Question

At $x=\dots .f\left(x\right)=12x^5-45x^4+40x^3+7$ is maximum

• A. 1
• B. 0
• C. 2
• D. -2

Answer 1

Answer: (A)

1

$f^{\prime }\left(x\right)=60x^4-180x^3+120x^2$
$f^{\prime }\left(x\right)=60x^2(x^2-3x+2)$
$=60x^2(x-2)(x-1)$
$f^{\prime }\left(x\right)>0$ If $x>2$ or $x<1$
$f^{\prime }\left(x\right)<0$ If $1<x<2$
So $f\left(x\right)$ is max at $x=1$