Exercise 5.1 Practice
Continuity of Functions: 34 Practice Questions with Detailed Solutions
Q1: Linear Function
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Prove that the function $f(x) = 5x - 3$ is continuous at $x = 0$, at $x = -3$,
and at $x = 5$.
Given: Function $f(x) = 5x - 3$.
Concept: A function is continuous at $x = c$ if
$\lim_{x \to c} f(x) = f(c)$.
1. At x = 0:
Limit: $\lim_{x \to 0} (5x - 3) = 5(0) - 3 = -3$.
Value: $f(0) = 5(0) - 3 = -3$.
Since $\lim_{x \to 0} f(x) = f(0)$, $f$ is continuous at $x = 0$.
Limit: $\lim_{x \to 0} (5x - 3) = 5(0) - 3 = -3$.
Value: $f(0) = 5(0) - 3 = -3$.
Since $\lim_{x \to 0} f(x) = f(0)$, $f$ is continuous at $x = 0$.
2. At x = -3:
Limit: $\lim_{x \to -3} (5x - 3) = 5(-3) - 3 = -15 - 3 = -18$.
Value: $f(-3) = 5(-3) - 3 = -18$.
Since $\lim_{x \to -3} f(x) = f(-3)$, $f$ is continuous at $x = -3$.
Limit: $\lim_{x \to -3} (5x - 3) = 5(-3) - 3 = -15 - 3 = -18$.
Value: $f(-3) = 5(-3) - 3 = -18$.
Since $\lim_{x \to -3} f(x) = f(-3)$, $f$ is continuous at $x = -3$.
3. At x = 5:
Limit: $\lim_{x \to 5} (5x - 3) = 5(5) - 3 = 25 - 3 = 22$.
Value: $f(5) = 5(5) - 3 = 22$.
Since $\lim_{x \to 5} f(x) = f(5)$, $f$ is continuous at $x = 5$.
Limit: $\lim_{x \to 5} (5x - 3) = 5(5) - 3 = 25 - 3 = 22$.
Value: $f(5) = 5(5) - 3 = 22$.
Since $\lim_{x \to 5} f(x) = f(5)$, $f$ is continuous at $x = 5$.
Conclusion: The function $f(x) = 5x - 3$ is continuous at $x = 0$, $x = -3$, and
$x = 5$.
Q2: Quadratic Function
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Examine the continuity of the function $f(x) = 2x^2 - 1$ at $x = 3$.
Given: $f(x) = 2x^2 - 1$. We check continuity at
$x = 3$.
1. Limit at x = 3:
$\lim_{x \to 3} f(x) = \lim_{x \to 3} (2x^2 - 1) = 2(3)^2 - 1 = 2(9) - 1 = 18 - 1 = 17$.
$\lim_{x \to 3} f(x) = \lim_{x \to 3} (2x^2 - 1) = 2(3)^2 - 1 = 2(9) - 1 = 18 - 1 = 17$.
2. Value of Function:
$f(3) = 2(3)^2 - 1 = 17$.
$f(3) = 2(3)^2 - 1 = 17$.
Since Limit = Value, $f(x)$ is continuous at $x = 3$.
Q3: General Continuity
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Examine the following functions for continuity:
(a) $f(x) = x - 5$
(b) $f(x) = \frac{1}{x-5}$
(c) $f(x) = \frac{x^2-25}{x+5}$
(d) $f(x) = |x-5|$
(a) $f(x) = x - 5$
(b) $f(x) = \frac{1}{x-5}$
(c) $f(x) = \frac{x^2-25}{x+5}$
(d) $f(x) = |x-5|$
(a) $f(x) = x - 5$:
This is a polynomial function. A polynomial function is defined and continuous for all real numbers.
$\therefore$ Continuous on $\mathbb{R}$.
This is a polynomial function. A polynomial function is defined and continuous for all real numbers.
$\therefore$ Continuous on $\mathbb{R}$.
(b) $f(x) = \frac{1}{x-5}$:
This is a rational function. It is undefined when denominator is zero, i.e., $x - 5 = 0 \Rightarrow x = 5$.
$\therefore$ Continuous for all $x \in \mathbb{R} - \{5\}$.
This is a rational function. It is undefined when denominator is zero, i.e., $x - 5 = 0 \Rightarrow x = 5$.
