Comprehensive solutions for Class 12 Maths (NCERT) Exercise 5.1 on Continuity and Differentiability. Learn to prove continuity, find points of discontinuity, and determine unknown constants ($k, a, b$) for continuous piecewise functions.
1. Function $f(x) = 5x – 3$
This is a polynomial function, and all polynomial functions are continuous everywhere.
- At $x = 0$:
- $f(0) = 5(0) – 3 = -3$
- $\lim_{x \to 0} f(x) = \lim_{x \to 0} (5x – 3) = 5(0) – 3 = -3$
- Since $\lim_{x \to 0} f(x) = f(0)$, $f(x)$ is continuous at $x=0$.
- At $x = -3$:
- $f(-3) = 5(-3) – 3 = -15 – 3 = -18$
- $\lim_{x \to -3} f(x) = \lim_{x \to -3} (5x – 3) = 5(-3) – 3 = -18$
- Since $\lim_{x \to -3} f(x) = f(-3)$, $f(x)$ is continuous at $x=-3$.
- At $x = 5$:
- $f(5) = 5(5) – 3 = 25 – 3 = 22$
- $\lim_{x \to 5} f(x) = \lim_{x \to 5} (5x – 3) = 5(5) – 3 = 22$
- Since $\lim_{x \to 5} f(x) = f(5)$, $f(x)$ is continuous at $x=5$.
2. Function $f(x) = 2x^2 – 1$ at $x = 3$
This is also a polynomial function, so it must be continuous at $x=3$.
- Function value: $f(3) = 2(3)^2 – 1 = 2(9) – 1 = 18 – 1 = 17$
- Limit: $\lim_{x \to 3} f(x) = \lim_{x \to 3} (2x^2 – 1) = 2(3)^2 – 1 = 17$
- Since $\lim_{x \to 3} f(x) = f(3)$, $f(x)$ is continuous at $x=3$.
3. Examine the following functions for continuity
(a) $f(x) = x – 5$
This is a polynomial function, which is continuous for all $x \in \mathbb{R}$.
(b) $f(x) = \frac{1}{x – 5}, x \neq 5$
This is a rational function. It is continuous at every point in its domain. The function is discontinuous only where the denominator is zero, i.e., $x-5 = 0 \Rightarrow x=5$. Since the domain is $x \neq 5$, the function is continuous for all $x \in \mathbb{R}$ except $x = 5$.
(c) $f(x) = \frac{x^2 – 25}{x + 5}, x \neq -5$
This is a rational function. It is defined and continuous for all $x \in \mathbb{R}$ except where the denominator is zero, i.e., $x+5 = 0 \Rightarrow x=-5$. The function is continuous for all $x \in \mathbb{R}$ except $x = -5$.
(Note: For $x \neq -5$, $f(x) = \frac{(x-5)(x+5)}{x+5} = x-5$. This is a line with a removable discontinuity at $x=-5$).
(d) $f(x) = |x – 5|$
This is a modulus function (absolute value function). The absolute value function, $g(x) = |x|$, is continuous everywhere. Since $f(x)$ is a composition of two continuous functions: $g(y)=|y|$ and $y=h(x)=x-5$, $f(x)$ is continuous for all $x \in \mathbb{R}$.
4. Function $f(x) = x^n$ at $x = n$ (n is a positive integer)
- Function value: $f(n) = n^n$
- Limit: $\lim_{x \to n} f(x) = \lim_{x \to n} x^n = n^n$
- Since $\lim_{x \to n} f(x) = f(n)$, the function $f(x) = x^n$ is continuous at $x = n$.(In general, $f(x) = x^n$ is a polynomial and is continuous for all $x \in \mathbb{R}$).
5. Function $f(x) = \begin{cases} x, & \text{if } x \leq 1 \\ 5, & \text{if } x > 1 \end{cases}$
At $x = 0$ (where $x \leq 1$):
- $f(0) = 0$
- $\lim_{x \to 0} f(x) = \lim_{x \to 0} x = 0$
- Continuous at $x=0$.
