Continuity and Differentiabilty: Chapter 5 Links
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In this NCERT exercise on Continuity 5.1, we will learn about continuity and discontinuity of functions, one of the most important concepts in calculus. Intuitively, a function is said to be continuous if its graph can be drawn without lifting the pen from the paper. In other words, there should be no breaks, jumps, or gaps in the graph at the point being considered.
To check whether a function is continuous at a point x = c, we compare the value of the function with its limit at that point. A function f(x) is continuous at x = c if
\displaystyle\lim_{x\to 2}x^2=4 and f(2)=4
Equivalently, the left-hand limit (LHL), right-hand limit (RHL), and the function value must all be equal. Functions that do not satisfy this condition are called discontinuous functions, and the corresponding values of x are known as points of discontinuity.
In this exercise, we will use limits to determine whether functions are continuous or discontinuous at given points and identify points of discontinuity wherever they exist.
Key Concepts
1. Continuity of a Function at a Point
A function f(x) is said to be continuous at a point x = c if the limit of the function at that point exists and equals its value.
The condition for continuity is
\displaystyle\lim_{x\to c}f(x)=f(c)
Example:
For f(x)=x^2,
\displaystyle\lim_{x\to 2}x^2=4 and f(2)=4
Hence, f(x) is continuous at x=2.
2. Equivalent Condition for Continuity
A function is continuous at x=c if the left-hand limit, right-hand limit and function value are all equal.
\displaystyle\lim_{x\to c^-}f(x)=\lim_{x\to c^+}f(x)=f(c)
Example:
If LHL = RHL = 5 and f(c)=5, then the function is continuous at x=c.
3. Discontinuous Functions
A function is said to be discontinuous at a point if it does not satisfy the condition of continuity.
This may happen when the limit does not exist or when the limit and function value are different.
Example:
f(x)=\frac{x^2-1}{x-1}
The function is not defined at x=1.
Therefore, it is discontinuous at x=1.
4. Points of Discontinuity
The values of x at which a function is not continuous are called points of discontinuity.
Example:
For f(x)=\frac{1}{x-3}, the function is not defined at x=3.
Thus, x=3 is a point of discontinuity.
5. Graphical Interpretation of Continuity
Graphically, a function is continuous if its graph can be drawn without lifting the pen from the paper throughout its domain.
A break, jump or gap in the graph indicates discontinuity.
Examples:
- Polynomial functions are continuous for all real values of x.
- \frac{1}{x} is discontinuous at x=0.
6. Standard Continuous Functions
The following functions are continuous on their respective domains:
- Constant functions: f(x)=c
- Linear functions: f(x)=ax+b
- Polynomial functions: f(x)=x^n
- Modulus functions: f(x)=|x|
- Trigonometric functions: \sin x,\ \cos x
- Exponential functions: a^x
- Logarithmic functions: \log x, for x>0
- Rational functions wherever the denominator is non-zero.
7. Algebra of Continuous Functions
If f and g are continuous at x=c, then the following functions are also continuous at x=c:
- f+g
- f-g
- f\times g
- \frac{f}{g}, provided g(c)\neq0
Example:
If f(x)=x and g(x)=x^2, then f(x)+g(x)=x+x^2 is continuous for all real values of x.
8. Composition of Continuous Functions
If f is continuous at x=c and g is continuous at f(c), then the composite function (g\circ f)(x)=g(f(x)) is continuous at x=c.
Examples:
- \cos(x^2)
- \sin|x|
- e^{x^2}
Let us now solve all the NCERT questions step by step in this exercise of Continuity 5.1.
Question 1: Continuity 5.1
1. Prove that the function f(x)=5x-3 is continuous at x=0,\ x=-3 and x=5.
Solution
Given,
f(x)=5x-3
A linear polynomial is continuous for all real values of x.
To prove continuity at a point x=c, we check whether
\displaystyle\lim_{x\to c}f(x)=f(c)
At x=0
\displaystyle\lim_{x\to 0}(5x-3)=-3 and f(0)=-3
Therefore, \displaystyle\lim_{x\to 0}f(x)=f(0).
Hence, f(x) is continuous at x=0.
At x=-3
\displaystyle\lim_{x\to -3}(5x-3)=-18 and f(-3)=-18
Therefore, \displaystyle\lim_{x\to -3}f(x)=f(-3).
Hence, f(x) is continuous at x=-3.
At x=5
\displaystyle\lim_{x\to 5}(5x-3)=22 and f(5)=22
Therefore, \displaystyle\lim_{x\to 5}f(x)=f(5).
Hence, f(x) is continuous at x=5.
Therefore, the function f(x)=5x-3 is continuous at x=0,\ x=-3 and x=5.
Question 2: Continuity Ex. 5.1
2. Examine the continuity of the function f(x)=2x^2-1 at x=3.
Solution
Given,
f(x)=2x^2-1
A polynomial function is continuous for all real values of x.
To examine continuity at x=3, we check whether
\displaystyle\lim_{x\to 3}f(x)=f(3)
Now,
\displaystyle\lim_{x\to 3}(2x^2-1)=2(3)^2-1=17
Also,
f(3)=2(3)^2-1=17
Thus,
\displaystyle\lim_{x\to 3}f(x)=f(3)=17
Hence, the function f(x)=2x^2-1 is continuous at x=3.
You might already be following Maths Better for NCERT Solutions for the topics like Matrices, Determinants, Relations and Functions and Inverse Trigonometric Functions. Similarly, this exercise on Continuity and Differentiability for Class 12 Maths is designed to strengthen your concepts and improve step-by-step problem-solving skills.
Now, let’s move on to the next question of Continuity and Differentiability 5.1.
Question 3: Continuity and Differentiability 5.1
3. Examine the following functions for continuity:
Solution
(a) Given, f(x)=x-5
A linear polynomial is continuous for all real values of x.
Since f(x)=x-5 is a linear polynomial, it is continuous for all real values of x.
To examine:
Let c be any real number.
Then,
\displaystyle\lim_{x\to c}(x-5)=c-5
Also,
f(c)=c-5
Thus, \displaystyle\lim_{x\to c}f(x)=f(c).
Hence, f(x)=x-5 is continuous for all real values of x.
(b) Given, f(x)=\frac{1}{x-5},\ x\ne 5
A rational function is continuous wherever its denominator is non-zero.
Here, the denominator x-5 becomes zero at x=5.
Therefore, the function is not defined at x=5.
To examine:
Let c be any real number such that c\ne 5.
Then,
\displaystyle\lim_{x\to c}\frac{1}{x-5}=\frac{1}{c-5}
Also,
f(c)=\frac{1}{c-5}
Thus, \displaystyle\lim_{x\to c}f(x)=f(c)
Hence, f(x)=\frac{1}{x-5} is continuous for all x\ne 5.
Since the function is not defined at x=5, it is discontinuous at x=5.
(c) Given,
f(x)=\frac{x^2-25}{x+5},\ x\ne -5
Factorising the numerator,
f(x)=\frac{(x-5)(x+5)}{x+5}
=x-5,\quad x\ne -5
A rational function is continuous wherever its denominator is non-zero.