$\therefore$ Continuous for all $x \in \mathbb{R} - \{5\}$.
(c) $f(x) = \frac{x^2-25}{x+5}$:
Defined for $x \neq -5$.
Simplifying: $f(x) = \frac{(x-5)(x+5)}{x+5} = x - 5$.
This acts as a linear polynomial for its domain.
$\therefore$ Continuous for all $x \neq -5$.
Defined for $x \neq -5$.
Simplifying: $f(x) = \frac{(x-5)(x+5)}{x+5} = x - 5$.
This acts as a linear polynomial for its domain.
$\therefore$ Continuous for all $x \neq -5$.
(d) $f(x) = |x-5|$:
Modulus functions are continuous everywhere.
Checking at vertex $x=5$:
$\lim_{x \to 5} |x-5| = 0$ and $f(5) = 0$.
$\therefore$ Continuous everywhere.
Modulus functions are continuous everywhere.
Checking at vertex $x=5$:
$\lim_{x \to 5} |x-5| = 0$ and $f(5) = 0$.
$\therefore$ Continuous everywhere.
Q4: Power Function
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Prove that the function $f(x) = x^n$ is continuous at $x = n$, where n is a
positive integer.
To Prove: Continuity at $x = n$.
1. Limit at x = n:
$\lim_{x \to n} f(x) = \lim_{x \to n} x^n = n^n$.
$\lim_{x \to n} f(x) = \lim_{x \to n} x^n = n^n$.
2. Value at x = n:
$f(n) = n^n$.
$f(n) = n^n$.
Since $\lim_{x \to n} f(x) = f(n)$, the function is continuous at $x = n$.
Q5: Piecewise Function
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Is the function $f$ defined by $f(x) = \begin{cases} x, & x \le 1 \\ 5, & x > 1
\end{cases}$ continuous at $x = 0$? At $x = 1$? At $x = 2$?
1. At x = 0:
$0 < 1$, so $f(x)=x$. Limit is 0, Value is 0. $\Rightarrow$ Continuous.
$0 < 1$, so $f(x)=x$. Limit is 0, Value is 0. $\Rightarrow$ Continuous.
2. At x = 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 $LHL \neq RHL$, the limit does not exist.
$\Rightarrow$ Discontinuous at $x=1$.
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 $LHL \neq RHL$, the limit does not exist.
$\Rightarrow$ Discontinuous at $x=1$.
3. At x = 2:
$2 > 1$, so $f(x) = 5$. Limit is 5, Value is 5. $\Rightarrow$ Continuous.
$2 > 1$, so $f(x) = 5$. Limit is 5, Value is 5. $\Rightarrow$ Continuous.
Q6: Discontinuity Point
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Find all points of discontinuity of $f$: $f(x) = \begin{cases} 2x+3, & x \le
2 \\ 2x-3, & x > 2 \end{cases}$.
Given: Breakpoint is at $x = 2$.
Left Hand Limit (LHL):
$\lim_{x \to 2^-} (2x+3) = 2(2)+3 = 7$.
$\lim_{x \to 2^-} (2x+3) = 2(2)+3 = 7$.
Right Hand Limit (RHL):
$\lim_{x \to 2^+} (2x-3) = 2(2)-3 = 1$.
$\lim_{x \to 2^+} (2x-3) = 2(2)-3 = 1$.
Conclusion:
Since $LHL \neq RHL$, the limit does not exist at $x = 2$.
Since $LHL \neq RHL$, the limit does not exist at $x = 2$.
Point of discontinuity is $x = 2$.
Q7: Discontinuity Points
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Find points of discontinuity: $f(x) = \begin{cases} |x|+3, & x \le -3 \\ -2x,
& -3 < x < 3 \\ 6x+2, & x \ge 3 \end{cases}$.
Check at x = -3:
LHL $= |-3|+3 = 6$.
RHL $= -2(-3) = 6$.
Value $= |-3|+3 = 6$.
$\therefore$ Continuous at $x=-3$.
LHL $= |-3|+3 = 6$.
RHL $= -2(-3) = 6$.
Value $= |-3|+3 = 6$.
$\therefore$ Continuous at $x=-3$.
Check at x = 3:
LHL $= -2(3) = -6$.
