MATHEMATICS CLASS- 12
CHAPTER-5
(CONTINUITY AND
DIFFERENTIABILITY)
Exercise 5.1
Q.1 Prove that the function $f(x)=5x-3$ is continuous at $x=0$, at $x=-3$ and at $x=5$.
Solution:
A function $f(x)$ is said to be continuous at $x=a$ if
$$ \lim_{x \to a}f(x)=f(a). $$Given,
$$ f(x)=5x-3. $$We verify the condition of continuity at each given point.
(i) At $x=0$
The value of the function is
$$ \begin{aligned} f(0) &=5(0)-3\\ &=-3. \end{aligned} $$Now,
$$ \begin{aligned} \lim_{x\to0}f(x) &=\lim_{x\to0}(5x-3)\\ &=5\left(\lim_{x\to0}x\right)-3\\ &=5(0)-3\\ &=-3. \end{aligned} $$Since
$$ \lim_{x\to0}f(x)=f(0)=-3, $$the function is continuous at $x=0$.
(ii) At $x=-3$
The value of the function is
$$ \begin{aligned} f(-3) &=5(-3)-3\\ &=-15-3\\ &=-18. \end{aligned} $$Now,
$$ \begin{aligned} \lim_{x\to-3}f(x) &=\lim_{x\to-3}(5x-3)\\ &=5(-3)-3\\ &=-18. \end{aligned} $$Hence,
$$ \lim_{x\to-3}f(x)=f(-3). $$Therefore, the function is continuous at $x=-3$.
(iii) At $x=5$
The value of the function is
$$ \begin{aligned} f(5) &=5(5)-3\\ &=25-3\\ &=22. \end{aligned} $$Now,
$$ \begin{aligned} \lim_{x\to5}f(x) &=\lim_{x\to5}(5x-3)\\ &=5(5)-3\\ &=22. \end{aligned} $$Thus,
$$ \lim_{x\to5}f(x)=f(5)=22. $$Hence, the function is continuous at $x=5$.
Conclusion:
Since
$$ \lim_{x\to a}f(x)=f(a) $$at each of the points $a=0,-3$ and $5$, the function
$$ f(x)=5x-3 $$is continuous at $x=0$, $x=-3$ and $x=5$.
Q.2 Examine the continuity of the function $f(x)=2x^2-1$ at $x=3$.
Solution:
A function $f(x)$ is continuous at $x=a$ if
$$ \lim_{x\to a}f(x)=f(a). $$Given,
$$ f(x)=2x^2-1. $$We shall verify the condition of continuity at $x=3$.
Step 1: Find the value of the function at $x=3$.
$$ \begin{aligned} f(3) &=2(3)^2-1\\ &=2(9)-1\\ &=18-1\\ &=17. \end{aligned} $$Step 2: Find the limit of the function as $x$ approaches $3$.
$$ \begin{aligned} \lim_{x\to3}f(x) &=\lim_{x\to3}(2x^2-1)\\ &=2\left(\lim_{x\to3}x\right)^2-1\\ &=2(3)^2-1\\ &=18-1\\ &=17. \end{aligned} $$Step 3: Compare the limit and the function value.
$$ \lim_{x\to3}f(x)=17=f(3). $$Since
$$ \lim_{x\to3}f(x)=f(3), $$the function satisfies the condition of continuity.
Conclusion:
Therefore, the function
$$ f(x)=2x^2-1 $$is continuous at $x=3$.
Q.3 Examine the following functions for continuity.
(a) $f(x)=x-5$
Solution:
A function $f(x)$ is continuous at $x=a$ if
$$ \lim_{x\to a}f(x)=f(a). $$Given,
$$ f(x)=x-5. $$Let $a$ be any real number.
The value of the function at $x=a$ is
$$ f(a)=a-5. $$Now,
$$ \begin{aligned} \lim_{x\to a}f(x) &=\lim_{x\to a}(x-5)\\ &=\lim_{x\to a}x-5\\ &=a-5. \end{aligned} $$Therefore,
$$ \lim_{x\to a}f(x)=a-5=f(a). $$Since the limit is equal to the function value for every real number $a$, the function is continuous for all real values of $x$.
Conclusion:
The function
$$ f(x)=x-5 $$is continuous for all real numbers.
(b) $f(x)=\dfrac{1}{x-5},\;x\ne5$
Solution:
Given,
$$ f(x)=\frac{1}{x-5}, \qquad x\ne5. $$Let $a\ne5$.
The value of the function is
$$ f(a)=\frac{1}{a-5}. $$Now,
$$ \begin{aligned} \lim_{x\to a}f(x) &=\lim_{x\to a}\frac{1}{x-5}\\ &=\frac{1}{a-5}\\ &=f(a). \end{aligned} $$Hence, the function is continuous at every point where it is defined.
At $x=5$,
$$ f(5) $$is not defined because the denominator becomes zero.
Conclusion:
The function is continuous for all $x\ne5$ and discontinuous at $x=5$.
(c) $f(x)=\dfrac{x^2-25}{x+5},\;x\ne-5$
Solution:
Given,
$$ f(x)=\frac{x^2-25}{x+5}, \qquad x\ne-5. $$Factorising the numerator,
$$ x^2-25=(x-5)(x+5). $$Therefore,
$$ \begin{aligned} f(x) &=\frac{(x-5)(x+5)}{x+5}\\ &=x-5,\qquad x\ne-5. \end{aligned} $$Let $a\ne-5$.
$$ \begin{aligned} \lim_{x\to a}f(x) &=\lim_{x\to a}(x-5)\\ &=a-5\\ &=f(a). \end{aligned} $$Hence, the function is continuous at every point except $x=-5$.
At $x=-5$,
$$ f(-5) $$is not defined.
Although
$$ \begin{aligned} \lim_{x\to-5}f(x) &=\lim_{x\to-5}(x-5)\\ &=-5-5\\ &=-10, \end{aligned} $$the function value does not exist.
Conclusion:
The function is continuous for all $x\ne-5$ and discontinuous at $x=-5$.
(d) $f(x)=|x-5|$
Solution:
Given,
$$ f(x)=|x-5|. $$The modulus function is continuous for every real number.
Let $a$ be any real number.
Then
$$ f(a)=|a-5|. $$Also,
$$ \begin{aligned} \lim_{x\to a}|x-5| &=|a-5|\\ &=f(a). \end{aligned} $$Since
$$ \lim_{x\to a}f(x)=f(a), $$the function satisfies the condition of continuity.
