Question 1

Prove that the function $f(x) = 5x - 3$ is continuous at $x = 0$, at $x = -3$ and at $x = 5$.

Accepted Answer

To prove that $f(x) = 5x - 3$ is continuous at $x = a$, we need to show that $\lim_{x \to a} f(x) = f(a)$.

For $x = 0$:
$f(0) = 5(0) - 3 = -3$.
$\lim_{x \to 0} f(x) = \lim_{x \to 0} (5x - 3) = 5(0) - 3 = -3$.
Since the limit equals the function value, $f$ is continuous at $x=0$.

For $x = -3$:
$f(-3) = 5(-3) - 3 = -15 - 3 = -18$.
$\lim_{x \to -3} f(x) = 5(-3) - 3 = -18$.
Hence continuous at $x = -3$.

For $x = 5$:
$f(5) = 5(5) - 3 = 25 - 3 = 22$.
$\lim_{x \to 5} f(x) = 5(5) - 3 = 22$.
Hence continuous at $x = 5$.

Therefore, $f(x) = 5x - 3$ is continuous at $x=0, -3, 5$. — The function $f(x) = 5x - 3$ is a polynomial, which is continuous everywhere. To verify continuity at specific points, we check if $\lim_{x \to a} f(x) = f(a)$. Substituting the values confirms the equality, hence continuity at those points.

Question 2

Examine the continuity of the function $f(x) = 2x^{2} - 1$ at $x = 3$.

Accepted Answer

We check if $\lim_{x \to 3} f(x) = f(3)$.

Calculate $f(3) = 2(3)^2 - 1 = 2(9) - 1 = 18 - 1 = 17$.

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

Since the limit equals the function value, $f$ is continuous at $x = 3$. — The function $f(x) = 2x^2 - 1$ is a polynomial, continuous everywhere. Checking the limit and function value at $x=3$ confirms continuity.

Question 3

Examine the following functions for continuity.

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

(b) $f(x) = \frac{1}{x - 5}, x \neq 5$

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

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

Accepted Answer

(a) $f(x) = x - 5$ is a polynomial function, continuous everywhere.

(b) $f(x) = \frac{1}{x - 5}$ is continuous for all $x \neq 5$. At $x=5$, the function is not defined, so discontinuous there.

(c) $f(x) = \frac{x^2 - 25}{x + 5} = \frac{(x-5)(x+5)}{x+5}$ for $x \neq -5$.
Simplifies to $x - 5$ for $x \neq -5$.
At $x = -5$, function is not defined, so discontinuous at $x = -5$.

(d) $f(x) = |x - 5|$ is continuous everywhere as absolute value functions are continuous. — Polynomials and absolute value functions are continuous everywhere. Rational functions are continuous where denominator is non-zero. Points where denominator is zero are points of discontinuity.

Question 4

Prove that the function $f(x) = x^{n}$ is continuous at $x = n$, where $n$ is a positive integer.

Accepted Answer

Since $f(x) = x^n$ is a polynomial function, it is continuous everywhere.

To prove continuity at $x = n$, check if $\lim_{x \to n} f(x) = f(n)$.

$f(n) = n^n$.

$\lim_{x \to n} x^n = n^n$ by direct substitution.

Hence, $f$ is continuous at $x = n$. — Polynomial functions are continuous everywhere. The limit at any point equals the function value at that point.

Question 5

Is the function $f$ defined by

$$
f(x) = \begin{cases} x, & \text{if } x \leq 1 \ 5, & \text{if } x > 1 \end{cases}
$$

continuous at $x = 0$? At $x = 1$? At $x = 2$?

Accepted Answer

Check continuity at each point:

At $x=0$:
$f(0) = 0$ (since $0 \leq 1$)
$\lim_{x \to 0^-} f(x) = 0$
$\lim_{x \to 0^+} f(x) = 0$ (since for $x > 1$ is not applicable near 0)
So, limit exists and equals $f(0)$. Continuous at $x=0$.

At $x=1$:
$f(1) = 1$
$\lim_{x \to 1^-} f(x) = 1$
$\lim_{x \to 1^+} f(x) = 5$
Left and right limits not equal, so discontinuous at $x=1$.

At $x=2$:
$f(2) = 5$
$\lim_{x \to 2^-} f(x) = 5$
$\lim_{x \to 2^+} f(x) = 5$
Continuous at $x=2$. — Continuity requires left and right limits to be equal and equal to function value. At $x=1$, left and right limits differ, so discontinuity occurs.

Question 6

Find all points of discontinuity of $f$, where $f$ is defined by

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

Accepted Answer

Check continuity at $x=2$:

$f(2) = 2(2) + 3 = 7$

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

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

Since left and right limits differ, $f$ is discontinuous at $x=2$.

No other points of discontinuity as both parts are linear functions continuous on their domains. — Discontinuity occurs where left and right limits differ or function is not defined. Here, only at $x=2$.

Question 7

Find all points of discontinuity of $f$, where $f$ is defined by

$$
f(x) = \begin{cases} |x| + 3, & \text{if } x \leq -3 \ -2x, & -3 < x < 3 \ 6x + 2, & x \geq 3 \end{cases}
$$

Accepted Answer

Check continuity at $x = -3$:

$f(-3) = |-3| + 3 = 3 + 3 = 6$

Left limit:
$\lim_{x \to -3^-} f(x) = |-3| + 3 = 6$

Right limit:
$\lim_{x \to -3^+} f(x) = -2(-3) = 6$

Limits equal and equal to function value, continuous at $x = -3$.

Check continuity at $x = 3$:

$f(3) = 6(3) + 2 = 18 + 2 = 20$

Left limit:
$\lim_{x \to 3^-} f(x) = -2(3) = -6$

Right limit:
$\lim_{x \to 3^+} f(x) = 20$

Left and right limits differ, so discontinuous at $x = 3$.

