^1({ax}^2+{bx}^2+cx)dx=6$
$\Rightarrow {[\frac{{ax}^4}{4}+\frac{{bx}^3}{3}+\frac{{cx}^2}{2}]}_{-1}^1=6$
$\Rightarrow [\frac{a}{4}+\frac{b}{3}+\frac{c}{2}-\frac{a}{4}+\frac{b}{3}-\frac{c}{2}]=16$
$\Rightarrow \frac{2b}{3}=6\Rightarrow b=9$ ...(1)
$f\left(x\right)={ax}^3+{bx}^2+cx{f}^{\prime }\left(x\right)={3ax}^2+bx+c$
$\therefore f^{\prime }\left(1\right)=0\Rightarrow 3a+bc+c=0$ ...(2)
$f^{\prime }\left(5\right)=0\Rightarrow 15a+5b+c=0$ ...(3)
From (1),(2) and (3)
$a=-1$