Conclusion:
The function
$$ f(x)=|x-5| $$is continuous for all real values of $x$.
Q.4 Prove that the function $f(x)=x^n$ is continuous at $x=n$, where $n$ is a positive integer.
Solution:
A function $f(x)$ is continuous at $x=a$ if
$$ \lim_{x\to a}f(x)=f(a). $$Given,
$$ f(x)=x^n, $$where $n$ is a positive integer.
We have to prove that the function is continuous at $x=n$.
Step 1: Find the value of the function at $x=n$.
$$ \begin{aligned} f(n) &=n^n. \end{aligned} $$Step 2: Find the limit of the function as $x$ approaches $n$.
$$ \begin{aligned} \lim_{x\to n}f(x) &=\lim_{x\to n}x^n\\ &=\left(\lim_{x\to n}x\right)^n\\ &=n^n. \end{aligned} $$Step 3: Compare the limit and the function value.
$$ \lim_{x\to n}f(x)=n^n=f(n). $$Since
$$ \lim_{x\to n}f(x)=f(n), $$the function satisfies the condition of continuity.
Conclusion:
Hence, the function
$$ f(x)=x^n $$is continuous at $x=n$, where $n$ is a positive integer.
Q.5 Is the function defined by
$$ f(x)= \begin{cases} x, & x\le1,\\ 5, & x>1 \end{cases} $$continuous at $x=0$, $x=1$ and $x=2$?
Solution:
A function $f(x)$ is continuous at $x=a$ if
$$ \lim_{x\to a}f(x)=f(a). $$We examine the continuity of the given function at each of the specified points.
(i) At $x=0$
Since $0\le1$, we have
$$ f(0)=0. $$As $x$ approaches $0$, all values of $x$ remain less than or equal to $1$. Therefore,
$$ \begin{aligned} \lim_{x\to0}f(x) &=\lim_{x\to0}x\\ &=0. \end{aligned} $$Hence,
$$ \lim_{x\to0}f(x)=0=f(0). $$Therefore, the function is continuous at $x=0$.
(ii) At $x=1$
Since $1\le1$,
$$ f(1)=1. $$The left-hand limit is
$$ \begin{aligned} \lim_{x\to1^-}f(x) &=\lim_{x\to1^-}x\\ &=1. \end{aligned} $$The right-hand limit is
$$ \begin{aligned} \lim_{x\to1^+}f(x) &=\lim_{x\to1^+}5\\ &=5. \end{aligned} $$Since
$$ \lim_{x\to1^-}f(x)\ne\lim_{x\to1^+}f(x), $$the limit of $f(x)$ at $x=1$ does not exist.
Therefore, the function is not continuous at $x=1$.
(iii) At $x=2$
Since $2>1$,
$$ f(2)=5. $$For all values of $x$ sufficiently close to $2$, we have $x>1$. Hence,
$$ \begin{aligned} \lim_{x\to2}f(x) &=\lim_{x\to2}5\\ &=5. \end{aligned} $$Thus,
$$ \lim_{x\to2}f(x)=5=f(2). $$Therefore, the function is continuous at $x=2$.
Conclusion:
| Point | Continuity |
|---|---|
| $x=0$ | Continuous |
| $x=1$ | Discontinuous |
| $x=2$ | Continuous |
Hence, the given function is continuous at $x=0$ and $x=2$, but discontinuous at $x=1$.
Q.6 Find all points of discontinuity of $f$, where $f$ is defined by
$$ f(x)= \begin{cases} 2x+3, & \text{if } x\le2,\\ 2x-3, & \text{if } x>2. \end{cases} $$Solution:
Both expressions $2x+3$ and $2x-3$ are polynomial functions. Therefore, they are continuous in their respective domains.
The only possible point of discontinuity is at the point where the definition changes, i.e., at $x=2$.
Step 1: Find the value of the function at $x=2$.
$$ \begin{aligned} f(2) &=2(2)+3\\ &=4+3\\ &=7. \end{aligned} $$Step 2: Find the left-hand limit.
$$ \begin{aligned} \lim_{x\to2^-}f(x) &=\lim_{x\to2^-}(2x+3)\\ &=2(2)+3\\ &=7. \end{aligned} $$Step 3: Find the right-hand limit.
$$ \begin{aligned} \lim_{x\to2^+}f(x) &=\lim_{x\to2^+}(2x-3)\\ &=2(2)-3\\ &=1. \end{aligned} $$Step 4: Compare the limits.
$$ \lim_{x\to2^-}f(x)\ne\lim_{x\to2^+}f(x). $$Since the left-hand limit and the right-hand limit are not equal, the limit at $x=2$ does not exist.
Conclusion:
Hence, the function is discontinuous at $x=2$.
Q.7 Find all points of discontinuity of $f$, where $f$ is defined by
$$ f(x)= \begin{cases} |x|+3, & \text{if } x\le-3,\\ -2x, & \text{if } -3Each expression is continuous in its own interval. Therefore, we only need to check the points where the definition changes, namely $x=-3$ and $x=3$.
(i) At $x=-3$
The value of the function is
$$ \begin{aligned} f(-3) &=|-3|+3\\ &=3+3\\ &=6. \end{aligned} $$Left-hand limit:
$$ \begin{aligned} \lim_{x\to-3^-}f(x) &=\lim_{x\to-3^-}(|x|+3)\\ &=6. \end{aligned} $$Right-hand limit:
$$ \begin{aligned} \lim_{x\to-3^+}f(x) &=\lim_{x\to-3^+}(-2x)\\ &=-2(-3)\\ &=6. \end{aligned} $$Since
$$ \lim_{x\to-3^-}f(x) = \lim_{x\to-3^+}f(x) = f(-3) = 6, $$the function is continuous at $x=-3$.
(ii) At $x=3$
The value of the function is
$$ \begin{aligned} f(3) &=6(3)+2\\ &=18+2\\ &=20. \end{aligned} $$Left-hand limit:
$$ \begin{aligned} \lim_{x\to3^-}f(x) &=\lim_{x\to3^-}(-2x)\\ &=-2(3)\\ &=-6. \end{aligned} $$Right-hand limit:
$$ \begin{aligned} \lim_{x\to3^+}f(x) &=\lim_{x\to3^+}(6x+2)\\ &=6(3)+2\\ &=20. \end{aligned} $$Since
$$ \lim_{x\to3^-}f(x)\ne\lim_{x\to3^+}f(x), $$the limit at $x=3$ does not exist.
Conclusion:
The function is continuous at $x=-3$ but discontinuous at $x=3$.