At $x = 2$ (where $x > 1$):
- $f(2) = 5$
- $\lim_{x \to 2} f(x) = \lim_{x \to 2} 5 = 5$
- Continuous at $x=2$.
At $x = 1$ (The boundary point):
- Function value: $f(1) = 1$ (using $x \leq 1$)
- LHL: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x = 1$
- RHL: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 5 = 5$
- Since $\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)$, $\lim_{x \to 1} f(x)$ does not exist.
- The function is discontinuous at $x=1$.
6. Function $f(x) = \begin{cases} 2x + 3, & \text{if } x \leq 2 \\ 2x – 3, & \text{if } x > 2 \end{cases}$
The boundary point to check is $x=2$.
- Function value: $f(2) = 2(2) + 3 = 7$ (using $x \leq 2$)
- LHL: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (2x + 3) = 2(2) + 3 = 7$
- RHL: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (2x – 3) = 2(2) – 3 = 4 – 3 = 1$
- Since $\lim_{x \to 2^-} f(x) \neq \lim_{x \to 2^+} f(x)$, the function is discontinuous at $x = 2$.
- For $x < 2$ and $x > 2$, $f(x)$ is a polynomial, thus continuous.
- Point of discontinuity: $x = 2$.
7. Function $f(x) = \begin{cases} |x| + 3, & \text{if } x \leq -3 \\ -2x, & \text{if } -3 < x < 3 \\ 6x + 2, & \text{if } x \geq 3 \end{cases}$
We must check the boundary points $x = -3$ and $x = 3$.
At $x = -3$
- Function value: $f(-3) = |-3| + 3 = 3 + 3 = 6$ (using $x \leq -3$)
- LHL: $\lim_{x \to -3^-} f(x) = \lim_{x \to -3^-} (|x| + 3) = |-3| + 3 = 6$
- RHL: $\lim_{x \to -3^+} f(x) = \lim_{x \to -3^+} (-2x) = -2(-3) = 6$
- Since $\lim_{x \to -3} f(x) = f(-3) = 6$, $f(x)$ is continuous at $x = -3$.
At $x = 3$
- Function value: $f(3) = 6(3) + 2 = 18 + 2 = 20$ (using $x \geq 3$)
- LHL: $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (-2x) = -2(3) = -6$
- RHL: $\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (6x + 2) = 6(3) + 2 = 20$
- Since $\lim_{x \to 3^-} f(x) \neq \lim_{x \to 3^+} f(x)$, the function is discontinuous at $x = 3$.
- Point of discontinuity: $x = 3$.
8. Function $f(x) = \begin{cases} \frac{|x|}{x}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$
We check the boundary point $x=0$.
Recall: $|x| = \begin{cases} x, & x \geq 0 \\ -x, & x < 0 \end{cases}$.
- Function value: $f(0) = 0$
- LHL: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{-x}{x} = \lim_{x \to 0^-} (-1) = -1$
- RHL: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{x}{x} = \lim_{x \to 0^+} (1) = 1$
- Since $\lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)$, the limit $\lim_{x \to 0} f(x)$ does not exist.
- Point of discontinuity: $x = 0$.
9. Function $f(x) = \begin{cases} \frac{x}{|x|}, & \text{if } x < 0 \\ -1, & \text{if } x \geq 0 \end{cases}$
We check the boundary point $x=0$.
- Function value: $f(0) = -1$ (using $x \geq 0$)
- LHL: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x}{|x|} = \lim_{x \to 0^-} \frac{x}{-x} = \lim_{x \to 0^-} (-1) = -1$
- RHL: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (-1) = -1$
- Since $\lim_{x \to 0} f(x) = f(0) = -1$, the function is continuous at $x = 0$.
- For $x < 0$, $f(x) = \frac{x}{-x} = -1$, which is continuous.
- For $x > 0$, $f(x) = -1$, which is continuous.
