Applications of Derivatives
Cubic polynomial; inverse trigonometric sum
MJMT_Full_Test_10
Grade 12
Question:
Let $f(x)$ be a cubic polynomial on $\mathbb{R}$ which increases on $(-\infty,0)$ and $(1,\infty)$, decreases on $(0,1)$. If $f'(2)=6$ and $f(2)=2$, then $\tan^{-1}(f(1)) + \tan^{-1}\!\left(f\!\left(\frac{3}{2}\right)\right) + \tan^{-1}(f(0))$ is equal to
$\tan^{-1}2$
$\cot^{-1}2$
$-\tan^{-1}2$
$-\cot^{-1}2$
Step-by-Step Solution
Key Concept: Determine $f(x)$ from critical points at $x=0,1$. Write $f'(x)=ax(x-1)$, use $f'(2)=6$ and $f(2)=2$ to find $a$ and constant.
$f(x)=x^2(x-3/2)$. Values: $f(1)=-1/2$, $f(3/2)=0$, $f(0)=0$. Sum $= -\tan^{-1}(1/2) = -\cot^{-1}2$.
Correct Answer: 4