The denominator becomes zero at x=-5. Therefore, the function is not defined at x=-5.
To examine:
Let c be any real number such that c\ne -5.
Then,
\displaystyle\lim_{x\to c}(x-5)=c-5
Also,
f(c)=c-5
Thus, \displaystyle\lim_{x\to c}f(x)=f(c)
Hence, f(x) is continuous for all x\ne -5.
Since the function is not defined at x=-5, it is discontinuous at x=-5.
Thus, the graph is the same as that of the line y=x-5 except for a missing point at x=-5.
(d) Given, f(x)=|x-5|
The modulus function is continuous for all real values of x.
To examine continuity, consider the following cases:
Case 1: c\ne 5
For c\ne 5, the function behaves like a linear function in a neighbourhood of c.
Therefore,
\displaystyle\lim_{x\to c}|x-5|=|c-5|
Also,
f(c)=|c-5|
Thus, \displaystyle\lim_{x\to c}f(x)=f(c).
Hence, f(x) is continuous at every point c\ne 5.
Case 2: c=5
Here,
f(5)=|5-5|=0
Also,
\displaystyle\lim_{x\to 5}|x-5|=0
Therefore,
\displaystyle\lim_{x\to 5}f(x)=f(5)
Hence, f(x)=|x-5| is continuous at x=5.
Thus, f(x)=|x-5| is continuous for all real values of x.
Question 4: Continuity Exercise 5.1
4. Prove that the function f(x)=x^n is continuous at x=n, where n is a positive integer.
Solution
Given,
f(x)=x^n
Every polynomial function is continuous for all real values of x.
To prove continuity at x=n, we check whether
\displaystyle\lim_{x\to n}f(x)=f(n)
Now,
\displaystyle\lim_{x\to n}x^n=n^n
Also,
f(n)=n^n
Thus,
\displaystyle\lim_{x\to n}f(x)=f(n)
Hence, the function f(x)=x^n is continuous at x=n.
Question 5: Continuity 5.1
5. Is the function
f(x)=\begin{cases} x, & x\le 1 \\ \\ 5, & x>1 \end{cases}
continuous at x=0? At x=1? At x=2?
Solution
Given,
f(x)=\begin{cases} x, & x\le 1 \\ \\ 5, & x>1 \end{cases}
At x=0
Since 0\lt 1, we use f(x)=x.
\displaystyle\lim_{x\to 0}f(x)=\lim_{x\to 0}x=0
Also, value of f(x) at x=0 (\le 1), is
f(0)=0
Thus, \displaystyle\lim_{x\to 0}f(x)=f(0)
Hence, the function is continuous at x=0.
At x=1
To check continuity at x=1, we find the left-hand limit and right-hand limit.
As x\to 1^-, we have x<1. Therefore, we use
f(x)=x
Thus,
\displaystyle\lim_{x\to 1^-}f(x)=\lim_{x\to 1^-}x=1
As x\to 1^+, we have x>1. Therefore, we use
f(x)=5
Thus,
\displaystyle\lim_{x\to 1^+}f(x)=\lim_{x\to 1^+}5=5
Since
\displaystyle\lim_{x\to 1^-}f(x)\ne\lim_{x\to 1^+}f(x)
the limit \displaystyle\lim_{x\to 1}f(x) does not exist.
Also, value of f(x) at x=1 (\le 1), is
f(1)=1.
Hence, the function is discontinuous at x=1.
At x=2
Since 2>1, we use f(x)=5.
\displaystyle\lim_{x\to 2}f(x)=\lim_{x\to 2}5=5
Also, value of f(x) at x=2 (\gt 1), is
f(2)=5
Thus, \displaystyle\lim_{x\to 2}f(x)=f(2)
Hence, the function is continuous at x=2.
While examining continuity, remember that a function is continuous at a point only when the limit exists and is equal to the function value at that point. For piecewise functions, always check the left-hand limit, right-hand limit and function value at the point where the definition changes.
Question 6: Continuity Ex 5.1
6. Find all points of discontinuity of
f(x)=\begin{cases} 2x+3, & \text{if } x\le 2 \\ \\ 2x-3, & \text{if } x>2 \end{cases}
Solution
For x<2, f(x)=2x+3, which is a linear polynomial and hence continuous.
For x>2, f(x)=2x-3, which is also a linear polynomial and hence continuous.
Therefore, the only possible point of discontinuity is x=2.
To examine continuity at x=2, we find the left-hand limit and right-hand limit.
As x\to 2^-, we have x<2. Therefore, we use
f(x)=2x+3
Thus,
\displaystyle\lim_{x\to 2^-}f(x)=\lim_{x\to 2^-}(2x+3)=7
As x\to 2^+, we have x>2. Therefore, we use
f(x)=2x-3
Thus,
\displaystyle\lim_{x\to 2^+}f(x)=\lim_{x\to 2^+}(2x-3)=1
Since
\displaystyle\lim_{x\to 2^-}f(x)\ne\lim_{x\to 2^+}f(x)
i.e., 7\ne 1
Also, value of f(x) at x=2 (\le 2), is
f(2)=2(2)+3=7
Since LHL ≠ RHL, the limit \displaystyle\lim_{x\to 2}f(x) does not exist.
Hence, the function is discontinuous at x=2.
Therefore, the point of discontinuity is x=2.
Question 7: Continuity and Discontinuity
7. Find all points of discontinuity of
f(x)=\begin{cases} |x|+3, & \text{if } x\le -3 \\ \\ -2x, & \text{if } -3<x<3 \\ \\ 6x+2, & \text{if } x\ge 3 \end{cases}
Solution
For x\lt -3, f(x)=|x|+3, which is a sum of two continuous functions namely modulus and constant and so it is also continuous.
For -3<x<3, f(x)=-2x, which is a linear polynomial and hence continuous.
And for x\gt 3, f(x)=6x+2, which is also a linear polynomial and hence continuous.
Therefore, the only possible points of discontinuity are x=-3 and x=3.
At x=-3
As x\to -3^-, we have x\lt -3. Therefore, we use
f(x)=|x|+3
Thus,
\displaystyle\lim_{x\to -3^-}f(x)=|-3|+3=6
As x\to -3^+, we have x>-3. Therefore, we use
f(x)=-2x
Thus,
\displaystyle\lim_{x\to -3^+}f(x)=-2(-3)=6
Since
\displaystyle\lim_{x\to -3^-}f(x)=\lim_{x\to -3^+}f(x)
i.e., 6=6
Also, value of f(x) at x=-3 (\le -3), is
f(-3)=|-3|+3=6
Thus, \displaystyle\lim_{x\to -3}f(x)=f(-3)
Hence, the function is continuous at x=-3.
At x=3
As x\to 3^-, we have x<3. Therefore, we use
f(x)=-2x
Thus,
\displaystyle\lim_{x\to 3^-}f(x)=-2(3)=-6
As x\to 3^+, we have x\gt 3. Therefore, we use
f(x)=6x+2
Thus,
\displaystyle\lim_{x\to 3^+}f(x)=6(3)+2=20
Since
\displaystyle\lim_{x\to 3^-}f(x)\ne\lim_{x\to 3^+}f(x)
i.e., -6\ne 20
Also, value of f(x) at x=3 (\ge 3), is
f(3)=6(3)+2=20
Since LHL ≠ RHL, the limit \displaystyle\lim_{x\to 3}f(x) does not exist.