- No points of discontinuity.
10. Function $f(x) = \begin{cases} x^2 + 1, & \text{if } x \geq 1 \\ x + 1, & \text{if } x < 1 \end{cases}$
We check the boundary point $x=1$.
- Function value: $f(1) = 1^2 + 1 = 2$ (using $x \geq 1$)
- LHL: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x + 1) = 1 + 1 = 2$
- RHL: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^2 + 1) = 1^2 + 1 = 2$
- Since $\lim_{x \to 1} f(x) = f(1) = 2$, the function is continuous at $x = 1$.
- No points of discontinuity.
11. Function $f(x) = \begin{cases} x^3 – 3, & \text{if } x \leq 2 \\ x^2 + 1, & \text{if } x > 2 \end{cases}$
We check the boundary point $x=2$.
- Function value: $f(2) = 2^3 – 3 = 8 – 3 = 5$ (using $x \leq 2$)
- LHL: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^3 – 3) = 2^3 – 3 = 5$
- RHL: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x^2 + 1) = 2^2 + 1 = 5$
- Since $\lim_{x \to 2} f(x) = f(2) = 5$, the function is continuous at $x = 2$.
- No points of discontinuity.
12. Function $f(x) = \begin{cases} x^{10} – 1, & \text{if } x \leq 1 \\ x^2, & \text{if } x > 1 \end{cases}$
We check the boundary point $x=1$.
- Function value: $f(1) = 1^{10} – 1 = 0$ (using $x \leq 1$)
- LHL: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^{10} – 1) = 1^{10} – 1 = 0$
- RHL: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} x^2 = 1^2 = 1$
- Since $\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)$, the function is discontinuous at $x = 1$.
- Point of discontinuity: $x = 1$.
13. Function $f(x) = \begin{cases} x + 5, & \text{if } x \leq 1 \\ x – 5, & \text{if } x > 1 \end{cases}$
We check the boundary point $x=1$.
- Function value: $f(1) = 1 + 5 = 6$ (using $x \leq 1$)
- LHL: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x + 5) = 1 + 5 = 6$
- RHL: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x – 5) = 1 – 5 = -4$
- Since $\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)$, the function is discontinuous at $x = 1$.
- No, the function is not a continuous function (it has one discontinuity).
14. Function $f(x) = \begin{cases} 3, & \text{if } 0 \leq x \leq 1 \\ 4, & \text{if } 1 < x < 3 \\ 5, & \text{if } 3 \leq x \leq 10 \end{cases}$
We check the boundary points $x = 1$ and $x = 3$.
At $x = 1$
- Function value: $f(1) = 3$ (using $0 \leq x \leq 1$)
- LHL: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 3 = 3$
- RHL: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 4 = 4$
- Since $\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)$, $f(x)$ is discontinuous at $x = 1$.
At $x = 3$
- Function value: $f(3) = 5$ (using $3 \leq x \leq 10$)
- LHL: $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} 4 = 4$
- RHL: $\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} 5 = 5$
- Since $\lim_{x \to 3^-} f(x) \neq \lim_{x \to 3^+} f(x)$, $f(x)$ is discontinuous at $x = 3$.
- Points of discontinuity: $x = 1$ and $x = 3$.
15. Function $f(x) = \begin{cases} 2x, & \text{if } x < 0 \\ 0, & \text{if } 0 \leq x \leq 1 \\ 4x, & \text{if } x > 1 \end{cases}$
We check the boundary points $x = 0$ and $x = 1$.
At $x = 0$
- Function value: $f(0) = 0$ (using $0 \leq x \leq 1$)
- LHL: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 2x = 2(0) = 0$
- RHL: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 0 = 0$
- Since $\lim_{x \to 0} f(x) = f(0) = 0$, $f(x)$ is continuous at $x = 0$.
At $x = 1$
- Function value: $f(1) = 0$ (using $0 \leq x \leq 1$)
- LHL: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 0 = 0$
- RHL: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 4x = 4(1) = 4$
- Since $\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)$, $f(x)$ is discontinuous at $x = 1$.