Hence, the only point of discontinuity is
$$ \boxed{x=3.} $$Q.8 Find all points of discontinuity of $f$, where $f$ is defined by
$$ f(x)= \begin{cases} \dfrac{|x|}{x}, & \text{if } x\ne0,\\ 0, & \text{if } x=0. \end{cases} $$Solution:
The function is defined differently at $x=0$. Therefore, we examine the continuity at $x=0$.
Step 1: Find the value of the function at $x=0$.
$$ f(0)=0. $$Step 2: Find the left-hand limit.
For $x<0$, we have $|x|=-x$. Hence,
$$ \frac{|x|}{x}=\frac{-x}{x}=-1. $$Therefore,
$$ \lim_{x\to0^-}f(x)=-1. $$Step 3: Find the right-hand limit.
For $x>0$, we have $|x|=x$. Hence,
$$ \frac{|x|}{x}=\frac{x}{x}=1. $$Therefore,
$$ \lim_{x\to0^+}f(x)=1. $$Step 4: Compare the limits.
$$ \lim_{x\to0^-}f(x)\ne\lim_{x\to0^+}f(x). $$Since the left-hand limit and the right-hand limit are not equal, the limit at $x=0$ does not exist.
Conclusion:
Hence, the function is discontinuous at $x=0$.
Q.9 Find all points of discontinuity of $f$, where $f$ is defined by
$$ f(x)= \begin{cases} \dfrac{x}{|x|}, & \text{if } x<0,\\ -1, & \text{if } x\ge0. \end{cases} $$Solution:
Both parts of the function are continuous in their respective intervals. Therefore, we only need to examine the point $x=0$.
Step 1: Find the value of the function at $x=0$.
$$ f(0)=-1. $$Step 2: Find the left-hand limit.
For $x<0$, we have $|x|=-x$. Therefore,
$$ \frac{x}{|x|} =\frac{x}{-x} =-1. $$Hence,
$$ \lim_{x\to0^-}f(x)=-1. $$Step 3: Find the right-hand limit.
For $x\ge0$,
$$ f(x)=-1. $$Therefore,
$$ \lim_{x\to0^+}f(x)=-1. $$Step 4: Compare the limits.
$$ \lim_{x\to0^-}f(x) = \lim_{x\to0^+}f(x) = f(0) = -1. $$Hence, the function satisfies the condition of continuity at $x=0$.
Conclusion:
The function is continuous at $x=0$.
Since both pieces are continuous in their respective domains and the function is also continuous at $x=0$, there is no point of discontinuity.
$$ \boxed{\text{The function is continuous for all real values of }x.} $$Q.10 Find all points of discontinuity of $f$, where $f$ is defined by
$$ f(x)= \begin{cases} x+1, & \text{if } x\ge1,\\ x^2+1, & \text{if } x<1. \end{cases} $$Solution:
The functions $x+1$ and $x^2+1$ are polynomials. Hence, they are continuous in their respective intervals.
The only possible point of discontinuity is at $x=1$.
Step 1: Find the value of the function at $x=1$.
$$ \begin{aligned} f(1) &=1+1\\ &=2. \end{aligned} $$Step 2: Find the left-hand limit.
$$ \begin{aligned} \lim_{x\to1^-}f(x) &=\lim_{x\to1^-}(x^2+1)\\ &=(1)^2+1\\ &=2. \end{aligned} $$Step 3: Find the right-hand limit.
$$ \begin{aligned} \lim_{x\to1^+}f(x) &=\lim_{x\to1^+}(x+1)\\ &=1+1\\ &=2. \end{aligned} $$Step 4: Compare the limits.
$$ \lim_{x\to1^-}f(x) = \lim_{x\to1^+}f(x) = f(1) = 2. $$Hence, the function is continuous at $x=1$.
Conclusion:
Since the function is continuous in each interval and also at $x=1$, there is no point of discontinuity.
$$ \boxed{\text{The function is continuous for all real values of }x.} $$Q.11 Find all points of discontinuity of $f$, where $f$ is defined by
$$ f(x)= \begin{cases} x^3-3, & \text{if } x\le2,\\ x^2+1, & \text{if } x>2. \end{cases} $$Solution:
Both $x^3-3$ and $x^2+1$ are polynomial functions. Therefore, they are continuous in their respective intervals.
The only possible point of discontinuity is at $x=2$.
Step 1: Find the value of the function at $x=2$.
$$ \begin{aligned} f(2) &=2^3-3\\ &=8-3\\ &=5. \end{aligned} $$Step 2: Find the left-hand limit.
$$ \begin{aligned} \lim_{x\to2^-}f(x) &=\lim_{x\to2^-}(x^3-3)\\ &=2^3-3\\ &=5. \end{aligned} $$Step 3: Find the right-hand limit.
$$ \begin{aligned} \lim_{x\to2^+}f(x) &=\lim_{x\to2^+}(x^2+1)\\ &=2^2+1\\ &=5. \end{aligned} $$Step 4: Compare the limits.
$$ \lim_{x\to2^-}f(x) = \lim_{x\to2^+}f(x) = f(2) = 5. $$Therefore, the function satisfies the condition of continuity at $x=2$.
Conclusion:
Since the function is continuous in each interval and also continuous at $x=2$, there is no point of discontinuity.
$$ \boxed{\text{The function is continuous for all real values of }x.} $$Q.12 Find all points of discontinuity of $f$, where $f$ is defined by
$$ f(x)= \begin{cases} x^{10}-1, & \text{if } x\le1,\\ x^2, & \text{if } x>1. \end{cases} $$Solution:
The functions $x^{10}-1$ and $x^2$ are polynomial functions. Hence, they are continuous in their respective intervals.
The only possible point of discontinuity is at $x=1$.
Step 1: Find the value of the function at $x=1$.
$$ \begin{aligned} f(1) &=1^{10}-1\\ &=1-1\\ &=0. \end{aligned} $$Step 2: Find the left-hand limit.
$$ \begin{aligned} \lim_{x\to1^-}f(x) &=\lim_{x\to1^-}(x^{10}-1)\\ &=1^{10}-1\\ &=0. \end{aligned} $$Step 3: Find the right-hand limit.
$$ \begin{aligned} \lim_{x\to1^+}f(x) &=\lim_{x\to1^+}x^2\\ &=1^2\\ &=1. \end{aligned} $$Step 4: Compare the limits.
$$ \lim_{x\to1^-}f(x)\ne\lim_{x\to1^+}f(x). $$Since the left-hand limit and the right-hand limit are not equal, the limit at $x=1$ does not exist.
