CUET Continuity Previous Year Question Paper and Solution with PDF

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CUET Previous Year Questions (PYQs)

CUET Continuity PYQ

Qus : 1 CUET PYQ

List I		List II
A.	Dog : Rabies :: Mosquito :	I.	Bacteria
B.	Amnesia : Memory :: Paralysis :	II.	Liver
C.	Meningitis : Brain :: Cirrhosis :	III.	Movement
D.	Influenza : Virus :: Typhoid :	IV.	Malaria

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CUET Previous Year PYQCUET CUET 2023 PYQ

Solution

Dog causes Rabies, Mosquito causes Malaria → A-IV

Amnesia affects Memory, Paralysis affects Movement → B-III

Meningitis affects Brain, Cirrhosis affects Liver → C-II

Influenza is caused by Virus, Typhoid is caused by Bacteria → D-I

Correct matching: A-IV, B-III, C-II, D-I

Qus : 2 CUET PYQ

The function $f(x)= \frac{[ln(1+ax)-ln(1-b x)]}{x}$ is not defined at $x=0$. What value may be assigned to $f$ at $x=0$, so that it is continuous?

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CUET Previous Year PYQ CUET CUET 2022 PYQ

Solution

Qus : 3 CUET PYQ

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: $f(x)=\tan^2 x$ is continuous at $x=\pi/2$.

Reason R: $g(x)=x^2$ is continuous at $x=\pi/2$.

In the light of the above statements, choose the correct answer from the options given below:

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CUET Previous Year PYQ CUET CUET 2023 PYQ

Solution

$f(x)=\tan^2 x$ is not defined at $x=\pi/2$, hence it is not continuous there.

So Assertion A is false.

$g(x)=x^2$ is a polynomial, hence continuous everywhere, including at $x=\pi/2$.

So Reason R is true.

Qus : 4 CUET PYQ

The point(s) at which the function $f$ given by $f(x)=\begin{cases} \dfrac{x}{|x|}, & x<0 \\ -1, & x\ge 0 \end{cases}$ is continuous is/are.

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CUET Previous Year PYQ CUET CUET 2023 PYQ

Solution

For $x<0$, $|x|=-x$ so $f(x)=\dfrac{x}{-x}=-1$ (constant) ⇒ continuous. For $x>0$, $f(x)=-1$ ⇒ continuous. At $x=0$: $\displaystyle \lim_{x\to0^-} f(x)=-1,\quad \lim_{x\to0^+} f(x)=-1,\quad f(0)=-1$ Hence $f$ is continuous at $x=0$. Therefore, $f$ is continuous for all real $x$.

Qus : 5 CUET PYQ

If $f(x)=\begin{cases}x\sin(\frac{1}{x}), & x\ne0 \\ 0, & x=0\end{cases}$, then $f(x)$ is

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CUET Previous Year PYQ CUET CUET 2025 PYQ

Solution

We have: \[ f(x) = \begin{cases} x \sin\!\left(\tfrac{1}{x}\right), & x \neq 0, \\[6pt] 0, & x = 0. \end{cases} \]

Step 1: Continuity at $x=0$

\[ \lim_{x \to 0} x \sin\!\left(\tfrac{1}{x}\right). \] Since $|\sin(1/x)| \leq 1$, \[ -|x| \;\leq\; x \sin\!\left(\tfrac{1}{x}\right) \;\leq\; |x|. \] By the squeeze theorem, \[ \lim_{x \to 0} f(x) = 0 = f(0). \] ✅ Thus, $f(x)$ is continuous everywhere.

Qus : 6 CUET PYQ

A function f(x) is defined as $$f(x)=\begin{cases}{\frac{1-\cos 4x}{{x}^2}} & {;x{\lt}0} \\ {a} & {;x=0} \\ {\frac{\sqrt[]{x}}{\sqrt[]{(16+\sqrt[]{x})-4}}} & {;x{\gt}0}\end{cases}$$ if the function f(x) is continuous at x = 0, then the value of a is:

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CUET Previous Year PYQ CUET CUET 2024 PYQ

Solution

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CUET Previous Year Questions (PYQs)

CUET Continuity PYQ

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Step 1: Continuity at \(x=0\)

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