- Point of discontinuity: $x = 1$.
16. Function $f(x) = \begin{cases} -2, & \text{if } x \leq -1 \\ 2x, & \text{if } -1 < x \leq 1 \\ 2, & \text{if } x > 1 \end{cases}$
We check the boundary points $x = -1$ and $x = 1$.
At $x = -1$
- Function value: $f(-1) = -2$ (using $x \leq -1$)
- LHL: $\lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (-2) = -2$
- RHL: $\lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} (2x) = 2(-1) = -2$
- Since $\lim_{x \to -1} f(x) = f(-1) = -2$, $f(x)$ is continuous at $x = -1$.
At $x = 1$
- Function value: $f(1) = 2(1) = 2$ (using $-1 < x \leq 1$)
- LHL: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (2x) = 2(1) = 2$
- RHL: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 2 = 2$
- Since $\lim_{x \to 1} f(x) = f(1) = 2$, $f(x)$ is continuous at $x = 1$.
- No points of discontinuity.
17. Find the relationship between $a$ and $b$
Function $f(x) = \begin{cases} ax + 1, & \text{if } x \leq 3 \\ bx + 3, & \text{if } x > 3 \end{cases}$ is continuous at $x = 3$.
For $f(x)$ to be continuous at $x=3$, we need $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)$.
- LHL: $\lim_{x \to 3^-} (ax + 1) = a(3) + 1 = 3a + 1$
- RHL: $\lim_{x \to 3^+} (bx + 3) = b(3) + 3 = 3b + 3$
- Function value: $f(3) = a(3) + 1 = 3a + 1$
Equating the LHL and RHL:
$$3a + 1 = 3b + 3$$
$$3a – 3b = 3 – 1$$
$$3(a – b) = 2$$
$$a – b = \frac{2}{3}$$
$$a = b + \frac{2}{3}$$
The relationship between $a$ and $b$ is $a = b + \frac{2}{3}$ or $3a – 3b = 2$.
18. Function $f(x) = \begin{cases} \lambda(x^2 – 2x), & \text{if } x \leq 0 \\ 4x + 1, & \text{if } x > 0 \end{cases}$
Continuity at $x = 0$
For continuity at $x=0$, we need $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$.
- LHL: $\lim_{x \to 0^-} \lambda(x^2 – 2x) = \lambda(0^2 – 2(0)) = \lambda(0) = 0$
- RHL: $\lim_{x \to 0^+} (4x + 1) = 4(0) + 1 = 1$
- Function value: $f(0) = \lambda(0^2 – 2(0)) = 0$
For the limit to exist, LHL must equal RHL:
$$0 = 1$$
This is a contradiction. Therefore, no value of $\lambda$ can make the function continuous at $x=0$.
Continuity at $x = 1$
At $x=1$, we use the definition $f(x) = 4x + 1$ (since $1 > 0$).
- Function value: $f(1) = 4(1) + 1 = 5$
- Limit: $\lim_{x \to 1} f(x) = \lim_{x \to 1} (4x + 1) = 4(1) + 1 = 5$
- Since $f(1) = \lim_{x \to 1} f(x)$, $f(x)$ is continuous at $x=1$ for all values of $\lambda$.
19. Function $g(x) = x – [x]$ (Greatest Integer Function)
The greatest integer function, $[x]$, is discontinuous at every integer.
Since $g(x) = x – [x]$, the points of discontinuity of $g(x)$ will be the points of discontinuity of $[x]$, as $x$ is continuous everywhere.
Let $a = k$, where $k$ is an integer.
