Question

Let $f\left( x \right) = x - {x^2}$ and g\left( x \right) = \left\\{ \begin{array}{l}\max f\left( t \right),0 \le t \le x,0 \le x \le 1\\\\\sin \pi x,x > 1\end{array} \right., then in the interval $\left[ {0,\left. \infty \right)} \right.$
A) $g\left( x \right)$ is everywhere continuous except at two points
B) $g\left( x \right)$ is everywhere differentiable except at two points
C) $g\left( x \right)$ is everywhere differentiable except at $x = 1$
D) None of these

Answer 1

Here, we have to find the differentiability and continuity of a function. We will use the condition for finding the value of the variable which satisfies the condition of maximal point. Then by using the maximal point we will prove the condition of differentiability and continuity at the given limits of the variable in the given functions to find the continual points and differentiable points.

Complete step by step solution:
Let $f\left( x \right) = x - {x^2}$
Now, substituting $x = t$ in the above equation, we get
$\Rightarrow f\left( t \right) = t - {t^2}$
Now, we will find the differentiation of the function $f\left( t \right)$ to find the maximum of $f\left( t \right)$ .
Differentiating both sides with respect to $t$, we get
$\Rightarrow f'\left( t \right) = 1 - 2t$
Now, to find the maximum of $f\left( t \right)$, we will apply the condition that $f'\left( t \right) = 0$. Therefore, we get
$f'\left( t \right) = 0$
$\Rightarrow 1 - 2t = 0$
By rewriting the equation, we get
$\Rightarrow 2t = 1$
Dividing both side by 2, we get
$\Rightarrow t = \dfrac{1}{2}$
Now, we will differentiate the function $f\left( t \right)$ with respect to $t$, we get
$\Rightarrow f''\left( t \right) = - 2 < 0$
We have a condition that if $f''\left( t \right) < 0$ , then $t$ is the maximum point and if $f''\left( t \right) > 0$ , then $t$ is the minimum point.
Therefore, $t = \dfrac{1}{2}$ is the maximum point.
So, the given function is maximum at $t = \dfrac{1}{2}$.
Now, we will substitute the value of $t$ in the function $f\left( t \right)$. Therefore, we get
$\Rightarrow f\left( {\dfrac{1}{2}} \right) = \dfrac{1}{2} - {\left( {\dfrac{1}{2}} \right)^2}$
Applying the exponent on the terms, we get
$\Rightarrow f\left( {\dfrac{1}{2}} \right) = \dfrac{1}{2} - \dfrac{1}{4}$
On taking LCM, we get
$\Rightarrow f\left( {\dfrac{1}{2}} \right) = \dfrac{1}{2} \times \dfrac{2}{2} - \dfrac{1}{4} = \dfrac{1}{4}$
Now, let us consider the function g\left( x \right) = \left\\{ \begin{array}{l}\max f\left( t \right),0 \le t \le x,0 \le x \le 1\\\\\sin \pi x,x > 1\end{array} \right.
g\left( x \right) = \left\\{ \begin{array}{l}\dfrac{1}{4};0 \le x \le 1\\\\\sin \pi x,x > 1\end{array} \right. at $x = 1$
Now, substituting the limits, we get
$\Rightarrow {\lim _{x \to {1^ - }}}g\left( x \right) = {\lim _{x \to {1^ - }}}\dfrac{1}{4}$
$\Rightarrow {\lim _{x \to {1^ - }}}g\left( x \right) = \dfrac{1}{4}$
Now writing the general equation, we get
$\Rightarrow {\lim _{x \to {1^ + }}}g\left( x \right) = {\lim _{x \to {1^ + }}}\sin \pi x$
$\Rightarrow {\lim _{x \to {1^ + }}}g\left( x \right) = \sin \pi = 0$
So, ${\lim _{x \to {1^ - }}}g\left( x \right) \ne {\lim _{x \to {1^ + }}}g\left( x \right)$
The given function is discontinuous at $x = 1$ and is not differentiable at $x = 1$.
So, at all other points $g\left( x \right)$ is continuous and differentiable.
Therefore, $g\left( x \right)$ is everywhere differentiable except at $x = 1$ in the interval $\left[ {0,\left. \infty \right)} \right.$.

Thus Option(C) is the correct answer.

Note:
We should know the condition of differentiability and continuity. If the right hand derivative is equal to the left hand derivative, then the given function is differentiable at the points. If the right hand derivative is not equal to the left hand derivative, then the given function is not differentiable at the points. If a function is differentiable then the function must be continuous at the same point.