Q92 / 180 · KCET 2024 · Mathematics
The value of $C$ in $(0,2)$ satisfying the mean value theorem for the function
$f(x)=x(x-1)^2, x \in[0,2]$ is equal to
- A$3 / 4$
- Correct answerB$4 / 3$
- C$1 / 3$
- D$2 / 3$
Solution
$f(x)=x(x-1)^2 ; x \in[0,2]$
$f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} ; f(2)=2, f(0)=0$
$\therefore f^{\prime}(x)=(x-1)^2+2 x(x-1)$
$=x^2+1-2 x+2 x^2-2 x$
$=3 x^2-4 x+1$
So, $f^{\prime}(c)=3 c^2-4 c+1$
Thus, $3 c^2-4 c+1=\frac{f(2)-(0)}{2-0}=\frac{2-0}{2-0}=1$
$\Rightarrow \quad c(3 c-4)=0$
Hence, $c=\frac{4}{3}$
Hence, the correct answer is option: 'B'.