If $f ( x )=\lfloor x\rfloor$ for $x \in(-1,2)$, then $f$ is discontinuous at
- A$x =-1,0,1,2$
- B$x=-1,0,1$
- ✓$x=0,1$
- D$x=2$
Answer: C.
View full solution →Question types
Continuity question types
237 questions across 7 question groups — pick any mix to generate a Maths paper with step-by-step answer keys.
237
Questions
7
Question groups
5
Question types
01
MCQ
178 Q→02Solve the following Question.(1 Marks)
6 Q→03Solve the Following Question.(2 Marks)
13 Q→04Solve the Following Question.(3 Marks)
22 Q→05Solve the Following Question.(4 Marks)
10 Q→06Solve the Following Question.(5 Marks)
7 Q→07Complete the following activities and rewrite it : (2M)
1 Q→Sample Questions
Continuity questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
If $f(x)=\left(\frac{4+5 x}{4-7 x}\right)^{\frac{4}{x}}$, for $x \neq 0$ and $f(0)=k$, is continuous at $x=0$, then $k$ is
- A$e ^7$
- B$e^3$
- ✓$e^{12}$
- D$e^{\frac{3}{4}}$
Answer: C.
View full solution →If $f(x)=\frac{12^x-4^{x^2-3^x+1}}{1-\cos 2 x}$, for $x \neq 0$ is continuous at $x=0$ then the value of $f(0)$ is
- A$\frac{\log 12}{2}$
- ✓$\log 2 \cdot \log 3$
- C$\frac{\log 2 \cdot \log 3}{2}$
- DNone of these
Answer: B.
View full solution →$f(x)=\frac{32^x-8^x-4^x+1}{4^x-2^{x+1}+1}, \text { for } x \neq 0$
$=k, \text { for } x=0$
is continuous at $x=0$, then value of ' $k$ ' is
$=k, \text { for } x=0$
is continuous at $x=0$, then value of ' $k$ ' is
- ✓$6$
- B$4$
- C$(\log 2)(\log 4)$
- D$3 \log 4$
Answer: A.
View full solution →$f(x)=\frac{\left(16^x-1\right)\left(9^x-1\right)}{\left(27^x-1\right)\left(32^x-1\right)}$, for $x \neq 0$ $=k_{\text {, }}$ for $x=0$
is continuous at $x=0$, then ' $k$ ' =
is continuous at $x=0$, then ' $k$ ' =
- A$\frac{8}{3}$
- ✓$\frac{8}{15}$
- C$-\frac{8}{15}$
- D$\frac{20}{3}$
Answer: B.
View full solution →Let $f(x) = ax + b$ (where a and b are unknown)
$= x^2 + 5$ for $x \in R$
Find the values of a and b, so that $f(x)$ is continuous at $x = 1$
View full solution →$= x^2 + 5$ for $x \in R$
Find the values of a and b, so that $f(x)$ is continuous at $x = 1$
Show that there is a root for the equation $x^3 – 3x = 0$ between $1$ and $2$.
View full solution →Show that there is a root for the equation $2x^3 – x – 16 = 0$ between $2$ and $3$.
View full solution →Examine the continuity of :
$f(x)=\frac{x^2-9}{x-3}$, for $x \neq 3$
$=8$ for $x=3$, at $x=3$.
View full solution →$f(x)=\frac{x^2-9}{x-3}$, for $x \neq 3$
$=8$ for $x=3$, at $x=3$.
Examine the continuity of :
$f(x)=\sin x$, for $x \leq \frac{\pi}{4}$ $=\cos x$, for $x>\frac{\pi}{4}$, at $x=\frac{\pi}{4}$
View full solution →$f(x)=\sin x$, for $x \leq \frac{\pi}{4}$ $=\cos x$, for $x>\frac{\pi}{4}$, at $x=\frac{\pi}{4}$
Solve using intermediate value theorem:
Show that $5^x-6 x=0$ has a root in $[1,2]$
View full solution →Show that $5^x-6 x=0$ has a root in $[1,2]$
Find $f(a)$, if $f$ is continuous at $x=a$ where, $f(x)=\frac{1-\cos [7(x-\pi)]}{5(x-\pi)^2}$, for $x \neq \pi$ at $a=\pi$
View full solution →Discuss the continuity of f on its domain, where
f(x) = |x + 1|, for -3 ≤ x ≤ 2
= |x – 5|, for 2 < x ≤ 7
f(x) = |x + 1|, for -3 ≤ x ≤ 2
= |x – 5|, for 2 < x ≤ 7
$ \text {If } \mathrm{f}(\mathrm{x})=\frac{\sin 2 x}{5 x}-\mathrm{a}, \text { for } \mathrm{x}>0$
$ =4 \text { for } \mathrm{x}=0 $
$ =\mathrm{x}^2+\mathrm{b}-3, \text { for } \mathrm{x}<0$
is continuous at $x=0$, find $a$ and $b$.