RHL $= 6(3)+2 = 20$.
$LHL \neq RHL$.
LHL $= -2(3) = -6$.
RHL $= 6(3)+2 = 20$.
$LHL \neq RHL$.
Function is discontinuous at $x = 3$.
Q8: Modulus Function
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Find points of discontinuity: $f(x) = \begin{cases} \frac{|x|}{x}, & x
\ne 0 \\ 0, & x = 0 \end{cases}$.
At x = 0:
LHL: $\lim_{x \to 0^-} \frac{-x}{x} = -1$.
RHL: $\lim_{x \to 0^+} \frac{x}{x} = 1$.
LHL: $\lim_{x \to 0^-} \frac{-x}{x} = -1$.
RHL: $\lim_{x \to 0^+} \frac{x}{x} = 1$.
Since LHL $\neq$ RHL, limit does not exist.
Discontinuous at $x = 0$.
Q9: Modulus Fraction
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Find points of discontinuity: $f(x) = \begin{cases} \frac{x}{|x|}, & x <
0 \\ -1, & x \ge 0 \end{cases}$.
For x < 0: $|x| = -x$, so $f(x) = x/(-x) = -1$.
At x = 0:
LHL $= -1$.
RHL $= -1$ (given).
Value $= -1$.
LHL $= -1$.
RHL $= -1$ (given).
Value $= -1$.
Continuous everywhere. No points of discontinuity.
Q10: Polynomial Split
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Find points of discontinuity: $f(x) = \begin{cases} x+1, & x \ge 1 \\
x^2+1, & x < 1 \end{cases}$.
At x = 1:
LHL: $\lim_{x \to 1^-} (x^2+1) = 1+1 = 2$.
RHL: $\lim_{x \to 1^+} (x+1) = 1+1 = 2$.
Value: $f(1) = 1+1 = 2$.
LHL: $\lim_{x \to 1^-} (x^2+1) = 1+1 = 2$.
RHL: $\lim_{x \to 1^+} (x+1) = 1+1 = 2$.
Value: $f(1) = 1+1 = 2$.
Continuous everywhere.
Q11: Cubic Split
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Find points of discontinuity: $f(x) = \begin{cases} x^3-3, & x
\le 2 \\ x^2+1, & x > 2 \end{cases}$.
At x = 2:
LHL: $\lim_{x \to 2^-} (x^3-3) = 8-3 = 5$.
RHL: $\lim_{x \to 2^+} (x^2+1) = 4+1 = 5$.
Value: $f(2) = 5$.
LHL: $\lim_{x \to 2^-} (x^3-3) = 8-3 = 5$.
RHL: $\lim_{x \to 2^+} (x^2+1) = 4+1 = 5$.
Value: $f(2) = 5$.
Continuous everywhere.
Q12: High Power
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Find points of discontinuity: $f(x) = \begin{cases} x^{10}-1, & x
\le 1 \\ x^2, & x > 1 \end{cases}$.
At x = 1:
LHL: $1^{10}-1 = 0$.
RHL: $1^2 = 1$.
LHL: $1^{10}-1 = 0$.
RHL: $1^2 = 1$.
Discontinuous at $x=1$.
Q13: Linear Split
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Is $f(x) = \begin{cases} x+5, & x \le 1 \\ x-5, & x > 1
\end{cases}$ continuous?
At x = 1:
LHL: $1+5=6$.
RHL: $1-5=-4$.
LHL: $1+5=6$.
RHL: $1-5=-4$.
Discontinuous at $x=1$.
Q14: Step Function
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Discuss continuity: $f(x) = \begin{cases} 3, & 0 \le x \le 1 \\
4, & 1 < x < 3 \\ 5, & 3 \le x \le 10 \end{cases}$.
At x = 1: LHL=3, RHL=4. Discontinuous.
At x = 3: LHL=4, RHL=5. Discontinuous.
Points of discontinuity: $x=1$ and $x=3$.
Q15: Intervals
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Discuss continuity: $f(x) = \begin{cases} 2x, & x < 0 \\ 0, &
0 \le x \le 1 \\ 4x, & x> 1 \end{cases}$.
At x = 0: LHL=$2(0)=0$, RHL=0. Continuous.
At x = 1: LHL=0, RHL=$4(1)=4$. Discontinuous.