Hence, the function is discontinuous at x=3.
Therefore, the only point of discontinuity is x=3.
NCERT Class 12 Maths includes several important exercises across both Part 1 and Part 2, and I’ll cover them one by one with clear explanations and step-by-step solutions. Many of these questions are also available in video format on my YouTube channel, @Mathsbetter, to help you understand the concepts more visually.
Now, let’s move to the next question.
Question 8: Continuity Ex 5.1
8. Find all points of discontinuity of
f(x)=\begin{cases} \frac{|x|}{x}, & \text{if } x\ne 0 \\ \\ 0, & \text{if } x=0 \end{cases}
Solution
For x>0,
\frac{|x|}{x}=\frac{x}{x}=1
which is a constant function and hence continuous.
For x<0
\frac{|x|}{x}=\frac{-x}{x}=-1
which is also a constant function and hence continuous.
Therefore, the only possible point of discontinuity is x=0.
To examine continuity at x=0, we find the left-hand limit and right-hand limit.
As x\to 0^-, we have x<0
So, f(x)=-1
Thus,
\displaystyle\lim_{x\to 0^-}f(x)=-1
As x\to 0^+, we have x>0. Therefore,
f(x)=1
Thus,
\displaystyle\lim_{x\to 0^+}f(x)=1
Since
\displaystyle\lim_{x\to 0^-}f(x)\ne\lim_{x\to 0^+}f(x)
i.e., -1\ne 1
Also, value of f(x) at x=0, is
f(0)=0
Since LHL ≠ RHL, the limit \displaystyle\lim_{x\to 0}f(x) does not exist.
Hence, the function is discontinuous at x=0.
Therefore, the only point of discontinuity is x=0.
Continuity is closely connected with the graphical behaviour of a function. If a graph can be traced near a point without any break, jump or hole, the function is likely to be continuous at that point.
Question 9: Continuity Ex 5.1
9. Find all points of discontinuity of
f(x)=\begin{cases} \frac{x}{|x|}, & \text{if } x<0 \\ \\ -1, & \text{if } x\ge 0 \end{cases}
Solution
For x<0, we have |x|=-x. Therefore,
\frac{x}{|x|}=\frac{x}{-x}=-1
Thus, for x<0, the function is a constant function and hence continuous.
For x\gt 0,
f(x)=-1
which is also a constant function and hence continuous.
Therefore, the only possible point of discontinuity is x=0.
To examine continuity at x=0, we find the left-hand limit and right-hand limit.
As x\to 0^-, we have x<0. Therefore,
f(x)=-1
Thus,
\displaystyle\lim_{x\to 0^-}f(x)=-1
As x\to 0^+, we have x\gt 0. Therefore,
f(x)=-1
Thus,
\displaystyle\lim_{x\to 0^+}f(x)=-1
Since
\displaystyle\lim_{x\to 0^-}f(x)=\lim_{x\to 0^+}f(x)
i.e., -1=-1
Also, value of f(x) at x=0 (\gt 0), is
f(0)=-1
Thus, \displaystyle\lim_{x\to 0}f(x)=f(0)
Hence, the function is continuous at x=0.
Therefore, the function has no points of discontinuity.
Question 10: Continuity 5.1
10. Find all points of discontinuity of
f(x)=\begin{cases} x+1, & \text{if } x\ge 1 \\ \\ x^2+1, & \text{if } x<1 \end{cases}
Solution
For x<1, f(x)=x^2+1, which is a polynomial function and hence continuous.
For x\gt 1, f(x)=x+1, which is a linear polynomial and hence continuous.
Therefore, the only possible point of discontinuity is x=1.
To examine continuity at x=1, we find the left-hand limit and right-hand limit.
As x\to 1^-, we have x<1. Therefore, we use
f(x)=x^2+1
Thus,
\displaystyle\lim_{x\to 1^-}f(x)=\lim_{x\to 1^-}(x^2+1)=2
As x\to 1^+, we have x\gt 1. Therefore, we use
f(x)=x+1
Thus,
\displaystyle\lim_{x\to 1^+}f(x)=\lim_{x\to 1^+}(x+1)=2
Since
\displaystyle\lim_{x\to 1^-}f(x)=\lim_{x\to 1^+}f(x)
i.e., 2=2
Also, value of f(x) at x=1 (\ge 1), is
f(1)=1+1=2
Thus,
\displaystyle\lim_{x\to 1}f(x)=f(1)=2
Hence, the function is continuous at x=1.
Therefore, the function has no points of discontinuity.
A good understanding of continuity is essential before studying differentiability, since every differentiable function is continuous, although the converse is not always true.
Here’s the next question.
Question 11: Continuity and Differentiability 5.1
11. Find all points of discontinuity of
f(x)=\begin{cases} x^3-3, & \text{if } x\le 2 \\ \\ x^2+1, & \text{if } x>2 \end{cases}
Solution
For x\lt 2, f(x)=x^3-3, which is a polynomial function and hence continuous.
For x>2, f(x)=x^2+1, which is also a polynomial function and hence continuous.
Therefore, the only possible point of discontinuity is x=2.
To examine continuity at x=2, we find the left-hand limit and right-hand limit.
As x\to 2^-, we have x\lt 2. Therefore, we use
f(x)=x^3-3
Thus,
\displaystyle\lim_{x\to 2^-}f(x)=\lim_{x\to 2^-}(x^3-3)=8-3=5
As x\to 2^+, we have x>2. Therefore, we use
f(x)=x^2+1
Thus,
\displaystyle\lim_{x\to 2^+}f(x)=\lim_{x\to 2^+}(x^2+1)=4+1=5
Since
\displaystyle\lim_{x\to 2^-}f(x)=\lim_{x\to 2^+}f(x)
i.e., 5=5
Also, value of f(x) at x=2 (\le 2), is
f(2)=2^3-3=8-3=5
Thus,
\displaystyle\lim_{x\to 2}f(x)=f(2)=5
Hence, the function is continuous at x=2.
Therefore, the function has no points of discontinuity.
Question 12: Continuity Ex. 5.1
12. Find all points of discontinuity of
f(x)=\begin{cases} x^{10}-1, & \text{if } x\le 1 \\ \\ x^2, & \text{if } x>1 \end{cases}
Solution
For x\lt 1, f(x)=x^{10}-1, which is a polynomial function and hence continuous.
For x>1, f(x)=x^2, which is also a polynomial function and hence continuous.
Therefore, the only possible point of discontinuity is x=1.
To examine continuity at x=1, we find the left-hand limit and right-hand limit.
As x\to 1^-, we have x\lt 1. Therefore, we use
f(x)=x^{10}-1
Thus,
\displaystyle\lim_{x\to 1^-}f(x)=\lim_{x\to 1^-}(x^{10}-1)=1-1=0
As x\to 1^+, we have x>1. Therefore, we use
f(x)=x^2
Thus,
\displaystyle\lim_{x\to 1^+}f(x)=\lim_{x\to 1^+}x^2=1
Since
\displaystyle\lim_{x\to 1^-}f(x)\ne\lim_{x\to 1^+}f(x)
i.e., 0\ne 1
Also, value of f(x) at x=1 (\le 1), is
f(1)=1^{10}-1=0
Since LHL ≠ RHL, the limit \displaystyle\lim_{x\to 1}f(x) does not exist.