View full solution →$ =4 \text { for } \mathrm{x}=0 $
$ =\mathrm{x}^2+\mathrm{b}-3, \text { for } \mathrm{x}<0$
is continuous at $x=0$, find $a$ and $b$.
If $f(x)=\frac{5^x+5^{-x}-2}{x^2}$, for $x \neq 0$
$=k_{\text {, }}$ for $x=0$
is continuous at $x=0$, then find $k$.
View full solution →$=k_{\text {, }}$ for $x=0$
is continuous at $x=0$, then find $k$.
Solve using intermediate value theorem:
Show that $x^3-5 x^2+3 x+6=0$ has at least two real root between $x=1$ and $x=5$
View full solution →Show that $x^3-5 x^2+3 x+6=0$ has at least two real root between $x=1$ and $x=5$
Discuss the continuity of the following function at the point or on the interval indicated against them. If the function is discontinuous, identify the type of discontinuity and state whether the discontinuity is removable. If it has a removable discontinuity, redefine the function so that it becomes continuous:
$f(x)=\frac{(x+3)\left(x^2-6 x+8\right)}{x^2-x-12}$
View full solution →$f(x)=\frac{(x+3)\left(x^2-6 x+8\right)}{x^2-x-12}$
Identify discontinuity for the following function as either a jump or a removable discontinuity on their respective domain:
$
\begin{aligned}
f(x) & =\frac{x^2+x+1}{x+1}, & & \text { for } x \in[0,3) \\
& =\frac{3 x+4}{x^2-5}, & & \text { for } x \in[3,6]
\end{aligned}
$
View full solution →$
\begin{aligned}
f(x) & =\frac{x^2+x+1}{x+1}, & & \text { for } x \in[0,3) \\
& =\frac{3 x+4}{x^2-5}, & & \text { for } x \in[3,6]
\end{aligned}
$
Identify discontinuity for the following function as either a jump or a removable discontinuity on their respective domain:
$
\begin{aligned}
& \mathrm{f}(\mathrm{x})=x^2+5 x+1, \text { for } 0 \leq x \leq 3 \\
& =x^3+x+5, \quad \text { for } 3<x \leq 6 \\
\end{aligned}
$
View full solution →$
\begin{aligned}
& \mathrm{f}(\mathrm{x})=x^2+5 x+1, \text { for } 0 \leq x \leq 3 \\
& =x^3+x+5, \quad \text { for } 3<x \leq 6 \\
\end{aligned}
$
Identify discontinuity for the following function as either a jump or a removable discontinuity on their respective domain:
$
\begin{aligned}
f(x) & =x^2+x-3 & & \text {, for } x \in[-5,-2) \\
& =x^2-5 & & \text {, for } x \in(-2,5]
\end{aligned}
$
View full solution →$
\begin{aligned}
f(x) & =x^2+x-3 & & \text {, for } x \in[-5,-2) \\
& =x^2-5 & & \text {, for } x \in(-2,5]
\end{aligned}
$
Find $f(a)$, if $f$ is continuous at $x=a$ where, $f(x)=\frac{1+\cos (\pi x)}{\pi(1-x)^2}$, for $x \neq 1$ and at $a =1$
View full solution →Find $k$ if the following function is continuous at the point indicated against them:
$
\left.\begin{array}{rlrl}
f(x) & =\left(\frac{5 x-8}{8-3 x}\right)^{\frac{3}{2 x-4}}, & & \text { for } x \neq 2 \\
& =k_1 & & \text { for } x=2
\end{array}\right\} \text { at } x=2
$
View full solution →$
\left.\begin{array}{rlrl}
f(x) & =\left(\frac{5 x-8}{8-3 x}\right)^{\frac{3}{2 x-4}}, & & \text { for } x \neq 2 \\
& =k_1 & & \text { for } x=2
\end{array}\right\} \text { at } x=2
$
Discuss the continuity of the following function at the point or on the interval indicated against them. If the function is discontinuous, identify the type of discontinuity and state whether the discontinuity is removable. If it has a removable discontinuity, redefine the function so that it becomes continuous:
$f(x)=\frac{(x+3)\left(x^2-6 x+8\right)}{x^2-x-12}$
View full solution →$f(x)=\frac{(x+3)\left(x^2-6 x+8\right)}{x^2-x-12}$
Discuss the continuity of the following function at the point(s) or on the interval indicated against them:
$
\begin{aligned}
& =\frac{x^2-3 x-10}{x-5}, & & \text { for } 3 \leq x \leq 6, x \neq 5 \\
f(x) & =10, & & \text { for } x=5 \\
& =\frac{x^2-3 x-10}{x-5}, & & \text { for } 6\end{aligned}
$
View full solution →$
\begin{aligned}
& =\frac{x^2-3 x-10}{x-5}, & & \text { for } 3 \leq x \leq 6, x \neq 5 \\
f(x) & =10, & & \text { for } x=5 \\
& =\frac{x^2-3 x-10}{x-5}, & & \text { for } 6\end{aligned}
$
For what values of $a$ and $b$ is the function
$ f(x)=\frac{x^2-4}{x-2} \text {, for } x<2 \\
=a x^2-b x+3, \text { for } 2 \leq x<3 \\
=2 x-a+b, \text { for } x \geq 3$
continuous for every $x$ on $R$ ?