Discontinuous at $x=1$.
Q16: Linear Steps
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Discuss continuity: $f(x) = \begin{cases} -2, & x \le -1 \\
2x, & -1 < x \le 1 \\ 2, & x> 1 \end{cases}$.
At x = -1: LHL=-2, RHL=$2(-1)=-2$.
Continuous.
At x = 1: LHL=$2(1)=2$, RHL=2. Continuous.
Continuous everywhere.
Q17: Find a, b
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Find relationship between $a$ and $b$ if $f(x) =
\begin{cases} ax+1, & x \le 3 \\ bx+3, & x > 3 \end{cases}$ is continuous at $x=3$.
LHL at 3: $a(3)+1 = 3a+1$.
RHL at 3: $b(3)+3 = 3b+3$.
For continuity, $3a+1 = 3b+3 \Rightarrow 3a-3b=2$.
Relationship: $a - b = 2/3$.
Q18: Find Lambda
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For what $\lambda$ is $f(x) = \begin{cases} \lambda(x^2-2x),
& x \le 0 \\ 4x+1, & x > 0 \end{cases}$ continuous at $x=0$?
LHL at 0: $\lambda(0-0) = 0$.
RHL at 0: $4(0)+1 = 1$.
Since $0 \neq 1$, LHL $\neq$ RHL.
No value of $\lambda$ can make it continuous at $x=0$.
Q19: Greatest Integer Function
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Show that $g(x) = x - [x]$ is discontinuous at all integral
points.
Let $c$ be an integer.
LHL: $\lim_{x \to c^-} (x - [x]) = c - (c-1) = 1$.
RHL: $\lim_{x \to c^+} (x - [x]) = c - c = 0$.
Since LHL $\neq$ RHL, it is discontinuous at all integers.
Q20: Trigonometric Polynomial
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Is the function $f(x) = x^2 - \sin x + 5$ continuous at $x =
\pi$?
Limit: $\pi^2 - \sin \pi + 5 = \pi^2 - 0 + 5 = \pi^2 + 5$.
Value: $f(\pi) = \pi^2 - \sin \pi + 5 = \pi^2 + 5$.
Yes, continuous at $x = \pi$.
Q21: Trig Operations
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Discuss continuity of: (a) $\sin x + \cos x$ (b) $\sin x -
\cos x$ (c) $\sin x \cdot \cos x$.
We know that $\sin x$ and $\cos x$ are continuous functions on
$\mathbb{R}$.
Algebra of continuous functions states that sum, difference,
and product of continuous functions are also continuous.
Therefore, all three functions are continuous.
Q22
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Discuss continuity of Cosine, Cosecant, Secant, Cotangent.
Cosine: Continuous everywhere.
Cosecant ($1/\sin x$): Continuous everywhere
except where $\sin x = 0$, i.e., $x = n\pi$.
Secant ($1/\cos x$): Continuous everywhere
except where $\cos x = 0$, i.e., $x = (2n+1)\pi/2$.
Cotangent ($\cos x/\sin x$): Continuous
everywhere except where $\sin x = 0$, i.e., $x = n\pi$.
Q23
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Find discontinuity points: $f(x) = \begin{cases} \frac{\sin
x}{x}, & x < 0 \\ x+1, & x \ge 0 \end{cases}$.
At x = 0:
LHL: $\lim_{x \to 0^-} \frac{\sin x}{x} = 1$ (Standard
Limit).
RHL: $\lim_{x \to 0^+} (x+1) = 0+1 = 1$.
LHL = RHL = f(0). Continuous everywhere.
Q24
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Determine if $f(x) = \begin{cases} x^2 \sin(1/x), & x \ne
0 \\ 0, & x = 0 \end{cases}$ is continuous.
Limit at $x \to 0$:
$-1 \le \sin(1/x) \le 1$.
Multiply by $x^2$: $-x^2 \le x^2 \sin(1/x) \le x^2$.
$-1 \le \sin(1/x) \le 1$.
Multiply by $x^2$: $-x^2 \le x^2 \sin(1/x) \le x^2$.
As $x \to 0$, both $-x^2$ and $x^2 \to 0$.
By Squeeze Theorem, $\lim_{x \to 0} x^2 \sin(1/x) = 0$.