Hence, the function is discontinuous at x=1.
Therefore, the point of discontinuity is x=1.
Now, let’s move on to the next question of Continuity and Differentiability.
Question 13: Continuity
13. Is the function
f(x)=\begin{cases} x+5, & \text{if } x\le 1 \\ \\ x-5, & \text{if } x>1 \end{cases}
a continuous function?
Solution
For x\lt 1, f(x)=x+5, which is a linear polynomial and hence continuous.
For x>1, f(x)=x-5, which is also a linear polynomial and hence continuous.
Therefore, we only need to examine continuity at x=1.
As x\to 1^-, we have x\lt 1. Therefore, we use
f(x)=x+5
Thus,
\displaystyle\lim_{x\to 1^-}f(x)=\lim_{x\to 1^-}(x+5)=6
As x\to 1^+, we have x>1. Therefore, we use
f(x)=x-5
Thus,
\displaystyle\lim_{x\to 1^+}f(x)=\lim_{x\to 1^+}(x-5)=-4
Since
\displaystyle\lim_{x\to 1^-}f(x)\ne\lim_{x\to 1^+}f(x)
i.e., 6\ne -4
Also, value of f(x) at x=1 (\le 1), is
f(1)=1+5=6
Since LHL ≠ RHL, the limit \displaystyle\lim_{x\to 1}f(x) does not exist.
Hence, the function is discontinuous at x=1.
Therefore, the given function is not continuous at x=1 and hence is not a continuous function.
Question 14: Continuity Ex 5.1
14. Discuss the continuity of the function
f(x)=\begin{cases} 3, & \text{if } 0\le x\le 1 \\ \\ 4, & \text{if } 1<x<3 \\ \\ 5, & \text{if } 3\le x\le 10 \end{cases}
Solution
For 0<x<1, f(x)=3, which is a constant function and hence continuous.
For 1<x<3, f(x)=4, which is a constant function and hence continuous.
And for 3<x<10, f(x)=5, which is a constant function and hence continuous.
Therefore, we only need to examine continuity at the points where the definition changes, namely x=1 and x=3.
At x=1
As x\to 1^-, we have x\lt 1. Therefore, we use
f(x)=3
Thus,
\displaystyle\lim_{x\to 1^-}f(x)=3
As x\to 1^+, we have x>1. Therefore, we use
f(x)=4
Thus,
\displaystyle\lim_{x\to 1^+}f(x)=4
Since
\displaystyle\lim_{x\to 1^-}f(x)\ne\lim_{x\to 1^+}f(x)
i.e., 3\ne4
Also, value of f(x) at x=1\in[0,1], is
f(1)=3
Since LHL ≠ RHL, the limit \displaystyle\lim_{x\to1}f(x) does not exist.
Hence, the function is discontinuous at x=1.
At x=3
As x\to 3^-, we have x<3. Therefore, we use
f(x)=4
Thus,
\displaystyle\lim_{x\to 3^-}f(x)=4
As x\to 3^+, we have x\gt 3. Therefore, we use
f(x)=5
Thus,
\displaystyle\lim_{x\to 3^+}f(x)=5
Since
\displaystyle\lim_{x\to 3^-}f(x)\ne\lim_{x\to 3^+}f(x)
i.e., 4\ne5
Also, value of f(x) at x=3\in[3,10], is
f(3)=5
Since LHL ≠ RHL, the limit \displaystyle\lim_{x\to3}f(x) does not exist.
Hence, the function is discontinuous at x=3.
Thus, the function is continuous on the given intervals but discontinuous at x=1 and x=3.
Practice is the key to mastering Continuity Ex 5.1. Solve questions step by step and try to understand the concept. If you have any doubt, feel free to leave a comment.
Question 15: Continuity 5.1
15. Discuss the continuity of the function
f(x)=\begin{cases} 2x, & \text{if } x<0 \\ \\ 0, & \text{if } 0\le x\le 1 \\ \\ 4x, & \text{if } x>1 \end{cases}
Solution
For x<0, f(x)=2x, which is a linear polynomial and hence continuous.
For 0\lt x\lt 1, f(x)=0, which is a constant function and hence continuous.
And for x>1, f(x)=4x, which is also a linear polynomial and hence continuous.
Therefore, we only need to examine continuity at x=0 and x=1.
At x=0
As x\to 0^-, we have x<0. Therefore, we use
f(x)=2x
Thus,
\displaystyle\lim_{x\to 0^-}f(x)=\lim_{x\to 0^-}(2x)=0
As x\to 0^+, we have x\gt 0. Therefore, we use
f(x)=0
Thus,
\displaystyle\lim_{x\to 0^+}f(x)=0
Since
\displaystyle\lim_{x\to 0^-}f(x)=\lim_{x\to 0^+}f(x)
i.e., 0=0
Also, value of f(x) at x=0\in[0,1], is
f(0)=0
Thus,
\displaystyle\lim_{x\to 0}f(x)=f(0)=0
Hence, the function is continuous at x=0.
At x=1
As x\to 1^-, we have x\lt 1. Therefore, we use
f(x)=0
Thus,
\displaystyle\lim_{x\to 1^-}f(x)=0
As x\to 1^+, we have x>1. Therefore, we use
f(x)=4x
Thus,
\displaystyle\lim_{x\to 1^+}f(x)=\lim_{x\to 1^+}(4x)=4
Since
\displaystyle\lim_{x\to 1^-}f(x)\ne\lim_{x\to 1^+}f(x)
i.e., 0\ne 4
Also, value of f(x) at x=1\in[0,1], is
f(1)=0
Since LHL ≠ RHL, the limit \displaystyle\lim_{x\to1}f(x) does not exist.
Hence, the function is discontinuous at x=1.
Thus, the function is continuous for all values in its domain except at x=1.
Question 16: Continuity and Functions
16. Discuss the continuity of the function
f(x)=\begin{cases} -2, & \text{if } x\le -1 \\ \\ 2x, & \text{if } -1<x\le 1 \\ \\ 2, & \text{if } x>1 \end{cases}
Solution
For x\lt -1, f(x)=-2, which is a constant function and hence continuous.
For -1<x\lt 1, f(x)=2xf, which is a linear polynomial and hence continuous.
And for x>1, f(x)=2, which is also a constant function and hence continuous.
Therefore, we only need to examine continuity at x=-1 and x=1.
At x=-1
As x\to -1^-, we have x\lt -1. Therefore, we use
f(x)=-2
Thus,
\displaystyle\lim_{x\to -1^-}f(x)=-2
As x\to -1^+, we have x\gt -1. Therefore, we use
f(x)=2x
Thus,
\displaystyle\lim_{x\to -1^+}f(x)=\lim_{x\to -1^+}(2x)=-2
Since
\displaystyle\lim_{x\to -1^-}f(x)=\lim_{x\to -1^+}f(x)
i.e., -2=-2
Also, value of f(x) at x=-1 (\le -1), is
f(-1)=-2
Thus,
\displaystyle\lim_{x\to -1}f(x)=f(-1)=-2
Hence, the function is continuous at x=-1.