View full solution →$ f(x)=\frac{x^2-4}{x-2} \text {, for } x<2 \\
=a x^2-b x+3, \text { for } 2 \leq x<3 \\
=2 x-a+b, \text { for } x \geq 3$
continuous for every $x$ on $R$ ?
Find $a$ and $b$ if the following function is continuous at the point or on the interval indicated against them:
$
\begin{aligned}
f(x) & =a x^2+b x+1, & & \text { for }|2 x-3| \geq 2 \\
& =3 x+2, & & \text { for } \frac{1}{2}\end{aligned}
$
View full solution →$
\begin{aligned}
f(x) & =a x^2+b x+1, & & \text { for }|2 x-3| \geq 2 \\
& =3 x+2, & & \text { for } \frac{1}{2}\end{aligned}
$
Find $a$ and $b$ if the following function is continuous at the point or on the interval indicated against them: $\begin{aligned} & =\frac{4 \tan x+5 \sin x}{a^2-1}, & & \text { for } x<0 \\ f(x) & =\frac{9}{\log 2}, & & \text { for } x=0 \\ & =\frac{11 x+7 x-\cos x}{b^2-1}, & & \text { for } x>0 \end{aligned}$
View full solution →Find $\mathrm{k}$ if the following function is continuous at the point indicated against them:
$
\left.\begin{array}{rl}
f(x)=\frac{45^x-9^2-5^x+1}{\left(k^2-1\right)\left(3^2-1\right)}, & \text { for } x \neq 0 \\
=\frac{2}{3}, & \text { for } x=0
\end{array}\right\} \text { at } \mathrm{x}=0
$
View full solution →$
\left.\begin{array}{rl}
f(x)=\frac{45^x-9^2-5^x+1}{\left(k^2-1\right)\left(3^2-1\right)}, & \text { for } x \neq 0 \\
=\frac{2}{3}, & \text { for } x=0
\end{array}\right\} \text { at } \mathrm{x}=0
$
Determine the values of p and q such that the following function is continuous on the entire real number line.
f(x) = x + 1, for 1 < x < 3
= x2 + px + q, for |x – 2| ≥ 1.
f(x) = x + 1, for 1 < x < 3
= x2 + px + q, for |x – 2| ≥ 1.
Discuss the continuity of $f(x)$ at $x=\frac{\pi}{4}$ where,
$f(x)=\frac{(\sin x+\cos x)^3-2 \sqrt{2}}{\sin 2 x-1}$, for $x \neq \frac{\pi}{4}$
$=\frac{3}{\sqrt{2}}$, for $x=\frac{\pi}{4}$
View full solution →$f(x)=\frac{(\sin x+\cos x)^3-2 \sqrt{2}}{\sin 2 x-1}$, for $x \neq \frac{\pi}{4}$
$=\frac{3}{\sqrt{2}}$, for $x=\frac{\pi}{4}$
Suppose $f(x)=p x+3$ for $a \leq x \leq b$
$=5 x^2-q \text { for } b<x \leq c$
Find the condition on $p, q$, so that $f(x)$ is continuous on $[a, c]$, by filling in the boxes
View full solution →$=5 x^2-q \text { for } b<x \leq c$
Find the condition on $p, q$, so that $f(x)$ is continuous on $[a, c]$, by filling in the boxes
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