Conclusion:
Hence, the function is discontinuous at $x=1$.
Q.13 Is the function defined by
$$ f(x)= \begin{cases} x+5, & \text{if } x\le1,\\ x-5, & \text{if } x>1 \end{cases} $$a continuous function?
Solution:
The functions $x+5$ and $x-5$ are polynomial functions. Therefore, they are continuous in their respective intervals.
The only possible point where discontinuity may occur is at $x=1$.
Step 1: Find the value of the function at $x=1$.
$$ \begin{aligned} f(1) &=1+5\\ &=6. \end{aligned} $$Step 2: Find the left-hand limit.
$$ \begin{aligned} \lim_{x\to1^-}f(x) &=\lim_{x\to1^-}(x+5)\\ &=1+5\\ &=6. \end{aligned} $$Step 3: Find the right-hand limit.
$$ \begin{aligned} \lim_{x\to1^+}f(x) &=\lim_{x\to1^+}(x-5)\\ &=1-5\\ &=-4. \end{aligned} $$Step 4: Compare the limits.
$$ \lim_{x\to1^-}f(x)\ne\lim_{x\to1^+}f(x). $$Since the left-hand limit and the right-hand limit are not equal, the limit at $x=1$ does not exist.
Hence,
$$ \lim_{x\to1}f(x)\ne f(1). $$Conclusion:
The given function is not a continuous function. It is discontinuous at $x=1$.
Q.14 Discuss the continuity of the function $f$, where $f$ is defined by
The function is constant in each of the intervals $[0,1]$, $(1,3)$ and $[3,10]$. Therefore, it is continuous in the interior of these intervals.
The possible points of discontinuity are the points where the definition changes, namely $x=1$ and $x=3$.
(i) At $x=1$
The value of the function is
$$ f(1)=3. $$The left-hand limit is
$$ \lim_{x\to1^-}f(x)=3. $$The right-hand limit is
$$ \lim_{x\to1^+}f(x)=4. $$Since
$$ \lim_{x\to1^-}f(x)\ne\lim_{x\to1^+}f(x), $$the limit at $x=1$ does not exist.
Hence, the function is discontinuous at $x=1$.
(ii) At $x=3$
The value of the function is
$$ f(3)=5. $$The left-hand limit is
$$ \lim_{x\to3^-}f(x)=4. $$The right-hand limit is
$$ \lim_{x\to3^+}f(x)=5. $$Since
$$ \lim_{x\to3^-}f(x)\ne\lim_{x\to3^+}f(x), $$the limit at $x=3$ does not exist.
Hence, the function is discontinuous at $x=3$.
Conclusion:
The function is continuous on
$$ [0,1),\ (1,3),\ (3,10], $$and discontinuous at $x=1$ and $x=3$.
Q.15 Discuss the continuity of the function $f$, where $f$ is defined by
$$ f(x)= \begin{cases} 2x, & \text{if } x<0,\\ 0, & \text{if } 0\le x\le1,\\ 4x, & \text{if } x>1. \end{cases} $$Solution:
The functions $2x$, $0$ and $4x$ are continuous in their respective intervals.
Therefore, we need to examine the points where the definition changes, namely $x=0$ and $x=1$.
(i) At $x=0$
The value of the function is
$$ f(0)=0. $$The left-hand limit is
$$ \begin{aligned} \lim_{x\to0^-}f(x) &=\lim_{x\to0^-}(2x)\\ &=0. \end{aligned} $$The right-hand limit is
$$ \lim_{x\to0^+}f(x)=0. $$Thus,
$$ \lim_{x\to0^-}f(x) = \lim_{x\to0^+}f(x) = f(0) = 0. $$Hence, the function is continuous at $x=0$.
(ii) At $x=1$
The value of the function is
$$ f(1)=0. $$The left-hand limit is
$$ \lim_{x\to1^-}f(x)=0. $$The right-hand limit is
$$ \begin{aligned} \lim_{x\to1^+}f(x) &=\lim_{x\to1^+}(4x)\\ &=4. \end{aligned} $$Since
$$ \lim_{x\to1^-}f(x)\ne\lim_{x\to1^+}f(x), $$the limit at $x=1$ does not exist.
Hence, the function is discontinuous at $x=1$.
Conclusion:
The function is continuous everywhere except at $x=1$.
Q.16 Discuss the continuity of the function $f$, where $f$ is defined by
$$ f(x)= \begin{cases} -2, & \text{if } x\le-1,\\ 2x, & \text{if } -1Solution:
The functions $-2$, $2x$ and $2$ are continuous in their respective intervals.
Hence, we only need to examine the points where the definition changes, namely $x=-1$ and $x=1$.
(i) At $x=-1$
The value of the function is
$$ f(-1)=-2. $$The left-hand limit is
$$ \lim_{x\to-1^-}f(x)=-2. $$The right-hand limit is
$$ \begin{aligned} \lim_{x\to-1^+}f(x) &=\lim_{x\to-1^+}(2x)\\ &=2(-1)\\ &=-2. \end{aligned} $$Thus,
$$ \lim_{x\to-1^-}f(x) = \lim_{x\to-1^+}f(x) = f(-1) = -2. $$Hence, the function is continuous at $x=-1$.
(ii) At $x=1$
The value of the function is
$$ f(1)=2(1)=2. $$The left-hand limit is
$$ \begin{aligned} \lim_{x\to1^-}f(x) &=\lim_{x\to1^-}(2x)\\ &=2. \end{aligned} $$The right-hand limit is
$$ \lim_{x\to1^+}f(x)=2. $$Therefore,
$$ \lim_{x\to1^-}f(x) = \lim_{x\to1^+}f(x) = f(1) = 2. $$Hence, the function is continuous at $x=1$.
Conclusion:
The function is continuous at both transition points $x=-1$ and $x=1$. Therefore, it is continuous for all real values of $x$.
Q.17 Find the relationship between $a$ and $b$ so that the function $f$ defined by
$$ f(x)= \begin{cases} ax+1, & \text{if } x\le3,\\ bx+3, & \text{if } x>3 \end{cases} $$is continuous at $x=3$.