- Function value: $g(k) = k – [k] = k – k = 0$
- LHL: $\lim_{x \to k^-} g(x) = \lim_{x \to k^-} (x – [x])$As $x \to k$ from the left, $x$ is slightly less than $k$, so $[x] = k – 1$.$$\lim_{x \to k^-} (x – (k – 1)) = k – (k – 1) = 1$$
- RHL: $\lim_{x \to k^+} g(x) = \lim_{x \to k^+} (x – [x])$As $x \to k$ from the right, $x$ is slightly greater than $k$, so $[x] = k$.$$\lim_{x \to k^+} (x – k) = k – k = 0$$
Since $\lim_{x \to k^-} g(x) = 1$ and $\lim_{x \to k^+} g(x) = 0$, the limit does not exist at any integer $k$.
Thus, $g(x) = x – [x]$ is discontinuous at all integral points.ShutterstockExplore
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20. Function $f(x) = x^2 – \sin x + 5$ at $x = \pi$
The function $f(x)$ is the difference of two continuous functions ($x^2 + 5$ is a polynomial, thus continuous; $\sin x$ is a trigonometric function, continuous everywhere).
- Function value: $f(\pi) = \pi^2 – \sin(\pi) + 5 = \pi^2 – 0 + 5 = \pi^2 + 5$
- Limit: $\lim_{x \to \pi} f(x) = \lim_{x \to \pi} (x^2 – \sin x + 5) = \pi^2 – \sin(\pi) + 5 = \pi^2 + 5$
- Since $\lim_{x \to \pi} f(x) = f(\pi)$, the function $f(x)$ is continuous at $x = \pi$.
21. Discuss the continuity of the following functions
We use the algebra of continuous functions:
- The function $g(x) = \sin x$ is continuous for all $x \in \mathbb{R}$.
- The function $h(x) = \cos x$ is continuous for all $x \in \mathbb{R}$.
(a) $f(x) = \sin x + \cos x$
This is the sum of two continuous functions. Therefore, $f(x)$ is continuous for all $x \in \mathbb{R}$.
(b) $f(x) = \sin x – \cos x$
This is the difference of two continuous functions. Therefore, $f(x)$ is continuous for all $x \in \mathbb{R}$.
(c) $f(x) = \sin x \cdot \cos x$
This is the product of two continuous functions. Therefore, $f(x)$ is continuous for all $x \in \mathbb{R}$.
22. Discuss the continuity of the cosine, cosecant, secant and cotangent functions
All trigonometric functions are continuous in their respective domains.
- Cosine function, $f(x) = \cos x$:
- Continuous for all $x \in \mathbb{R}$ (all real numbers).
- Cosecant function, $f(x) = \csc x = \frac{1}{\sin x}$:
- Discontinuous when $\sin x = 0$.
- Discontinuous at $x = n\pi$, where $n$ is any integer.
- Continuous in its domain: $\mathbb{R} \setminus \{n\pi, n \in \mathbb{Z}\}$.
- Secant function, $f(x) = \sec x = \frac{1}{\cos x}$:
- Discontinuous when $\cos x = 0$.
- Discontinuous at $x = (2n + 1)\frac{\pi}{2}$, where $n$ is any integer.
- Continuous in its domain: $\mathbb{R} \setminus \{(2n + 1)\frac{\pi}{2}, n \in \mathbb{Z}\}$.
- Cotangent function, $f(x) = \cot x = \frac{\cos x}{\sin x}$:
- Discontinuous when $\sin x = 0$.
- Discontinuous at $x = n\pi$, where $n$ is any integer.
- Continuous in its domain: $\mathbb{R} \setminus \{n\pi, n \in \mathbb{Z}\}$.
23. Function $f(x) = \begin{cases} \frac{\sin x}{x}, & \text{if } x < 0 \\ x + 1, & \text{if } x \geq 0 \end{cases}$
We check the boundary point $x=0$.
- Function value: $f(0) = 0 + 1 = 1$ (using $x \geq 0$)
- LHL: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin x}{x}$Using the standard limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:$$\lim_{x \to 0^-} \frac{\sin x}{x} = 1$$
- RHL: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x + 1) = 0 + 1 = 1$
- Since $\lim_{x \to 0} f(x) = f(0) = 1$, the function is continuous at $x = 0$.
