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CUET UG Mathematics Applied Previous Year Questions (PYQs)
CUET UG Mathematics Applied Continuity PYQ
CUET UG Mathematics Applied PYQ
Go to Discussion
CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ
Identify the correct option(s):
A. A modulus function is continuous at every point in its domain.
B. A modulus function may or may not be continuous at every point in its domain.
C. Every rational function is continuous in its domain.
D. If a function $f$ is differentiable at a point then it is also continuous at that point.
E. If a function $f$ is continuous at a point then it is also differentiable at that point.
Choose the correct answer from the options given below:
Go to Discussion
CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ
Solution
A. A modulus function $|x|$ is continuous at every point in its domain.
Hence A is true.
B. Since modulus function is continuous everywhere in its domain, this statement is false.
C. A rational function $\frac{p(x)}{q(x)}$ is continuous at every point where $q(x)\neq0$.
Thus it is continuous in its domain. Hence C is true.
D. If a function is differentiable at a point, then it must be continuous at that point.
Hence D is true.
E. Continuity does not necessarily imply differentiability (example: $f(x)=|x|$ at $x=0$).
Hence E is false.
Correct statements are A, C and D.
CUET UG Mathematics Applied PYQ
Go to Discussion
CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ
If
$f(x)=
\begin{cases}
\frac{\sqrt{1-\cos2x}}{x\sqrt2}, & x\ne0 \\
k, & x=0
\end{cases}$
then the value of $k$ will make function $f$ continuous at $x=0$ is:
Go to Discussion
CUET UG Mathematics Applied Previous Year PYQ CUET UG Mathematics Applied CUET UG 2022 Applied Mathematics PYQ
Solution
For continuity at $x=0$,
$\lim_{x\to0}f(x)=f(0)=k$
So evaluate limit:
$\lim_{x\to0}\frac{\sqrt{1-\cos2x}}{x\sqrt2}$
Using identity:
$1-\cos2x=2\sin^2x$
So,
$\sqrt{1-\cos2x}=\sqrt{2\sin^2x}=\sqrt2|\sin x|$
Hence,
$\frac{\sqrt{1-\cos2x}}{x\sqrt2}=\frac{\sqrt2|\sin x|}{x\sqrt2}=\frac{|\sin x|}{x}$
As $x\to0$,
$\frac{|\sin x|}{x}\to1$
Therefore,
$k=1$
CUET UG Mathematics Applied
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