Solution:
For continuity at $x=3$,
$$ \lim_{x\to3^-}f(x) = \lim_{x\to3^+}f(x) = f(3). $$Step 1: Find the value of the function at $x=3$.
$$ \begin{aligned} f(3) &=3a+1. \end{aligned} $$Step 2: Find the left-hand limit.
$$ \begin{aligned} \lim_{x\to3^-}f(x) &=\lim_{x\to3^-}(ax+1)\\ &=3a+1. \end{aligned} $$Step 3: Find the right-hand limit.
$$ \begin{aligned} \lim_{x\to3^+}f(x) &=\lim_{x\to3^+}(bx+3)\\ &=3b+3. \end{aligned} $$Step 4: Apply the condition of continuity.
$$ 3a+1=3b+3. $$Rearranging,
$$ \begin{aligned} 3a-3b &=2\\ a-b &=\frac23. \end{aligned} $$Conclusion:
The required relationship between $a$ and $b$ is
$$ \boxed{a-b=\frac23.} $$Q.18 For what value of $\lambda$ is the function defined by
$$ f(x)= \begin{cases} \lambda(x^2-2x), & \text{if } x\le0,\\ 4x+1, & \text{if } x>0 \end{cases} $$continuous at $x=0$? What about continuity at $x=1$?
Solution:
The functions $\lambda(x^2-2x)$ and $4x+1$ are polynomials. Hence, they are continuous in their respective intervals.
We examine the continuity at $x=0$ and $x=1$.
(i) Continuity at $x=0$
For continuity at $x=0$,
$$ \lim_{x\to0^-}f(x) = \lim_{x\to0^+}f(x) = f(0). $$Step 1: Find the value of the function at $x=0$.
$$ \begin{aligned} f(0) &=\lambda(0^2-2\times0)\\ &=\lambda(0)\\ &=0. \end{aligned} $$Step 2: Find the left-hand limit.
$$ \begin{aligned} \lim_{x\to0^-}f(x) &=\lim_{x\to0^-}\lambda(x^2-2x)\\ &=\lambda(0^2-2\times0)\\ &=0. \end{aligned} $$Step 3: Find the right-hand limit.
$$ \begin{aligned} \lim_{x\to0^+}f(x) &=\lim_{x\to0^+}(4x+1)\\ &=1. \end{aligned} $$For continuity,
$$ 0=1, $$which is impossible.
Conclusion:
There is no value of $\lambda$ for which the function is continuous at $x=0$.
(ii) Continuity at $x=1$
Since $1>0$, the function near $x=1$ is
$$ f(x)=4x+1. $$This is a polynomial function and is continuous for all real values of $x$.
The value of the function is
$$ \begin{aligned} f(1) &=4(1)+1\\ &=5. \end{aligned} $$Also,
$$ \begin{aligned} \lim_{x\to1}f(x) &=\lim_{x\to1}(4x+1)\\ &=5. \end{aligned} $$Hence,
$$ \lim_{x\to1}f(x)=f(1). $$Conclusion:
- There is no value of $\lambda$ for which the function is continuous at $x=0$.
- The function is continuous at $x=1$ for every value of $\lambda$.
Q.19 Show that the function defined by $g(x)=x-[x]$ is discontinuous at all integral points. Here $[x]$ denotes the greatest integer less than or equal to $x$.
Solution:
Given,
$$ g(x)=x-[x], $$where $[x]$ denotes the greatest integer less than or equal to $x$.
The function $g(x)$ represents the fractional part of $x$.
Let $n$ be any integer. We examine the continuity of the function at $x=n$.
Step 1: Find the value of the function at $x=n$.
Since $[n]=n$,
$$ \begin{aligned} g(n) &=n-[n]\\ &=n-n\\ &=0. \end{aligned} $$Step 2: Find the left-hand limit.
For $x$ approaching $n$ from the left,
$$ [n]=n-1. $$Therefore,
$$ \begin{aligned} g(x) &=x-(n-1). \end{aligned} $$Hence,
$$ \begin{aligned} \lim_{x\to n^-}g(x) &=n-(n-1)\\ &=1. \end{aligned} $$Step 3: Find the right-hand limit.
For $x$ approaching $n$ from the right,
$$ [n]=n. $$Therefore,
$$ \begin{aligned} g(x) &=x-n. \end{aligned} $$Hence,
$$ \begin{aligned} \lim_{x\to n^+}g(x) &=n-n\\ &=0. \end{aligned} $$Step 4: Compare the limits.
$$ \lim_{x\to n^-}g(x)=1 \ne 0=\lim_{x\to n^+}g(x). $$Since the left-hand limit and the right-hand limit are not equal, the limit at $x=n$ does not exist.
Conclusion:
Hence, the function
$$ g(x)=x-[x] $$is discontinuous at every integral point.
Q.20 Is the function defined by $f(x)=x^2-\sin x+5$ continuous at $x=\pi$?
Solution:
Given,
$$ f(x)=x^2-\sin x+5. $$The functions $x^2$, $\sin x$ and the constant $5$ are continuous for all real values of $x$.
Therefore, their sum and difference are also continuous for all real values of $x$.
Step 1: Find the value of the function at $x=\pi$.
$$ \begin{aligned} f(\pi) &=\pi^2-\sin\pi+5\\ &=\pi^2-0+5\\ &=\pi^2+5. \end{aligned} $$Step 2: Find the limit as $x\to\pi$.
$$ \begin{aligned} \lim_{x\to\pi}f(x) &=\lim_{x\to\pi}(x^2-\sin x+5)\\ &=\pi^2-\sin\pi+5\\ &=\pi^2+5. \end{aligned} $$Thus,
$$ \lim_{x\to\pi}f(x)=f(\pi). $$Conclusion:
Hence, the function
$$ f(x)=x^2-\sin x+5 $$is continuous at $x=\pi$.
Q.21 Examine the continuity of the following functions at the indicated points.
(a) $f(x)=7x-11$ at $x=4$
Solution:
A function $f(x)$ is continuous at $x=a$ if
$$ \lim_{x\to a}f(x)=f(a). $$Given,
$$ f(x)=7x-11. $$Step 1: Find the value of the function at $x=4$.
$$ \begin{aligned} f(4) &=7(4)-11\\ &=28-11\\ &=17. \end{aligned} $$Step 2: Find the limit as $x$ approaches $4$.
$$ \begin{aligned} \lim_{x\to4}f(x) &=\lim_{x\to4}(7x-11)\\ &=7(4)-11\\ &=17. \end{aligned} $$Therefore,
$$ \lim_{x\to4}f(x)=f(4)=17. $$Conclusion:
Hence, the function is continuous at $x=4$.