- For $x < 0$, $f(x)$ is a rational function of continuous functions (and the denominator is not zero), so it is continuous.
- For $x > 0$, $f(x)$ is a polynomial, thus continuous.
- No points of discontinuity.
24. Function $f(x) = \begin{cases} x^2 \sin \left(\frac{1}{x}\right), & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$
We check the boundary point $x=0$.
- Function value: $f(0) = 0$
- Limit at $x=0$: $\lim_{x \to 0} x^2 \sin \left(\frac{1}{x}\right)$We know that $-1 \leq \sin \left(\frac{1}{x}\right) \leq 1$.Multiplying by $x^2$ (which is positive for $x \neq 0$):$$-x^2 \leq x^2 \sin \left(\frac{1}{x}\right) \leq x^2$$Applying the limit:$$\lim_{x \to 0} (-x^2) \leq \lim_{x \to 0} x^2 \sin \left(\frac{1}{x}\right) \leq \lim_{x \to 0} x^2$$$$0 \leq \lim_{x \to 0} x^2 \sin \left(\frac{1}{x}\right) \leq 0$$By the Sandwich Theorem (or Squeeze Theorem), $\lim_{x \to 0} f(x) = 0$.
- Since $\lim_{x \to 0} f(x) = f(0) = 0$, the function is continuous at $x = 0$.
- For $x \neq 0$, $f(x)$ is the product of continuous functions ($x^2$ and $\sin(1/x)$), so it is continuous.
- Yes, $f(x)$ is a continuous function.
25. Function $f(x) = \begin{cases} \frac{\sin x – \cos x}{x}, & \text{if } x \neq 0 \\ -1, & \text{if } x = 0 \end{cases}$
Correction: The question in the textbook for continuity should likely be $f(x) = \begin{cases} \frac{\sin x – \cos x}{x}, & \text{if } x \neq \mathbf{0} \\ -\mathbf{1}, & \text{if } x = \mathbf{0} \end{cases}$ (as this form makes more sense for $x=0$ check), but since the function is written as $f(x) = \begin{cases} \frac{\sin x – \cos x}{x}, & \text{if } x \neq 0 \\ -1, & \text{if } x = 0 \end{cases}$ in the prompt, let’s proceed with this.
Assumption: The function given in the textbook is $f(x) = \begin{cases} \frac{\sin x – \cos x}{x}, & \text{if } x \neq 0 \\ -1, & \text{if } x = 0 \end{cases}$.
We check the boundary point $x=0$.
- Function value: $f(0) = -1$
- Limit at $x=0$: $\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin x – \cos x}{x}$$$\lim_{x \to 0} \left( \frac{\sin x}{x} – \frac{\cos x}{x} \right) = \lim_{x \to 0} \frac{\sin x}{x} – \lim_{x \to 0} \frac{\cos x}{x}$$$$1 – \lim_{x \to 0} \frac{\cos x}{x}$$Since $\lim_{x \to 0} \cos x = 1$ and $\lim_{x \to 0} x = 0$, the second term $\lim_{x \to 0} \frac{\cos x}{x}$ is of the form $\frac{1}{0}$, which means the limit does not exist (it approaches $\pm \infty$).Therefore, the $\lim_{x \to 0} f(x)$ does not exist.
- Conclusion: The function is discontinuous at $x = 0$.
(Alternative Interpretation based on common textbook typos: If the function was $f(x) = \begin{cases} \frac{\sin x – \cos x}{\mathbf{x – \pi/4}}, & \text{if } x \neq \mathbf{\pi/4} \\ \sqrt{2}, & \text{if } x = \mathbf{\pi/4} \end{cases}$ it would be continuous. If the function was $f(x) = \begin{cases} \frac{\sin x – \cos x}{\mathbf{x – \pi/2}}, & \text{if } x \neq \mathbf{\pi/2} \\ 1, & \text{if } x = \mathbf{\pi/2} \end{cases}$ it would be continuous. Based on the prompt as written, the result is discontinuity.)
