A function f(x) is defined by : f(x) = cases x + 2, & if x 0 cases Which of the following is true ? Option 1: f(x) is discontinuous at x = 0. Option 2: f(x) is not a function as f(0) is not defined. Option 3: f(x) is a continuous function on its domain. Option 4: f(x) is a discontinuous function on its domain.

Answer: f(x) is a continuous function on its domain. Solution: The domain of f excludes x = 0, so the domain is R \0\. On x 0: f(x) = -x + 2 is a polynomial, continuous. Since 0 is not in the domain, f is continuous on its domain.

Calculus

Continuity & Differentiability

Medium

A function $f(x)$ is defined by :

$f(x) = \begin{cases} x + 2, & \text{if } x < 0 \\ -x + 2, & \text{if } x > 0 \end{cases}$

Which of the following is true ?

$f(x)$ is discontinuous at $x = 0$ .

$f(x)$ is not a function as $f(0)$ is not defined.

$f(x)$ is a continuous function on its domain.

$f(x)$ is a discontinuous function on its domain.

✅ Correct Option: 3

The domain of $f$ excludes $x = 0$ , so the domain is $\mathbb{R} \setminus \{0\}$ .

On $x < 0$ : $f(x) = x + 2$ is a polynomial, continuous.

On $x > 0$ : $f(x) = -x + 2$ is a polynomial, continuous.

Since $0$ is not in the domain, $f$ is continuous on its domain.

2026: 18 May Shift 1

Match List-I with List-II

List-I	List-II
Parametric Curve	$\dfrac{dy}{dx} =$
(A) $x = t, y = t^2$	(I) $\dfrac{3}{2}t$
(B) $x = t^2, y = t^3$	(II) $2t$
(C) $x = t^3, y = t^2$	(III) $\dfrac{1}{2t}$
(D) $x = t^2, y = t$	(IV) $\dfrac{2}{3t}$

Choose the correct answer from the options given below:

2025: 29 May Shift 2

$\frac{d^2}{dx^2} \left\{ \det \begin{bmatrix} x^3 & x \\ 2 & e^x \end{bmatrix} \right\}$ equals

2025: 13 May Shift 2

If $x = \frac{1-t}{1+t}$ and $y = \frac{3t}{1+t}$ , then $\frac{d^2y}{dx^2}$ is equal to

2025: 3 June Shift 1

Match List-I with List-II

List-I	List-II
Function	Points of discontinuity
(A) $f(x) = \frac{x^2 + 1}{x}$	(I) $x = 4$
(B) $f(x) = \frac{\vert x - 1 \vert}{x - 1}$	(II) $x = 2$
(C) $f(x) = \begin{cases} x - 1, & x < 2 \\ x + 1, & x \ge 2 \end{cases}$	(III) $x = 0$
(D) $f(x) = \frac{1 - x}{(x - 4)}$	(IV) $x = 1$

Choose the correct answer from the options given below:

2025: 2 June Shift 1

If $y = \sin^{-1}x$ , then $(1-x^2)\frac{d^2y}{dx^2}$ is equal to

2025: 16 May Shift 1

If $y = \left(\log\left(x + \sqrt{x^2+a^2}\right)\right)^2$ and $x \neq \frac{1-a^2}{2}$ , then $(x^2+a^2)\frac{d^2y}{dx^2} + x\frac{dy}{dx}$ is equal to:

A function f(x)f(x)f(x) is defined by : f(x)={x+2,if x<0−x+2,if x>0f(x) = \begin{cases} x + 2, & \text{if } x < 0 \\ -x + 2, & \text{if } x > 0 \end{cases}f(x)={x+2,−x+2,​if x<0if x>0​ Which of the following is true ?

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

A function $f(x)$ is defined by :

$f(x) = \begin{cases} x + 2, & \text{if } x < 0 \\ -x + 2, & \text{if } x > 0 \end{cases}$

Which of the following is true ?