At x=1
As x\to 1^-, we have x\lt 1. Therefore, we use
f(x)=2x
Thus,
\displaystyle\lim_{x\to 1^-}f(x)=\lim_{x\to 1^-}(2x)=2
As x\to 1^+, we have x>1. Therefore, we use
f(x)=2
Thus,
\displaystyle\lim_{x\to 1^+}f(x)=2
Since
\displaystyle\lim_{x\to 1^-}f(x)=\lim_{x\to 1^+}f(x)
i.e., 2=2
Also, value of f(x) at x=1\in(-1,1], is
f(1)=2(1)=2
Thus,
\displaystyle\lim_{x\to 1}f(x)=f(1)=2
Hence, the function is continuous at x=1.
Therefore, the function is continuous throughout its domain.
You may already be following Maths Better for NCERT Solutions for various topics. Likewise, this exercise on Continuity and Differentiability for Class 12 Maths will strengthen your concepts and improve step-by-step problem-solving skills.
Now, let’s proceed to the next question of Continuity Ex 5.1.
Question 17: Continuity 5.1
17. Find the relationship between a and b so that the function
f(x)=\begin{cases} ax+1, & \text{if } x\le 3 \\ \\ bx+3, & \text{if } x>3 \end{cases}
is continuous at x=3.
Solution
Both ax+1 and bx+3 are linear polynomials and hence continuous.
Therefore, we only need to examine continuity at x=3, where the definition changes.
For continuity at x=3,
\displaystyle\lim_{x\to 3^-}f(x)=\lim_{x\to 3^+}f(x)=f(3)
As x\to 3^-, we have x\lt 3. Therefore, we use
f(x)=ax+1
Thus,
\displaystyle\lim_{x\to 3^-}f(x)=3a+1
As x\to 3^+, we have x>3. Therefore, we use
f(x)=bx+3
Thus,
\displaystyle\lim_{x\to 3^+}f(x)=3b+3
Since the function is continuous at x=3,
3a+1=3b+3
Therefore,
3a-3b=2
or
a-b=\frac{2}{3}
Hence, the required relationship between a and b is
a-b=\frac{2}{3}
Question 18: Exercise 5.1
18. For what value of \lambda is the function
f(x)=\begin{cases} \lambda(x^2-2x), & \text{if } x\le 0 \\ \\ 4x+1, & \text{if } x>0 \end{cases}
continuous at x=0? What about continuity at x=1?
Solution
Continuity at x=0
Both pieces are polynomial functions and hence continuous on their respective intervals.
Therefore, we only need to examine continuity at x=0, where the definition changes.
For continuity at x=0,
\displaystyle\lim_{x\to 0^-}f(x)=\lim_{x\to 0^+}f(x)=f(0)
As x\to 0^-, we have x\lt 0. Therefore, we use
f(x)=\lambda(x^2-2x)
Thus,
\displaystyle\lim_{x\to 0^-}f(x)=\lambda(0^2-2\cdot0)=0
As x\to 0^+, we have x>0. Therefore, we use
f(x)=4x+1
Thus,
\displaystyle\lim_{x\to 0^+}f(x)=4(0)+1=1
Since
\displaystyle\lim_{x\to 0^-}f(x)\ne\lim_{x\to 0^+}f(x)
i.e., 0\ne1
the limit \displaystyle\lim_{x\to 0}f(x) does not exist.
Hence, the function is discontinuous at x=0 for every value of \lambda.
Therefore, there is no value of \lambda for which the function is continuous at x=0.
Continuity at x=1
Since 1>0, there is no change in the definition of the function near x=1.
Therefore, in a neighbourhood of x=1,
f(x)=4x+1
Thus,
\displaystyle\lim_{x\to 1}f(x)=\lim_{x\to 1}(4x+1)=4(1)+1=5
Also, value of f(x) at x=1(\gt 0), is
f(1)=4(1)+1=5
Thus,
\displaystyle\lim_{x\to 1}f(x)=f(1)=5
Hence, the function is continuous at x=1 for every value of \lambda.
Question 19: Continuity 5.1
19. Show that the function g(x)=x-[x] is discontinuous at all integral points, where [x] denotes the greatest integer less than or equal to x.
Solution
The function g(x)=x-[x] is called the fractional part function.
To show that g(x) is discontinuous at all integral points, let n be any integer.
We shall examine continuity at x=n.
As x\to n^-, we have
n-1<x<n
Therefore,
[x]=n-1
Hence,
g(x)=x-(n-1)
Thus,
\displaystyle\lim_{x\to n^-}g(x)=n-(n-1)=1
As x\to n^+, we have
n\lt x<n+1
Therefore,
[x]=n
Hence,
g(x)=x-n
Thus,
\displaystyle\lim_{x\to n^+}g(x)=n-n=0
Since
\displaystyle\lim_{x\to n^-}g(x)\ne\lim_{x\to n^+}g(x)
i.e., 1\ne0
Also, value of g(x) at x=n, is
g(n)=n-[n]=n-n=0
Since LHL ≠ RHL, the limit \displaystyle\lim_{x\to n}g(x) does not exist.
Hence, g(x)=x-[x] is discontinuous at x=n.
Since n is an arbitrary integer, the function is discontinuous at every integral point.
The fractional part function x-[x] is continuous for all non-integral values of x and discontinuous at every integer.
Question 20: Continuity at a Point
20. Is the function f(x)=x^2-\sin x+5 continuous at x=\pi?
Solution
Polynomial and trigonometric functions are continuous for all real values of x.
Since x^2, \sin x and the constant function 5 are continuous for all real values of x, their sum and difference are also continuous.
Therefore, f(x)=x^2-\sin x+5 is continuous for all real values of x.
To examine continuity at x=\pi,
\displaystyle\lim_{x\to\pi}f(x)=\lim_{x\to\pi}(x^2-\sin x+5)
=\pi^2-\sin\pi+5
=\pi^2+5
Also, value of f(x) at x=\pi, is
f(\pi)=\pi^2-\sin\pi+5=\pi^2+5
Thus,
\displaystyle\lim_{x\to\pi}f(x)=f(\pi)=\pi^2+5
Hence, the function is continuous at x=\pi.
Question 21: Continuous Functions
21. Discuss the continuity of the following functions:
(a) Given, f(x)=\sin x+\cos x
The sum of two continuous functions is continuous.
Since \sin x and \cos x are continuous for all real values of x, their sum is also continuous.
To examine, let c be any real number.
Then,
\displaystyle\lim_{x\to c}(\sin x+\cos x)=\sin c+\cos c
Also,
f(c)=\sin c+\cos c
Thus, \displaystyle\lim_{x\to c}f(x)=f(c).
Hence, f(x)=\sin x+\cos x is continuous for all real values of x.
(b) Given, f(x)=\sin x-\cos x
The difference of two continuous functions is continuous.
Since \sin x and \cos x are continuous for all real values of x, their difference is also continuous.