(b) $f(x)=x^2+2x-3$ at $x=1$
Solution:
Given,
$$ f(x)=x^2+2x-3. $$Step 1: Find the value of the function at $x=1$.
$$ \begin{aligned} f(1) &=(1)^2+2(1)-3\\ &=1+2-3\\ &=0. \end{aligned} $$Step 2: Find the limit as $x$ approaches $1$.
$$ \begin{aligned} \lim_{x\to1}f(x) &=\lim_{x\to1}(x^2+2x-3)\\ &=(1)^2+2(1)-3\\ &=0. \end{aligned} $$Thus,
$$ \lim_{x\to1}f(x)=f(1)=0. $$Conclusion:
Hence, the function is continuous at $x=1$.
(c) $f(x)=\sqrt{x+1}$ at $x=0$
Solution:
Given,
$$ f(x)=\sqrt{x+1}. $$The function $\sqrt{x+1}$ is defined for
$$ x\ge-1. $$Step 1: Find the value of the function at $x=0$.
$$ \begin{aligned} f(0) &=\sqrt{0+1}\\ &=\sqrt{1}\\ &=1. \end{aligned} $$Step 2: Find the limit as $x$ approaches $0$.
$$ \begin{aligned} \lim_{x\to0}f(x) &=\lim_{x\to0}\sqrt{x+1}\\ &=\sqrt{0+1}\\ &=1. \end{aligned} $$Hence,
$$ \lim_{x\to0}f(x)=f(0)=1. $$Conclusion:
Therefore, the function
$$ f(x)=\sqrt{x+1} $$is continuous at $x=0$.
Q.22 Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
Solution:
The continuity of each trigonometric function is discussed below.
(i) Cosine Function
The cosine function is
$$ f(x)=\cos x. $$Since the cosine function is defined for every real number and has no breaks or jumps,
$$ \lim_{x\to a}\cos x=\cos a $$for every real number $a$.
Hence, the cosine function is continuous for all real values of $x$.
$$ \boxed{\text{Domain: }(-\infty,\infty)} $$(ii) Cosecant Function
The cosecant function is
$$ f(x)=\csc x=\frac{1}{\sin x}. $$Since $\sin x=0$ when
$$ x=n\pi,\qquad n\in\mathbb{Z}, $$the function $\csc x$ is not defined at these points.
Therefore, $\csc x$ is continuous at every point of its domain and discontinuous at
$$ x=n\pi,\qquad n\in\mathbb{Z}. $$ $$ \boxed{\text{Continuous for }x\ne n\pi,\; n\in\mathbb{Z}.} $$(iii) Secant Function
The secant function is
$$ f(x)=\sec x=\frac{1}{\cos x}. $$Since $\cos x=0$ when
$$ x=\frac{(2n+1)\pi}{2},\qquad n\in\mathbb{Z}, $$the function $\sec x$ is not defined at these points.
Hence, $\sec x$ is continuous at every point of its domain and discontinuous at
$$ x=\frac{(2n+1)\pi}{2},\qquad n\in\mathbb{Z}. $$ $$ \boxed{\text{Continuous for }x\ne\frac{(2n+1)\pi}{2},\; n\in\mathbb{Z}.} $$(iv) Cotangent Function
The cotangent function is
$$ f(x)=\cot x=\frac{\cos x}{\sin x}. $$Since $\sin x=0$ when
$$ x=n\pi,\qquad n\in\mathbb{Z}, $$the function $\cot x$ is not defined at these points.
Therefore, $\cot x$ is continuous at every point of its domain and discontinuous at
$$ x=n\pi,\qquad n\in\mathbb{Z}. $$ $$ \boxed{\text{Continuous for }x\ne n\pi,\; n\in\mathbb{Z}.} $$Conclusion:
| Function | Continuous On | Discontinuous At |
|---|---|---|
| $\cos x$ | All real numbers | None |
| $\csc x$ | $x\ne n\pi$ | $x=n\pi,\; n\in\mathbb{Z}$ |
| $\sec x$ | $x\ne\dfrac{(2n+1)\pi}{2}$ | $x=\dfrac{(2n+1)\pi}{2},\; n\in\mathbb{Z}$ |
| $\cot x$ | $x\ne n\pi$ | $x=n\pi,\; n\in\mathbb{Z}$ |
Q.23 Find all points of discontinuity of $f$, where
$$ f(x)= \begin{cases} \dfrac{\sin x}{x}, & \text{if } x<0,\\ x+1, & \text{if } x\ge0. \end{cases} $$Solution:
The functions
$$ \frac{\sin x}{x} $$and
$$ x+1 $$are continuous in their respective domains. Therefore, the only possible point of discontinuity is at $x=0$.
Step 1: Find the value of the function at $x=0$.
Since $0\ge0$,
$$ \begin{aligned} f(0) &=0+1\\ &=1. \end{aligned} $$Step 2: Find the left-hand limit.
$$ \begin{aligned} \lim_{x\to0^-}f(x) &=\lim_{x\to0^-}\frac{\sin x}{x}\\ &=1. \end{aligned} $$(Using the standard result $\displaystyle \lim_{x\to0}\frac{\sin x}{x}=1$.)
Step 3: Find the right-hand limit.
$$ \begin{aligned} \lim_{x\to0^+}f(x) &=\lim_{x\to0^+}(x+1)\\ &=0+1\\ &=1. \end{aligned} $$Step 4: Compare the limits.
$$ \lim_{x\to0^-}f(x) = \lim_{x\to0^+}f(x) = f(0) = 1. $$Hence,
$$ \lim_{x\to0}f(x)=f(0). $$Conclusion:
The function is continuous at $x=0$.
Since both parts are continuous in their respective intervals and the function is also continuous at $x=0$, there is no point of discontinuity.
$$ \boxed{\text{The function is continuous for all real values of }x.} $$Q.24 Determine if the function defined by
$$ f(x)= \begin{cases} x^2\sin\frac{1}{x}, & \text{if } x\ne0,\\ 0, & \text{if } x=0 \end{cases} $$is a continuous function?
Solution:
The function
$$ x^2\sin\frac{1}{x} $$is continuous for every $x\ne0$, since it is the product of continuous functions.
Therefore, we only need to examine the continuity at $x=0$.
Step 1: Find the value of the function at $x=0$.
$$ f(0)=0. $$Step 2: Find the limit as $x\to0$.
We know that
$$ -1\le\sin\frac{1}{x}\le1. $$Multiplying throughout by $x^2\;(x^2\ge0)$, we get
$$ -x^2\le x^2\sin\frac{1}{x}\le x^2. $$Now,
$$ \lim_{x\to0}(-x^2)=0 \qquad\text{and}\qquad \lim_{x\to0}x^2=0. $$Therefore, by the Squeeze Theorem,
$$ \lim_{x\to0}x^2\sin\frac{1}{x}=0. $$Step 3: Compare the limit and the function value.
$$ \lim_{x\to0}f(x)=0=f(0). $$Hence,
$$ \lim_{x\to0}f(x)=f(0). $$Therefore, the function is continuous at $x=0$.