By Squeeze Theorem, $\lim_{x \to 0} x^2 \sin(1/x) = 0$.
Since Limit = Value ($0$), function is continuous.
Q25
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Examine continuity: $f(x) = \begin{cases} \sin x - \cos
x, & x \ne 0 \\ -1, & x = 0 \end{cases}$.
Limit at $x \to 0$: $\sin 0 - \cos 0 = 0 - 1 = -1$.
Value at $x = 0$: $-1$.
Continuous at $x=0$.
Q26
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Find $k$ if continuous at $x = \pi/2$: $f(x) =
\begin{cases} \frac{k \cos x}{\pi - 2x}, & x \ne \pi/2 \\ 3, & x = \pi/2
\end{cases}$.
Limit calculation using substitution $x = \pi/2 + h$:
$\lim_{h \to 0} \frac{k \cos(\pi/2 + h)}{\pi - 2(\pi/2 + h)} = \lim_{h \to 0} \frac{k(-\sin h)}{-2h}$.
$\lim_{h \to 0} \frac{k \cos(\pi/2 + h)}{\pi - 2(\pi/2 + h)} = \lim_{h \to 0} \frac{k(-\sin h)}{-2h}$.
$= \frac{k}{2} \lim_{h \to 0} \frac{\sin h}{h} =
\frac{k}{2}$.
Equating to value: $\frac{k}{2} = 3 \Rightarrow k = 6$.
Value of k is 6.
Q27
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Find $k$ if continuous at $x=2$: $f(x) = \begin{cases}
kx^2, & x \le 2 \\ 3, & x > 2 \end{cases}$.
LHL: $k(2)^2 = 4k$.
RHL: $3$.
$4k = 3 \Rightarrow k = 3/4$.
k = 3/4.
Q28
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Find $k$ if continuous at $x=\pi$: $f(x) = \begin{cases}
kx+1, & x \le \pi \\ \cos x, & x > \pi \end{cases}$.
LHL: $k\pi + 1$.
RHL: $\cos \pi = -1$.
$k\pi + 1 = -1 \Rightarrow k\pi = -2$.
k = -2/π.
Q29
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Find $k$ if continuous at $x=5$: $f(x) = \begin{cases}
kx+1, & x \le 5 \\ 3x-5, & x > 5 \end{cases}$.
LHL: $5k+1$.
RHL: $3(5)-5 = 10$.
$5k+1 = 10 \Rightarrow 5k = 9$.
k = 9/5.
Q30
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Find values of a and b such that the function defined by
$f(x) = \begin{cases} 5, & x \le 2 \\ ax+b, & 2 < x < 10 \\ 21, & x \ge 10
\end{cases}$ is continuous.
At x = 2: LHL=5, RHL=$2a+b$.
$2a+b=5$ ... (i)
$2a+b=5$ ... (i)
At x = 10: LHL=$10a+b$,
RHL=21.
$10a+b=21$ ... (ii)
$10a+b=21$ ... (ii)
Subtract (i) from (ii): $8a = 16 \Rightarrow a =
2$.
Substitute in (i): $2(2)+b=5 \Rightarrow b=1$.
a = 2, b = 1.
Q31
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Show that $f(x) = \cos(x^2)$ is continuous.
Let $g(x) = \cos x$ (Continuous) and $h(x) = x^2$
(Continuous).
$f(x) = g(h(x)) = (g \circ h)(x)$.
Composition of two continuous functions is always
continuous.
Hence, $f(x)$ is continuous everywhere.
Q32
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Show that $f(x) = |\cos x|$ is continuous.
Let $g(x) = |x|$ and $h(x) = \cos x$.
$f(x) = g(h(x))$.
Both modulus and cosine functions are continuous.
Their composition is continuous.
Q33
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Examine that $\sin|x|$ is continuous.
Let $g(x) = \sin x$ and $h(x) = |x|$.
$f(x) = g(h(x)) = \sin|x|$.
Both are continuous, so their composition is
continuous.
Q34
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Find points of discontinuity for $f(x) = |x| -
|x+1|$.
Let $g(x) = |x|$ and $h(x) = |x+1|$. Both are
continuous functions.
The difference of two continuous functions ($g - h$)
is also continuous.
Therefore, there are no points of discontinuity.