To examine, let c be any real number.
Then,
\displaystyle\lim_{x\to c}(\sin x-\cos x)=\sin c-\cos c
Also,
f(c)=\sin c-\cos c
Thus, \displaystyle\lim_{x\to c}f(x)=f(c).
Hence, f(x)=\sin x-\cos x is continuous for all real values of x.
(c) Given, f(x)=\sin x\cdot\cos x
The product of two continuous functions is continuous.
Since \sin x and \cos x are continuous for all real values of x, their product is also continuous.
To examine, let c be any real number.
Then,
\displaystyle\lim_{x\to c}(\sin x\cdot\cos x)=\sin c\cdot\cos c
Also,
f(c)=\sin c\cdot\cos c
Thus, \displaystyle\lim_{x\to c}f(x)=f(c).
Hence, f(x)=\sin x\cdot\cos x is continuous for all real values of x.
NCERT Class 12 Maths includes 61 exercises across both Part 1 and Part 2, and I’ll cover them one by one with clear explanations and step-by-step solutions. Many of these questions are also available in video format on my YouTube channel, @Mathsbetter, to help you understand the concepts more visually.
Now, let’s move to the next question.
Question 22: Continuity 5.1
22. Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
Solution
(a) Cosine Function
Given, f(x)=\cos x
The cosine function is continuous for all real values of x.
Let c be any real number.
Then,
\displaystyle\lim_{x\to c}\cos x=\cos c
Also,
f(c)=\cos c
Thus, \displaystyle\lim_{x\to c}f(x)=f(c).
Hence, \cos x is continuous for all real values of x.
(b) Cosecant Function
Given, f(x)=\cosec x=\frac{1}{\sin x}
A quotient of continuous functions is continuous wherever the denominator is non-zero.
Since \sin x is continuous for all real values of x, \cosec x is continuous wherever \sin x\ne0.
Now, \sin x=0 when x=n\pi, where n\in\mathbb{Z}.
Therefore, \cosec x is continuous for all x\ne n\pi and discontinuous at x=n\pi.
(c) Secant Function
Given, f(x)=\sec x=\frac{1}{\cos x}
A quotient of continuous functions is continuous wherever the denominator is non-zero.
Since \cos x is continuous for all real values of x, \sec x is continuous wherever \cos x\ne0.
Now, \cos x=0 when
x=\frac{(2n+1)\pi}{2},\quad n\in\mathbb{Z}
Therefore, \sec x is continuous for all x\ne\frac{(2n+1)\pi}{2} and discontinuous at these points.
(d) Cotangent Function
Given, f(x)=\cot x=\frac{\cos x}{\sin x}
A quotient of continuous functions is continuous wherever the denominator is non-zero.
Since \sin x and \cos x are continuous for all real values of x, \cot x is continuous wherever \sin x\ne0.
Now, \sin x=0 when x=n\pi, where n\in\mathbb{Z}.
Therefore, \cot x is continuous for all x\ne n\pi and discontinuous at x=n\pi.
\cos x is continuous for all real values of x, whereas \cosec x, \sec x and \cot x are continuous wherever they are defined.
Question 23: Points of Discontinuity
23. Find all points of discontinuity of
f(x)=\begin{cases} \dfrac{\sin x}{x}, & \text{if } x<0 \\ \\ x+1, & \text{if } x\ge 0 \end{cases}
Solution
For x<0, f(x)=\dfrac{\sin x}{x}.
Since \sin x and x are continuous functions and x\ne 0 for x<0, the quotient \dfrac{\sin x}{x} is continuous for all x<0.
For x\gt 0, f(x)=x+1, which is a linear polynomial and hence continuous.
Therefore, the only possible point of discontinuity is x=0.
\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1
To examine continuity at x=0, we find the left-hand limit and right-hand limit.
As x\to 0^-, we have x<0. Therefore, we use
f(x)=\dfrac{\sin x}{x}
Using the standard limit
\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1
we get
\displaystyle\lim_{x\to 0^-}f(x)=1
As x\to 0^+, we have x\gt 0. Therefore, we use
f(x)=x+1
Thus,
\displaystyle\lim_{x\to 0^+}f(x)=0+1=1
Since
\displaystyle\lim_{x\to 0^-}f(x)=\lim_{x\to 0^+}f(x)
i.e., 1=1
Also, value of f(x) at x=0, is
f(0)=0+1=1
Thus,
\displaystyle\lim_{x\to 0}f(x)=f(0)=1
Hence, the function is continuous at x=0.
Therefore, the function has no points of discontinuity.
Question 24: Continuity 5.1
24. Determine if
f(x)=\begin{cases} x^2\sin\frac{1}{x}, & \text{if } x\ne 0 \\ \\ 0, & \text{if } x=0 \end{cases}
is a continuous function?
Solution
For x\ne 0, the functions x^2 and \sin\frac{1}{x} are continuous.
Therefore, their product
f(x)=x^2\sin\frac{1}{x}
is continuous for all x\ne 0.
Hence, the only possible point of discontinuity is x=0.
To examine continuity at x=0, we find
\displaystyle\lim_{x\to 0}x^2\sin\frac{1}{x}
For every real number \theta, -1\le \sin\theta \le 1.
Therefore, we have
-1\le\sin\frac{1}{x}\le1
multiplying throughout by x^2 gives
-x^2\le x^2\sin\frac{1}{x}\le x^2
Now,
\displaystyle\lim_{x\to0}(-x^2)=0
and
\displaystyle\lim_{x\to0}(x^2)=0
Therefore,
\displaystyle\lim_{x\to0}x^2\sin\frac{1}{x}=0
Also, value of f(x) at x=0, is
f(0)=0
Thus,
\displaystyle\lim_{x\to 0}f(x)=f(0)=0
Hence, the function is continuous at x=0.
Since f(x) is continuous for all x\ne 0 and also continuous at x=0, the function is continuous for all real values of x.
Therefore, f(x) is a continuous function.
Squeeze Theorem (Sandwich Theorem): If g(x)\le f(x)\le h(x) for all x near a and
\displaystyle\lim_{x\to a}g(x)=\lim_{x\to a}h(x)=L
then
\displaystyle\lim_{x\to a}f(x)=L
Alternative Method (By Sandwich/Squeeze Theorem):
Now, since -1\le \sin\frac{1}{x}\le 1, we have
-x^2\le x^2\sin\frac{1}{x}\le x^2
Also,
\displaystyle\lim_{x\to0}(-x^2)=0 and \displaystyle\lim_{x\to0}(x^2)=0
Therefore, by the Squeeze Theorem,
\displaystyle\lim_{x\to0}x^2\sin\frac{1}{x}=0
And now proceed as above.
Here’s the next question of this exercise.
Question 25: Continuity Exercise 5.1
25. Examine the continuity of
f(x)=\begin{cases} \sin x-\cos x, & \text{if } x\ne 0 \\ \\ -1, & \text{if } x=0 \end{cases}
Solution
For x\ne 0, \sin x and \cos x are continuous functions.
Therefore, their difference
f(x)=\sin x-\cos x
is continuous for all x\ne 0.
Hence, the only possible point of discontinuity is x=0.