Conclusion:
The function is continuous for all $x\ne0$ and is also continuous at $x=0$.
Hence, the given function is a continuous function on the entire real line.
$$ \boxed{\text{The function }f(x)\text{ is continuous for all real values of }x.} $$Q.25 Examine the continuity of $f$, where $f$ is defined by
$$ f(x)= \begin{cases} \sin x-\cos x, & \text{if } x\ne0,\\ -1, & \text{if } x=0. \end{cases} $$Solution:
The function $\sin x-\cos x$ is continuous for all real values of $x$. Therefore, the only possible point of discontinuity is at $x=0$.
Step 1: Find the value of the function at $x=0$.
$$ f(0)=-1. $$Step 2: Find the limit as $x$ approaches $0$.
$$ \begin{aligned} \lim_{x\to0}f(x) &=\lim_{x\to0}(\sin x-\cos x)\\ &=\sin0-\cos0\\ &=0-1\\ &=-1. \end{aligned} $$Step 3: Compare the limit and the function value.
$$ \lim_{x\to0}f(x)=-1=f(0). $$Since
$$ \lim_{x\to0}f(x)=f(0), $$the function satisfies the condition of continuity at $x=0$.
Conclusion:
The function is continuous for all $x\ne0$ and is also continuous at $x=0$.
Hence, the given function is a continuous function for all real values of $x$.
$$ \boxed{\text{The function }f(x)\text{ is continuous for all real values of }x.} $$Q.26 Find the value of $k$ so that the function $f$ is continuous at $x=\dfrac{\pi}{2}$, where
$$ f(x)= \begin{cases} \dfrac{k\cos x}{\pi-2x}, & \text{if } x\ne\dfrac{\pi}{2},\\ 3, & \text{if } x=\dfrac{\pi}{2}. \end{cases} $$Solution:
For continuity at
$$ x=\frac{\pi}{2}, $$the following condition must be satisfied:
$$ \lim_{x\to\frac{\pi}{2}}f(x)=f\left(\frac{\pi}{2}\right). $$Step 1: Find the value of the function at $x=\dfrac{\pi}{2}$.
$$ f\left(\frac{\pi}{2}\right)=3. $$Step 2: Find the limit as $x\to\dfrac{\pi}{2}$.
Substituting $x=\dfrac{\pi}{2}$ gives the indeterminate form
$$ \frac{0}{0}. $$Using the standard limit
$$ \lim_{\theta\to0}\frac{\sin\theta}{\theta}=1, $$or equivalently,
$$ \lim_{x\to\frac{\pi}{2}} \frac{\cos x}{\pi-2x} =\frac12, $$we obtain
$$ \begin{aligned} \lim_{x\to\frac{\pi}{2}} \frac{k\cos x}{\pi-2x} &= k\left(\frac12\right)\\ &=\frac{k}{2}. \end{aligned} $$Step 3: Apply the condition of continuity.
$$ \frac{k}{2}=3. $$Therefore,
$$ \begin{aligned} k &=6. \end{aligned} $$Conclusion:
$$ \boxed{k=6.} $$Q.27 Find the value of $k$ so that the function $f$ is continuous at $x=2$, where
$$ f(x)= \begin{cases} kx^2, & \text{if } x\le2,\\ 3, & \text{if } x>2. \end{cases} $$Solution:
For continuity at $x=2$,
$$ \lim_{x\to2}f(x)=f(2). $$Step 1: Find the value of the function at $x=2$.
$$ \begin{aligned} f(2) &=k(2)^2\\ &=4k. \end{aligned} $$Step 2: Find the left-hand limit.
$$ \begin{aligned} \lim_{x\to2^-}f(x) &=\lim_{x\to2^-}(kx^2)\\ &=4k. \end{aligned} $$Step 3: Find the right-hand limit.
$$ \lim_{x\to2^+}f(x)=3. $$Step 4: Apply the condition of continuity.
For continuity,
$$ 4k=3. $$Therefore,
$$ \begin{aligned} k &=\frac34. \end{aligned} $$Conclusion:
$$ \boxed{k=\frac34.} $$Q.28 Find the value of $k$ so that the function $f$ is continuous at $x=\pi$, where
$$ f(x)= \begin{cases} kx+1, & \text{if } x\le\pi,\\ \cos x, & \text{if } x>\pi. \end{cases} $$Solution:
For continuity at $x=\pi$, the following condition must be satisfied:
$$ \lim_{x\to\pi}f(x)=f(\pi). $$Step 1: Find the value of the function at $x=\pi$.
$$ \begin{aligned} f(\pi) &=k\pi+1. \end{aligned} $$Step 2: Find the left-hand limit.
$$ \begin{aligned} \lim_{x\to\pi^-}f(x) &=\lim_{x\to\pi^-}(kx+1)\\ &=k\pi+1. \end{aligned} $$Step 3: Find the right-hand limit.
$$ \begin{aligned} \lim_{x\to\pi^+}f(x) &=\lim_{x\to\pi^+}\cos x\\ &=\cos\pi\\ &=-1. \end{aligned} $$Step 4: Apply the condition of continuity.
For continuity,
$$ k\pi+1=-1. $$Therefore,
$$ \begin{aligned} k\pi &=-2\\ k &=-\frac{2}{\pi}. \end{aligned} $$Conclusion:
$$ \boxed{k=-\frac{2}{\pi}.} $$Q.29 Find the value of $k$ so that the function $f$ is continuous at $x=5$, where
$$ f(x)= \begin{cases} kx+1, & \text{if } x\le5,\\ 3x-5, & \text{if } x>5. \end{cases} $$Solution:
For continuity at $x=5$,
$$ \lim_{x\to5}f(x)=f(5). $$Step 1: Find the value of the function at $x=5$.
$$ \begin{aligned} f(5) &=5k+1. \end{aligned} $$Step 2: Find the left-hand limit.
$$ \begin{aligned} \lim_{x\to5^-}f(x) &=\lim_{x\to5^-}(kx+1)\\ &=5k+1. \end{aligned} $$Step 3: Find the right-hand limit.
$$ \begin{aligned} \lim_{x\to5^+}f(x) &=\lim_{x\to5^+}(3x-5)\\ &=3(5)-5\\ &=15-5\\ &=10. \end{aligned} $$Step 4: Apply the condition of continuity.
