Limit Continuity And Differentiability Ques 55
- Let $f(x)=x^3-x^2-x+1$
and $\quad g(x)\left\{\begin{array}{l}=\max \{f(t) ; 0 \leq t \leq x\}, 0 \leq x \leq 1 \\ =3-x, 1<x \leq 2\end{array}\right.$
Discuss the continuity and differentiability of the function $g(x)$ in the interval $(0,2)$.
(1985, 5M)
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Answer:
Correct Answer: 55.$g(x)$ is continuous for all $x \in(0,2)-\{1]$ and
$g(x)$ is differentiable for all $x \in(0,2)-\{1\}$
Solution: Given, $f(x)=x^3-x^2-x+1$
$\Rightarrow \quad f^{\prime}(x)=3 x^2-2 x-1=(3 x+1)(x-1)$
$\therefore \quad f(x)$ is increasing for $x \in(-\infty,-1 / 3) \cup(1, \infty)$
and decreasing for $x \in(-1 / 3,1)$
Also, given $g(x)=\left\{\begin{array}{cc}\max \{f(t) ; 0 \leq t \leq x\}, & 0 \leq x \leq 1 \\ 3-x, & 1<x \leq 2\end{array}\right.$
$\Rightarrow \quad g(x)= \begin{cases}f(x), & 0 \leq x \leq 1 \\ 3-x, & 1<x \leq 2\end{cases}$
$\Rightarrow \quad g(x)=\left\{\begin{array}{cc}x^3-x^2-x+1, & 0 \leq x \leq 1 \\ 3-x, & 1<x \leq 2\end{array}\right.$
At $x=1$,
$ \mathrm{RHL}=\lim _{x \rightarrow 1}(3-x)=2 $
and $\quad \mathrm{LHL}=\lim _{x \rightarrow 1} \left(x^3-x^2-x+1\right)=0$
$\therefore \quad$ It is discontinuous at $x=1$.
Also, $\quad g^{\prime}(x)=\left\{\begin{array}{cc}3 x^2-2 x-1, & 0 \leq x \leq 1 \\ -1, & 1<x \leq 2\end{array}\right.$
$\Rightarrow \quad g^{\prime}\left(1^{+}\right)=-1$
and $g^{\prime}\left(1^{-}\right)=3-2-1=0$
$\therefore \quad g(x)$ is continuous for all $x \in(0,2)-\{1\}$ and
$g(x)$ is differentiable for all $x \in(0,2)-\{1\}$.
BETA