To examine continuity at x=0, we find
\displaystyle\lim_{x\to 0}f(x)
Since \sin x and \cos x are continuous at x=0,
\displaystyle\lim_{x\to 0}(\sin x-\cos x)=\sin 0-\cos 0
=0-1
=-1
Also, for the value of the function
As x\to 0, we have x\ne 0. Therefore, we use
f(x)=\sin x-\cos x
f(0)=-1
Thus,
\displaystyle\lim_{x\to 0}f(x)=f(0)=-1
Hence, the function is continuous at x=0.
Since f(x) is continuous for all x\ne 0 and also continuous at x=0, the function is continuous for all real values of x.
Therefore, f(x) is a continuous function.
Question 26: Exercise 5.1
26. Find the value of k so that the function
f(x)=\begin{cases} \dfrac{k\cos x}{\pi-2x}, & \text{if } x\ne \dfrac{\pi}{2} \\ \\ 3, & \text{if } x=\dfrac{\pi}{2} \end{cases}
is continuous at x=\dfrac{\pi}{2}.
Solution
For continuity at
x=\dfrac{\pi}{2}
we must have
\displaystyle\lim_{x\to \frac{\pi}{2}}f(x)=f\left(\frac{\pi}{2}\right)
Since x\ne\dfrac{\pi}{2} near the point \dfrac{\pi}{2}, we use
f(x)=\dfrac{k\cos x}{\pi-2x}
Thus,
\displaystyle\lim_{x\to \frac{\pi}{2}}f(x)=\displaystyle\lim_{x\to \frac{\pi}{2}}\frac{k\cos x}{\pi-2x}
=k\displaystyle\lim_{x\to \frac{\pi}{2}}\frac{\cos x}{\pi-2x}
Since
\pi-2x=-2\left(x-\frac{\pi}{2}\right)
we get
\displaystyle\lim_{x\to \frac{\pi}{2}}f(x)=-\frac{k}{2}\displaystyle\lim_{x\to \frac{\pi}{2}}\frac{\cos x}{x-\frac{\pi}{2}}
Let
h=x-\frac{\pi}{2}
Then, as x\to\frac{\pi}{2}, we have h\to0.
\cos x=\cos\left(\frac{\pi}{2}+h\right)=-\sin h
and
\pi-2x=\pi-2\left(\frac{\pi}{2}+h\right)=-2h
Therefore,
\displaystyle\lim_{x\to \frac{\pi}{2}}f(x)=k\displaystyle\lim_{h\to0}\frac{-\sin h}{-2h}
=\frac{k}{2}\displaystyle\lim_{h\to0}\frac{\sin h}{h}
=\frac{k}{2}\qquad\left(\because\ \displaystyle\lim_{h\to0}\frac{\sin h}{h}=1\right)
Also, value of f(x) at x=\frac{\pi}{2}, is
f\left(\frac{\pi}{2}\right)=3
For continuity,
\displaystyle\lim_{x\to\frac{\pi}{2}}f(x)=f\left(\frac{\pi}{2}\right)
\frac{k}{2}=3
k=6
Therefore, the function is continuous at x=\frac{\pi}{2} when k=6.
Question 27: Continuity Ex 5.1
27. Find the value of k so that the function
f(x)=\begin{cases} kx^2, & \text{if } x\le 2 \\ \\ 3, & \text{if } x>2 \end{cases}
is continuous at x=2.
Solution
Since both kx^2 and 3 are polynomial and constant functions resp., so will be continuous on their respective intervals.
Therefore, we only need to examine continuity at x=2, where the definition changes.
For continuity at x=2,
\displaystyle\lim_{x\to 2^-}f(x)=\lim_{x\to 2^+}f(x)=f(2)
As x\to 2^-, we have x\lt 2. Therefore, we use
f(x)=kx^2
Thus,
\displaystyle\lim_{x\to 2^-}f(x)=k(2)^2=4k
As x\to 2^+, we have x>2. Therefore, we use
f(x)=3
Thus,
\displaystyle\lim_{x\to 2^+}f(x)=3
Since the function is continuous at x=2,
4k=3
Therefore,
k=\frac{3}{4}
Also, value of f(x) at x=2(\lt 2), is
f(2)=k(2)^2=4k=4\left(\frac{3}{4}\right)=3
Hence, the function is continuous at x=2 when
k=\frac{3}{4}
Question 28: Continuity and Differentiability
28. Find the value of k so that the function
f(x)=\begin{cases} kx+1, & \text{if } x\le \pi \\ \\ \cos x, & \text{if } x>\pi \end{cases}
is continuous at x=\pi.
Solution
Both kx+1 and \cos x being polynomial and cosine functions resp. So they are continuous on their respective intervals.
Therefore, we only need to examine continuity at x=\pi, where the definition changes.
For continuity at x=\pi,
\displaystyle\lim_{x\to \pi^-}f(x)=\lim_{x\to \pi^+}f(x)=f(\pi)
As x\to \pi^-, we have x\lt \pi. Therefore, we use
f(x)=kx+1
Thus,
\displaystyle\lim_{x\to \pi^-}f(x)=k\pi+1
As x\to \pi^+, we have x>\pi. Therefore, we use
f(x)=\cos x
Thus,
\displaystyle\lim_{x\to \pi^+}f(x)=\cos\pi=-1
Since the function is continuous at x=\pi,
k\pi+1=-1
Therefore,
k\pi=-2
k=-\frac{2}{\pi}
Also, value of f(x) at x=\pi(\le \pi), is
f(\pi)=k\pi+1
=\left(-\frac{2}{\pi}\right)\pi+1
=-2+1=-1
Thus,
\displaystyle\lim_{x\to \pi}f(x)=f(\pi)=-1
Hence, the function is continuous at x=\pi when
k=-\frac{2}{\pi}
Question 29: Continuity 5.1
29. Find the value of k so that the function
f(x)=\begin{cases} kx+1, & \text{if } x\le 5 \\ \\ 3x-5, & \text{if } x>5 \end{cases}
is continuous at x=5.
Solution
Both kx+1 and 3x-5 are linear polynomials and hence continuous on their respective intervals.
Therefore, we only need to examine continuity at x=5, where the definition changes.
For continuity at x=5,
\displaystyle\lim_{x\to 5^-}f(x)=\lim_{x\to 5^+}f(x)=f(5)
As x\to 5^-, we have x\lt 5. Therefore, we use
f(x)=kx+1
Thus,
\displaystyle\lim_{x\to 5^-}f(x)=5k+1
As x\to 5^+, we have x>5. Therefore, we use
f(x)=3x-5
Thus,
\displaystyle\lim_{x\to 5^+}f(x)=3(5)-5=10
Since the function is continuous at x=5,
5k+1=10
Therefore,
5k=9
k=\frac{9}{5}
Also, value of f(x) at x=5(\le 5), is
f(5)=5k+1
=5\left(\frac{9}{5}\right)+1
=9+1=10
Thus,
\displaystyle\lim_{x\to 5}f(x)=f(5)=10
Hence, the function is continuous at x=5 when
k=\frac{9}{5}
NCERT Class 12 Maths includes several important exercises across both Part 1 and Part 2, and I’ll cover them one by one with clear explanations and step-by-step solutions. Many of these questions are also available in video format on my YouTube channel, @Mathsbetter, to help you understand the concepts more visually.
