ISI2015-MMA-69

Question

Consider the function $$f(x) = \begin{cases} \int_0^x \{5+ \mid 1-y \mid \} dy & \text{ if } x>2 \\ 5x+2 & \text{ if } x \leq 2 \end{cases}$$ Then

• A. $f$ is not continuous at $x=2$
• B. $f$ is continuous and differentiable everywhere
• C. $f$ is continuous everywhere but not differentiable at $x=1$
• D. $f$ is continuous everywhere but not differentiable at $x=2$

Answer 1

$$f(x)=\begin{cases}
 \displaystyle{\int_{0}^{x}5+|1-y| \ \ \mathrm{d}y} & \text { if } x>2 \\[2ex]
5 x+{2} & \text { if } x \leq 2
\end{cases}$$

$$f'(x)=\begin{cases}
 5+|1-x| & \text { if } x>2 \\[2ex]
5  & \text { if } x \leq 2
\end{cases}$$
Using Newton-Leibnitz rule for differentiation
  

for checking the  differentiability: $$f'(2^+)=\lim_\limits{h \to 0} 5+|1-(2+h)|=6 $$
And
$$f'(2^-)=5$$
Hence,

the function is not differentiable ($\because f'(2^+)\neq f'(2^-) $) .

Comments

The first task should have been to check whether the function is continuous. Indeed, it is continuous. Hence, D is the correct answer.

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It is not continuous, after solving for x>2 condition we get  $((x^2)/2)+4x+1$  then f($2^+$) =11,

f($2^-$) =$5x+2$=12,

f($2^+$)$≠$f($2^-$)

So it is not continuous.

It would have been continuous if the f($2^-$)=$5x+1$, but it is not

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Function is not even continuous.

Answer 2

$|1-y| =\left\{\begin{matrix} y-1, & y>1\\ 1-y,&y \leq1 \end{matrix}\right.$

We have to evaluate $\int_{0}^x(5 + |1-y|)dy$ $=\int_{0}^1(5 + (1-y))dy + \int_{1}^x(5 + (y-1))dy$

                                                                          $=\int_{0}^1(6-y)dy + \int_{1}^x(4+y)dy$ 

                                                                          $=6\int_{0}^1 dy - \int_{0}^1y$ $dy  + 4\int_{1}^xdy + \int_{1}^xy$ $dy$

                                                                          $=6[$ $y$ $]_0^1 - \left[\dfrac{y^2}{2}\right]_0^1 + 4[$ $x$ $]_1^x + \left[\dfrac{y^2}{2}\right]_1^x$

                                                                          $=(6-0) - \left[\dfrac{1}{2} - 0\right] + [4x -4] + \left[\dfrac{x^2}{2} - \dfrac{1}{2}\right]$ 

                                                                          $= 6 - 4 -1 +4x +\dfrac{x^2}{2}$

                                                                          $=\dfrac{x^2}{2} + 4x + 1$

 Now  $f(x) =\left\{\begin{matrix} \dfrac{x^2}{2} + 4x + 1, & x>2\\ 5x+2,&x \leq2 \end{matrix}\right.$

$ $

Continuity at $x =2$ :-

$f(2) = 5(2) + 2 = 12$

Left hand limit:- $\lim_{x \to 2^-} f(x) =\lim_{x \to 2^-} 5x + 2 = 5(2) + 2 = 12 = f(2)$

Right hand limt:  $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \dfrac{x^2}{2} + 4x + 1 =\dfrac{2^2}{2} + 4(2) + 1 = 2 + 8 + 1 = 11 \neq f(2)$

So $f(x)$ is not continuous at $x=0$, option A is correct one.