For continuity,
$$ 5k+1=10. $$Therefore,
$$ \begin{aligned} 5k &=9\\ k &=\frac95. \end{aligned} $$Conclusion:
$$ \boxed{k=\frac95.} $$Q.30 Find the values of $a$ and $b$ such that the function defined by
Solution:
For the function to be continuous, it must be continuous at the points where its definition changes, namely at $x=2$ and $x=10$.
Step 1: Apply the continuity condition at $x=2$.
The value of the function at $x=2$ is
$$ f(2)=5. $$The right-hand limit is
$$ \begin{aligned} \lim_{x\to2^+}(ax+b) &=2a+b. \end{aligned} $$Since the function is continuous at $x=2$,
$$ 2a+b=5. $$Step 2: Apply the continuity condition at $x=10$.
The value of the function at $x=10$ is
$$ f(10)=21. $$The left-hand limit is
$$ \begin{aligned} \lim_{x\to10^-}(ax+b) &=10a+b. \end{aligned} $$Since the function is continuous at $x=10$,
$$ 10a+b=21. $$Step 3: Solve the simultaneous equations.
Subtracting the first equation from the second,
$$ \begin{aligned} (10a+b)-(2a+b) &=21-5\\ 8a &=16\\ a &=2. \end{aligned} $$Substituting $a=2$ into
$$ 2a+b=5, $$we get
$$ \begin{aligned} 2(2)+b &=5\\ 4+b &=5\\ b &=1. \end{aligned} $$Verification:
At $x=2$,
$$ 2a+b=2(2)+1=5=f(2). $$At $x=10$,
$$ 10a+b=10(2)+1=21=f(10). $$Thus, the function is continuous at both transition points.
Conclusion:
$$ \boxed{a=2,\qquad b=1.} $$Q.31 Show that the function defined by $f(x)=\cos(x^2)$ is a continuous function.
Solution:
We know that the polynomial function
$$ x^2 $$is continuous for all real values of $x$.
Also, the cosine function
$$ \cos x $$is continuous for all real values of $x$.
The given function is
$$ f(x)=\cos(x^2), $$which is the composition of the continuous functions $\cos x$ and $x^2$.
Let $a$ be any real number.
Then,
$$ \begin{aligned} \lim_{x\to a}f(x) &=\lim_{x\to a}\cos(x^2)\\ &=\cos\left(\lim_{x\to a}x^2\right)\\ &=\cos(a^2). \end{aligned} $$Since
$$ f(a)=\cos(a^2), $$we have
$$ \lim_{x\to a}f(x)=f(a). $$Hence, the function satisfies the condition of continuity at every real number.
Conclusion:
$$ \boxed{\text{The function }f(x)=\cos(x^2)\text{ is continuous for all real values of }x.} $$Q.32 Show that the function defined by $f(x)=|\cos x|$ is a continuous function.
Solution:
We know that the cosine function
$$ \cos x $$is continuous for all real values of $x$.
Also, the modulus function
$$ |x| $$is continuous for all real values of $x$.
The given function is
$$ f(x)=|\cos x|, $$which is obtained by applying the modulus function to the continuous function $\cos x$.
Let $a$ be any real number.
Then,
$$ \begin{aligned} \lim_{x\to a}f(x) &=\lim_{x\to a}|\cos x|\\ &=\left|\lim_{x\to a}\cos x\right|\\ &=|\cos a|. \end{aligned} $$Since
$$ f(a)=|\cos a|, $$we obtain
$$ \lim_{x\to a}f(x)=f(a). $$Hence, the function is continuous at every real number.
Conclusion:
$$ \boxed{\text{The function }f(x)=|\cos x|\text{ is continuous for all real values of }x.} $$Q.33 Show that the function defined by $f(x)=\sin|x|$ is a continuous function.
Solution:
We know that the modulus function
$$ |x| $$is continuous for all real values of $x$.
Also, the sine function
$$ \sin x $$is continuous for all real values of $x$.
The given function is
$$ f(x)=\sin|x|, $$which is the composition of the continuous functions $|x|$ and $\sin x$.
Let $a$ be any real number.
Then,
$$ \begin{aligned} \lim_{x\to a}f(x) &=\lim_{x\to a}\sin|x|\\ &=\sin\left(\lim_{x\to a}|x|\right)\\ &=\sin|a|. \end{aligned} $$Since
$$ f(a)=\sin|a|, $$we have
$$ \lim_{x\to a}f(x)=f(a). $$Hence, the function satisfies the condition of continuity at every real number.
Conclusion:
$$ \boxed{\text{The function }f(x)=\sin|x|\text{ is continuous for all real values of }x.} $$Q.34 Find all the points of discontinuity of $f$ defined by $f(x)=|x|-|x+1|$.
Solution:
The modulus functions $|x|$ and $|x+1|$ are continuous for all real values of $x$.
Since the difference of two continuous functions is also continuous, the function
$$ f(x)=|x|-|x+1| $$is continuous for every real value of $x$.
Alternatively, we can write the function in piecewise form.
Case 1: $x<-1$
$$ \begin{aligned} f(x) &=(-x)-(-(x+1))\\ &=-x+x+1\\ &=1. \end{aligned} $$Case 2: $-1\le x<0$
$$ \begin{aligned} f(x) &=(-x)-(x+1)\\ &=-2x-1. \end{aligned} $$Case 3: $x\ge0$
$$ \begin{aligned} f(x) &=x-(x+1)\\ &=-1. \end{aligned} $$The only possible points of discontinuity are $x=-1$ and $x=0$.
At $x=-1$
$$ \begin{aligned} \lim_{x\to-1^-}f(x) &=1,\\ \lim_{x\to-1^+}f(x) &=-2(-1)-1\\ &=1,\\ f(-1) &=|-1|-|0|\\ &=1. \end{aligned} $$Hence,
$$ \lim_{x\to-1}f(x)=f(-1)=1. $$Therefore, the function is continuous at $x=-1$.
At $x=0$
$$ \begin{aligned} \lim_{x\to0^-}f(x) &=-2(0)-1\\ &=-1,\\ \lim_{x\to0^+}f(x) &=-1,\\ f(0) &=|0|-|1|\\ &=-1. \end{aligned} $$Thus,
$$ \lim_{x\to0}f(x)=f(0)=-1. $$Therefore, the function is continuous at $x=0$.
Conclusion:
The function is continuous at every real number.
$$ \boxed{\text{There is no point of discontinuity.}} $$