Now, let’s move to the next question.
Question 30: Continuity 5.1
30. Find the values of a and b such that the function
f(x)=\begin{cases} 5, & \text{if } x\le 2 \\ \\ ax+b, & \text{if } 2<x<10 \\ \\ 21, & \text{if } x\ge 10 \end{cases}
is a continuous function.
Solution
Each piece of the function is continuous on its respective interval because they are either constant or polynomial.
Therefore, we only need to examine continuity at the points where the definition changes, namely x=2 and x=10.
At x=2
As x\to 2^-, we have x\lt 2. Therefore, we use
f(x)=5
Thus,
\displaystyle\lim_{x\to 2^-}f(x)=5
As x\to 2^+, we have x>2. Therefore, we use
f(x)=ax+b
Thus,
\displaystyle\lim_{x\to 2^+}f(x)=2a+b
Since the function is continuous at x=2,
2a+b=5
At x=10
As x\to 10^-, we have x<10. Therefore, we use
f(x)=ax+b
Thus,
\displaystyle\lim_{x\to 10^-}f(x)=10a+b
As x\to 10^+, we have x\gt 10. Therefore, we use
f(x)=21
Thus,
\displaystyle\lim_{x\to 10^+}f(x)=21
Since the function is continuous at x=10,
10a+b=21
We now solve the equations
2a+b=5
and
10a+b=21
Subtracting,
8a=16
a=2
Substituting in 2a+b=5,
2(2)+b=5
4+b=5
b=1
Hence,
a=2,\qquad b=1
Therefore, the function is continuous when a=2 and b=1.
Question 31: Continuity Ex 5.1
31. Show that the function f(x)=\cos(x^2) is a continuous function.
Solution
If f and g are continuous, then the composite function (g\circ f)(x) is also continuous.
Since x^2 is a polynomial function, it is continuous for all real values of x.
Also, \cos x is continuous for all real values of x.
Therefore, the composite function
f(x)=\cos(x^2)
is continuous for all real values of x.
To examine, let c be any real number.
Then,
\displaystyle\lim_{x\to c}\cos(x^2)=\cos\left(\displaystyle\lim_{x\to c}x^2\right)
=\cos(c^2)
Also,
f(c)=\cos(c^2)
Thus,
\displaystyle\lim_{x\to c}f(x)=f(c)
Hence, f(x)=\cos(x^2) is continuous for all real values of x.
Question 32: Exercise 5.1
32. Show that the function f(x)=|\cos x| is a continuous function.
Solution
If f and g are continuous, then the composite function (g\circ f)(x) is also continuous.
Since \cos x is continuous for all real values of x.
Also, the modulus function |x| is continuous for all real values of x.
Therefore, the composite function
f(x)=|\cos x|
is continuous for all real values of x.
To examine, let c be any real number.
Then,
\displaystyle\lim_{x\to c}|\cos x|=\left|\displaystyle\lim_{x\to c}\cos x\right|
=|\cos c|
Also,
f(c)=|\cos c|
Thus,
\displaystyle\lim_{x\to c}f(x)=f(c)
Hence, f(x)=|\cos x| is continuous for all real values of x.
Question 33: NCERT Ex. 5.1
33. Examine that f(x)=\sin|x| is a continuous function.
Solution
If f and g are continuous, then the composite function (g\circ f)(x) is also continuous.
Since the modulus function |x| is continuous for all real values of x.
Also, \sin x is continuous for all real values of x.
Therefore, the composite function
f(x)=\sin|x|
is continuous for all real values of x.
To examine, let c be any real number.
Then,
\displaystyle\lim_{x\to c}\sin|x|=\sin\left(\displaystyle\lim_{x\to c}|x|\right)
=\sin|c|
Also,
f(c)=\sin|c|
Thus,
\displaystyle\lim_{x\to c}f(x)=f(c)
Hence, f(x)=\sin|x| is continuous for all real values of x.
Question 34: Continuity 5.1
34. Find all the points of discontinuity of f(x)=|x|-|x+1|
Solution
The difference of two continuous functions is continuous.
The modulus function |x| is continuous for all real values of x.
Also, |x+1| is continuous for all real values of x because x+1 is a polynomial and the modulus function is continuous.
Therefore,
f(x)=|x|-|x+1|
is the difference of two continuous functions and hence is continuous for all real values of x.
Hence, the function has no points of discontinuity.
Note: The expressions inside the modulus signs change sign at x=-1 and x=0. However, modulus functions are continuous at these points. Therefore, no discontinuity occurs.
Common Mistakes to Avoid
- Checking continuity only by looking at the formula: A function may appear simple but can still be discontinuous at points where it is not defined or where its definition changes.
- Ignoring the function value: Many students check only the limit and forget to compare it with f(c). For continuity at x=c, both must be equal.
- Not verifying continuity at boundary points of piecewise functions: When a function is defined differently on different intervals, continuity must be checked at the points where the definition changes.
- Confusing continuity with existence of a limit: A limit may exist at a point, but the function is continuous only if the function is also defined there and the limit equals the function value.
- Forgetting to check LHL and RHL separately: For piecewise functions, both the left-hand limit and right-hand limit must be calculated before concluding whether the limit exists.
- Assuming rational functions are continuous everywhere: Rational functions are continuous only where their denominators are non-zero. Points where the denominator becomes zero must always be checked separately.
- Ignoring points where a function is not defined: If f(c) does not exist, then the function cannot be continuous at x=c, even if the limit exists.
- Using the wrong branch while finding LHL or RHL: In piecewise functions, the correct expression must be chosen according to whether x approaches the point from the left or from the right.
- Not checking all possible points of discontinuity: In functions involving multiple intervals, every point where the rule changes should be examined.
- Ignoring continuity theorems: Once basic functions are known to be continuous, results such as sum, difference, product, quotient and composition of continuous functions can be used directly to simplify the solution.
Continue Learning
After completing Exercise 5.1 of Continuity and Differentiability, you should now be comfortable with the concept of continuity of a function, identifying points of discontinuity and applying continuity conditions to piecewise and parameter-based functions.
To strengthen your understanding further, make sure that you revise:
- Meaning of continuity of a function at a point
- Condition for continuity: \displaystyle\lim_{x\to c}f(x)=f(c)
- Relationship between LHL, RHL and continuity
- Difference between continuous and discontinuous functions
- Identifying points of discontinuity in piecewise functions
- Continuity of polynomial, rational, modulus, trigonometric, exponential and logarithmic functions
- Algebra of continuous functions such as sum, difference, product and quotient
- Composition of continuous functions and composite functions like \cos(x^2) and \sin|x|
- Finding unknown constants using continuity conditions
- Special limits involving trigonometric functions and their use in continuity problems
Explore More
To master the concept of continuity, practise examining different types of functions and determine whether they are continuous or discontinuous at given points. Regular practice with limits, piecewise functions and continuity conditions will help build a strong foundation for differentiability in the next exercise.
All the best and keep learning 👍
