Continuity and Differentiability — NCERT Solutions

For any real number $c$,

$$\lim_{x\to c}(5x-3)=5c-3=f(c).$$

So $f(x)=5x-3$ is continuous at every real number. Hence it is continuous at $x=0$, $x=-3$ and $x=5$.

At $x=3$,

$$f(3)=2(3)^2-1=18-1=17.$$

Also,

$$\lim_{x\to 3}(2x^2-1)=2(3)^2-1=17.$$

Since $\lim_{x\to 3}f(x)=f(3)$, the function is continuous at $x=3$.

The function is a polynomial:

$$f(x)=x-5.$$

Every polynomial function is continuous at every real number. So $f$ is continuous everywhere.

The function is

$$f(x)=\frac{1}{x-5},\quad x\ne 5.$$

This is a rational function, so it is continuous wherever it is defined. Since it is not defined at $x=5$, it is continuous for all $x\ne 5$.

For $x\ne -5$,

$$f(x)=\frac{x^2-25}{x+5}=\frac{(x-5)(x+5)}{x+5}=x-5.$$

So the function agrees with the polynomial $x-5$ wherever it is defined. Hence it is continuous for all $x\ne -5$.

We know

$$f(x)=|x-5|$$

and the modulus function is continuous everywhere. Since $x-5$ is a polynomial and continuous everywhere, $|x-5|$ is also continuous everywhere.

Let $p(x)=x^n$, where $n$ is a positive integer. A polynomial function is continuous at every real number. Therefore $p(x)=x^n$ is continuous at $x=n$.

For $x\le 1$, $f(x)=x$ and for $x>1$, $f(x)=5$.

• At $x=0$, the function is given by $f(x)=x$, so $f(0)=0$ and

$$\lim_{x\to 0}f(x)=\lim_{x\to 0}x=0=f(0).$$
Hence it is continuous at $x=0$.

• At $x=1$,

$$f(1)=1,$$
but
$$\lim_{x\to 1^-}f(x)=\lim_{x\to 1^-}x=1, \qquad \lim_{x\to 1^+}f(x)=\lim_{x\to 1^+}5=5.$$
Since the left and right limits are not equal, the function is discontinuous at $x=1$.

• At $x=2$, since $2>1$, $f(x)=5$ in a neighbourhood of 2. Thus

$$\lim_{x\to 2}f(x)=5=f(2),$$
so it is continuous at $x=2$.

So the function is continuous at $x=0$ and $x=2$, and discontinuous at $x=1$.

Given

$$f(x)=\begin{cases}2x+3,& x\le 2\\2x-3,& x>2\end{cases}$$

For $x<2$, $f$ is linear, so continuous. For $x>2$, it is also linear, so continuous.

Check $x=2$:

• Left value: $f(2)=2(2)+3=7$
• Right hand limit: $\lim_{x\to 2^+}(2x-3)=1$

Since these are not equal, the function is discontinuous at $x=2$.

So the only point of discontinuity is $x=2$.

The function is

$$f(x)=\begin{cases}|x|+3,& x\le -3\\-2x,& -3<x<3\\6x+2,& x\ge 3\end{cases}$$

Each branch is continuous on its interval. So only the boundary points need checking.

At $x=-3$:

• $f(-3)=|-3|+3=6$
• Right hand limit from $-2x$ is $-2(-3)=6$

So it is continuous at $x=-3$.

At $x=3$:

• Left hand limit from $-2x$ is $-2(3)=-6$
• $f(3)=6(3)+2=20$

These do not match, so it is discontinuous at $x=3$.

Hence the only point of discontinuity is $x=3$.

For $x>0$, $\frac{|x|}{x}=1$; for $x<0$, $\frac{|x|}{x}=-1$.

At $x=0$, the function value is defined as $0$.

The left hand limit at $0$ is $-1$ and the right hand limit at $0$ is $1$, so the limits are not equal. Therefore the function is discontinuous at $x=0$.

For $x<0$,

$$\frac{x}{|x|}=\frac{x}{-x}=-1.$$

For $x\ge 0$, the function is also $-1$.

So the function is constant equal to $-1$ on its whole domain, hence continuous wherever it is defined. Since it is defined for all real numbers, it is continuous everywhere.

The function is

$$f(x)=\begin{cases}x+1,& x\ge 1\\x^2+1,& x<1\end{cases}$$

Each branch is continuous on its interval. Check the joining point $x=1$:

• Left hand limit: $\lim_{x\to 1^-}(x^2+1)=2$
• Right hand limit: $\lim_{x\to 1^+}(x+1)=2$
• Value: $f(1)=1+1=2$

Since all three are equal, the function is continuous at $x=1$ and hence everywhere.