Find the values of $k$ (Exercises 26-29)
26. Function $f(x) = \begin{cases} \frac{k \cos x}{\pi – 2x}, & \text{if } x \neq \frac{\pi}{2} \\ 3, & \text{if } x = \frac{\pi}{2} \end{cases}$ at $x = \frac{\pi}{2}$
For continuity, $\lim_{x \to \pi/2} f(x) = f(\pi/2)$.
- Function value: $f(\frac{\pi}{2}) = 3$
- Limit: $\lim_{x \to \pi/2} \frac{k \cos x}{\pi – 2x}$This is in the indeterminate form $\frac{0}{0}$ ($\cos(\pi/2) = 0$, $\pi – 2(\pi/2) = 0$).Let $x = \frac{\pi}{2} + h$. As $x \to \frac{\pi}{2}$, $h \to 0$.$$\lim_{h \to 0} \frac{k \cos(\frac{\pi}{2} + h)}{\pi – 2(\frac{\pi}{2} + h)} = \lim_{h \to 0} \frac{k (-\sin h)}{\pi – \pi – 2h}$$$$= \lim_{h \to 0} \frac{-k \sin h}{-2h} = \lim_{h \to 0} \frac{k}{2} \cdot \frac{\sin h}{h}$$$$= \frac{k}{2} \cdot 1 = \frac{k}{2}$$
Equating the limit and function value:
$$\frac{k}{2} = 3$$
$$k = 6$$
27. Function $f(x) = \begin{cases} kx^2, & \text{if } x \leq 2 \\ 3, & \text{if } x > 2 \end{cases}$ at $x = 2$
For continuity, $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)$.
- LHL: $\lim_{x \to 2^-} kx^2 = k(2^2) = 4k$
- RHL: $\lim_{x \to 2^+} 3 = 3$
- Function value: $f(2) = k(2^2) = 4k$
Equating the LHL and RHL:
$$4k = 3$$
$$k = \frac{3}{4}$$
28. Function $f(x) = \begin{cases} kx + 1, & \text{if } x \leq \pi \\ \cos x, & \text{if } x > \pi \end{cases}$ at $x = \pi$
For continuity, $\lim_{x \to \pi^-} f(x) = \lim_{x \to \pi^+} f(x) = f(\pi)$.
- LHL: $\lim_{x \to \pi^-} (kx + 1) = k\pi + 1$
- RHL: $\lim_{x \to \pi^+} \cos x = \cos \pi = -1$
- Function value: $f(\pi) = k\pi + 1$
Equating the LHL and RHL:
$$k\pi + 1 = -1$$
$$k\pi = -2$$
$$k = -\frac{2}{\pi}$$
29. Function $f(x) = \begin{cases} kx + 1, & \text{if } x \leq 5 \\ 3x – 5, & \text{if } x > 5 \end{cases}$ at $x = 5$
For continuity, $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) = f(5)$.
- LHL: $\lim_{x \to 5^-} (kx + 1) = k(5) + 1 = 5k + 1$
- RHL: $\lim_{x \to 5^+} (3x – 5) = 3(5) – 5 = 15 – 5 = 10$
- Function value: $f(5) = k(5) + 1 = 5k + 1$
Equating the LHL and RHL:
$$5k + 1 = 10$$
$$5k = 9$$
$$k = \frac{9}{5}$$
30. Function $f(x) = \begin{cases} 5, & \text{if } x \leq 2 \\ ax + b, & \text{if } 2 < x < 10 \\ 21, & \text{if } x \geq 10 \end{cases}$
The function is continuous everywhere, so we check the boundary points $x = 2$ and $x = 10$.