No other points of discontinuity. — Continuity requires matching left and right limits and function value. Discontinuity at $x=3$ due to mismatch.

Question 8

Find all points of discontinuity of $f$, where $f$ is defined by

$$
f(x) = \begin{cases} \frac{|x|}{x}, & x \neq 0 \ 0, & x = 0 \end{cases}
$$

Accepted Answer

Check continuity at $x=0$:

$f(0) = 0$

Left limit:
$\lim_{x \to 0^-} \frac{|x|}{x} = \lim_{x \to 0^-} \frac{-x}{x} = -1$

Right limit:
$\lim_{x \to 0^+} \frac{|x|}{x} = \lim_{x \to 0^+} \frac{x}{x} = 1$

Left and right limits differ, so discontinuous at $x=0$.

For $x \neq 0$, function is continuous. — Function jumps from -1 to 1 at zero, but is defined as 0 at zero, causing discontinuity.

Question 9

Find all points of discontinuity of $f$, where $f$ is defined by

$$
f(x) = \begin{cases} \frac{x}{|x|}, & x < 0 \ -1, & x \geq 0 \end{cases}
$$

Accepted Answer

Check continuity at $x=0$:

$f(0) = -1$

Left limit:
$\lim_{x \to 0^-} \frac{x}{|x|} = \lim_{x \to 0^-} \frac{x}{-x} = -1$

Right limit:
$\lim_{x \to 0^+} f(x) = -1$

Left and right limits equal and equal to function value, so continuous at $x=0$.

For $x < 0$, $f(x) = -1$, continuous.

For $x \geq 0$, $f(x) = -1$, continuous.

Hence, $f$ is continuous everywhere. — Function is constant -1 on both sides including at zero, so continuous everywhere.

Question 10

Find all points of discontinuity of $f$, where $f$ is defined by

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

Accepted Answer

Check continuity at $x=1$:

$f(1) = 1 + 1 = 2$

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

Right limit:
$\lim_{x \to 1^+} f(x) = 1 + 1 = 2$

Limits equal and equal to function value, so continuous at $x=1$.

Function is continuous elsewhere as both parts are continuous on their domains. — Continuity at boundary point verified by matching limits and function value.

Question 11

Find all points of discontinuity of $f$, where $f$ is defined by

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

Accepted Answer

Check continuity at $x=2$:

$f(2) = 2^3 - 3 = 8 - 3 = 5$

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

Right limit:
$\lim_{x \to 2^+} f(x) = 2^2 + 1 = 4 + 1 = 5$

Limits equal and equal to function value, so continuous at $x=2$.

Function continuous elsewhere as both parts are polynomials. — Continuity at boundary point verified by matching limits and function value.

Question 12

Find all points of discontinuity of $f$, where $f$ is defined by

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

Accepted Answer

Check continuity at $x=1$:

$f(1) = 1^{10} - 1 = 1 - 1 = 0$

Left limit:
$\lim_{x \to 1^-} f(x) = 0$

Right limit:
$\lim_{x \to 1^+} f(x) = 1^2 = 1$

Left and right limits differ, so discontinuous at $x=1$.

Function continuous elsewhere as both parts are polynomials. — Discontinuity at $x=1$ due to mismatch of left and right limits.

Question 13

Is the function defined by

$$
f(x) = \begin{cases} x + 5, & x \leq 1 \ x - 5, & x > 1 \end{cases}
$$
a continuous function?

Accepted Answer

Check continuity at $x=1$:

$f(1) = 1 + 5 = 6$

Left limit:
$\lim_{x \to 1^-} f(x) = 6$

Right limit:
$\lim_{x \to 1^+} f(x) = 1 - 5 = -4$

Since left and right limits differ, function is discontinuous at $x=1$.

Hence, $f$ is not continuous. — Discontinuity at $x=1$ due to jump in function values.

Question 14

Discuss the continuity of the function $f$, where $f$ is defined by

$$
f(x) = \begin{cases} 3, & 0 \leq x \leq 1 \ 4, & 1 < x < 3 \ 5, & 3 \leq x \leq 10 \end{cases}
$$

Accepted Answer

Check continuity at $x=1$:

$f(1) = 3$

Left limit:
$\lim_{x \to 1^-} f(x) = 3$

Right limit:
$\lim_{x \to 1^+} f(x) = 4$

Discontinuous at $x=1$.

Check continuity at $x=3$:

$f(3) = 5$

Left limit:
$\lim_{x \to 3^-} f(x) = 4$

Right limit:
$\lim_{x \to 3^+} f(x) = 5$

Discontinuous at $x=3$.

Function is constant on intervals, continuous there. — Discontinuities at points where function value jumps between intervals.

Question 15

Discuss the continuity of the function $f$, where $f$ is defined by

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

Accepted Answer

Check continuity at $x=0$:

$f(0) = 0$

Left limit:
$\lim_{x \to 0^-} 2x = 0$

Right limit:
$\lim_{x \to 0^+} 0 = 0$

Continuous at $x=0$.

Check continuity at $x=1$:

$f(1) = 0$

Left limit:
$\lim_{x \to 1^-} 0 = 0$

Right limit:
$\lim_{x \to 1^+} 4(1) = 4$

Discontinuous at $x=1$.

Continuous elsewhere as functions are continuous on their domains. — Continuity at boundary points requires matching limits and function value; discontinuity at $x=1$ due to jump.

Continuity and Differentiability

Continuity and Differentiability — Study Notes

Continuity of a Function

Continuity of Composite Functions

Differentiability

Practice Questions — Continuity and Differentiability

All 6 Chapters in Mathematics Part-I