The function is

$$f(x)=\begin{cases}x^3-3,& x\le 2\\x^2+1,& x>2\end{cases}$$

Each branch is continuous on its interval. Check $x=2$:

• Left value: $f(2)=2^3-3=5$
• Right hand limit: $\lim_{x\to 2^+}(x^2+1)=5$

They are equal, so the function is continuous at $x=2$.

Therefore the function is continuous everywhere.

The function is

$$f(x)=\begin{cases}x^{10}-1,& x\le 1\\x^2,& x>1\end{cases}$$

Each piece is continuous on its interval. Check at $x=1$:

• Left value: $1^{10}-1=0$
• Right hand limit: $\lim_{x\to 1^+}x^2=1$

Since these are not equal, the function is not continuous at $x=1$.

So it is continuous at all points except $x=1$.

We check the point where the definition changes, i.e. $x=1$.

For $x\le 1$, $f(x)=x+5$, so
$$f(1)=1+5=6.$$
Now,
$$\lim_{x\to 1^-}f(x)=\lim_{x\to 1^-}(x+5)=6,$$
and
$$\lim_{x\to 1^+}f(x)=\lim_{x\to 1^+}(x-5)=-4.$$
These two limits are not equal, so the function is not continuous at $x=1$.

Everywhere else each part is a linear function, so it is continuous there. Therefore the function is not a continuous function.

The function is
$$f(x)=\begin{cases} 3, & 0\le x\le 1\\ 4, & 1<x<3\\ 5, & 3\le x\le 10 \end{cases}$$
Its domain is $[0,10]$.

• On each open interval where it is constant, the function is continuous.
• At $x=0$, continuity is checked by the right-hand limit:

$$\lim_{x\to 0^+}f(x)=3=f(0).$$

• At $x=10$, continuity is checked by the left-hand limit:

$$\lim_{x\to 10^-}f(x)=5=f(10).$$

• At $x=1$,

$$f(1)=3,\quad \lim_{x\to 1^-}f(x)=3,\quad \lim_{x\to 1^+}f(x)=4.$$
Since the one-sided limits are different, it is discontinuous at $x=1$.

• At $x=3$,

$$\lim_{x\to 3^-}f(x)=4,\quad f(3)=5,\quad \lim_{x\to 3^+}f(x)=5.$$
Since the left-hand limit is not equal to the value, it is discontinuous at $x=3$.

Hence the function is continuous everywhere in its domain except at $x=1$ and $x=3$.

The function is
$$f(x)=\begin{cases} 2x, & x<0\\ 0, & 0\le x\le 1\\ 4x, & x>1 \end{cases}$$

• For $x<0$, $f(x)=2x$, a linear function, so it is continuous there.
• For $0<x<1$, $f(x)=0$, a constant function, so it is continuous there.
• For $x>1$, $f(x)=4x$, a linear function, so it is continuous there.

Now check the junctions.

At $x=0$:
$$\lim_{x\to 0^-}f(x)=\lim_{x\to 0^-}2x=0, \qquad \lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}0=0,$$
and
$$f(0)=0.$$
So the function is continuous at $x=0$.

At $x=1$:
$$f(1)=0, \qquad \lim_{x\to 1^-}f(x)=0, \qquad \lim_{x\to 1^+}f(x)=\lim_{x\to 1^+}4x=4.$$
Since the left and right limits are different, the function is discontinuous at $x=1$.

So the function is continuous for all $x\neq 1$ and discontinuous at $x=1$.

The function is
$$f(x)=\begin{cases} -2, & x\le -1\\ 2x, & -1<x\le 1\\ 2, & x>1 \end{cases}$$

• For $x<-1$, it is constant, so continuous.
• For $-1<x<1$, it is linear, so continuous.
• For $x>1$, it is constant, so continuous.

Now check the boundary points.

At $x=-1$:
$$f(-1)=-2, \qquad \lim_{x\to -1^-}f(x)=-2, \qquad \lim_{x\to -1^+}f(x)=\lim_{x\to -1^+}2x=-2.$$
Hence the function is continuous at $x=-1$.

At $x=1$:
$$f(1)=2, \qquad \lim_{x\to 1^-}f(x)=\lim_{x\to 1^-}2x=2, \qquad \lim_{x\to 1^+}f(x)=2.$$
Hence the function is continuous at $x=1$.

So, in fact, the function is continuous at all points in its domain, i.e. everywhere.

For continuity at $x=3$, the left-hand value and right-hand limit must be equal.

From the first part,
$$f(3)=3a+1.$$
From the second part,
$$\lim_{x\to 3^+}f(x)=3b+3.$$
Set them equal:
$$3a+1=3b+3.$$
So
$$3a-3b=2 \quad\Rightarrow\quad 3(a-b)=2 \quad\Rightarrow\quad a-b=\frac{2}{3}.$$
Thus the relationship is
$$\boxed{a-b=\frac{2}{3}}.$$