At $x = 2$
For continuity at $x=2$: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)$
$$\lim_{x \to 2^-} 5 = \lim_{x \to 2^+} (ax + b)$$
$$5 = a(2) + b$$
(Equation 1): $2a + b = 5$
At $x = 10$
For continuity at $x=10$: $\lim_{x \to 10^-} f(x) = \lim_{x \to 10^+} f(x)$
$$\lim_{x \to 10^-} (ax + b) = \lim_{x \to 10^+} 21$$
$$a(10) + b = 21$$
(Equation 2): $10a + b = 21$
Solve the system of equations:
Subtract (1) from (2):
$$(10a + b) – (2a + b) = 21 – 5$$
$$8a = 16$$
$$a = 2$$
Substitute $a=2$ into (1):
$$2(2) + b = 5$$
$$4 + b = 5$$
$$b = 1$$
The values are $a = 2$ and $b = 1$.
31. Show that $f(x) = \cos (x^2)$ is continuous
The function $f(x)$ is a composition of two functions:
- $g(x) = x^2$ (an algebraic function, continuous everywhere).
- $h(y) = \cos y$ (a trigonometric function, continuous everywhere).Since $f(x) = h(g(x))$, the composition of two continuous functions is continuous.Therefore, $f(x) = \cos(x^2)$ is continuous for all $x \in \mathbb{R}$.
32. Show that $f(x) = | \cos x |$ is continuous
The function $f(x)$ is a composition of two functions:
- $g(x) = \cos x$ (a trigonometric function, continuous everywhere).
- $h(y) = |y|$ (the absolute value function, continuous everywhere).Since $f(x) = h(g(x))$, the composition of two continuous functions is continuous.Therefore, $f(x) = | \cos x |$ is continuous for all $x \in \mathbb{R}$.
33. Examine that $\sin | x |$ is a continuous function
The function $f(x) = \sin |x|$ is a composition of two functions:
- $g(x) = |x|$ (the absolute value function, continuous everywhere).
- $h(y) = \sin y$ (a trigonometric function, continuous everywhere).Since $f(x) = h(g(x))$, the composition of two continuous functions is continuous.Therefore, $f(x) = \sin |x|$ is continuous for all $x \in \mathbb{R}$.
34. Find all the points of discontinuity of $f(x) = | x | – | x + 1 |$
The function is the difference of two absolute value functions:
- $g(x) = |x|$, which is continuous everywhere.
- $h(x) = |x+1|$, which is continuous everywhere.The difference of two continuous functions is continuous.
Therefore, $f(x) = |x| – |x+1|$ is continuous for all $x \in \mathbb{R}$.
No points of discontinuity.
❓ Frequently Asked Questions (FAQs) on Continuity
What are the three conditions for a function to be continuous at a point $x=a$?
A function $f(x)$ is continuous at $x=a$ if and only if all three conditions are met:
$f(a)$ is defined (the function exists at $x=a$).
$\lim_{x \to a} f(x)$ exists (the left-hand limit and right-hand limit are equal).
$\lim_{x \to a} f(x) = f(a)$.
Are all polynomial functions continuous?
Yes, all polynomial functions (like $f(x) = 5x-3$ or $f(x) = 2x^2-1$) are continuous for all real numbers $x \in \mathbb{R}$.
Where are rational functions, like $f(x) = \frac{1}{x-5}$, discontinuous?
Rational functions, $f(x) = \frac{p(x)}{q(x)}$, are continuous everywhere except at the points where the denominator $q(x)$ is zero. For $f(x) = \frac{1}{x-5}$, the discontinuity is at $x=5$.
How do you check for continuity in a piecewise function?
For a piecewise function, you must check for continuity:
Within each defined interval (where the function is usually a polynomial or a standard continuous function).
At the boundary points (the points where the function’s definition changes), by comparing the Left-Hand Limit (LHL), Right-Hand Limit (RHL), and the function value $f(a)$.
What is the continuity rule for composite functions?
If a function g is continuous at x=a and a function f is continuous at g(a), then the composite function (f∘g)(x)=f(g(x)) is continuous at x=a. This is why ∣cosx∣ and cos(x2) are continuous everywhere.