The condition on a and b, such that for y = a/x - b/x^2, dy/dx = 0 at x=1 is : Option 1: b = 2a Option 2: b = -2b Option 3: a = 2b Option 4: b = -2a

Answer: a = 2b. Solution: dy/dx = -a/x^2 + 2b/x^3. At x=1: dy/dx = -a + 2b = 0. This gives a = 2b.

Calculus

Continuity & Differentiability

Easy

common

The condition on a and b, such that for $y = \frac{a}{x} - \frac{b}{x^2}$ , $\frac{dy}{dx} = 0$ at $x=1$ is :

$b = 2a$

$b = -2b$

$a = 2b$

$b = -2a$

✅ Correct Option: 3

$\frac{dy}{dx} = -\frac{a}{x^2} + \frac{2b}{x^3}$ .

At $x=1$ : $\frac{dy}{dx} = -a + 2b = 0$ .

This gives $a = 2b$ .

2025: 16 May Shift 1

Match List-I with List-II

List-I	List-II
(A) $f(x) = x \sin x$	(I) is not continuous at $x = -3$
(B) $f(x) = \frac{\vert x\vert }{x}, x \neq 0$ and $f(x) = 1 \text{ at } x = 0$	(II) is continuous everywhere
(C) $f(x) = x - [x]$ , $[x]$ denotes greatest integer function	(III) is not differentiable at $x = 1$
(D) $f(x) = e^{\vert x - 1\vert }$	(IV) is not continuous at $x = 0$

Choose the correct answer from the options given below:

2026: 22 May Shift 1

If $y=e^{\pi x}$ and $t=x^{\pi}$ , then which of following statements is/are TRUE ?

(A) $\frac{d^2y}{dt^2}=\frac{y}{t^2}\left(\frac{\pi(x-1)+1}{\pi}\right)x$

(B) $\frac{d^2y}{dt^2}=y\left(\frac{\pi(x-1)+1}{\pi}\right)x$

(D) $\frac{d^2y}{dt^2}=\frac{e^{\pi}}{\pi}$ at $x=1$

Choose the correct answer from the options given below:

2026: 22 May Shift 2

If $y^2 = ax + b, a,b \in \mathbb{R}$ where $\mathbb{R}$ is the set of real numbers, then $y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2$ is equal to:

2023: 15 June Shift 2

If $f(x) = \begin{cases} \frac{x^2 - 9}{x - 3}, & x \neq 3 \\ 5, & x = 3 \end{cases}$ then $f(x)$ :

2026: 26 May Shift 2

If $y = x^3\log(3x^2+6)$ , then $\dfrac{dy}{dx} =$

2025: 13 May Shift 1

Match List-I with List-II

List-I	List-II
Function	Property
(A) $f(x) = \begin{cases} \frac{x}{\vert x \vert} & : x \neq 0 \\ 0 & : x = 0 \end{cases}$	(I) continuous but not differentiable at $x= 0$
(B) $f(x) = \vert x \vert$	(II) continuous but not differentiable at $x=1$
(C) $f(x) = \vert x^2 - 1 \vert$	(III) discontinuous at $x = 0$
(D) $f(x) = \vert x - 1 \vert$	(IV) continuous but not differentiable at $x = 1, -1$

Choose the correct answer from the options given below:

The condition on a and b, such that for y=ax−bx2y = \frac{a}{x} - \frac{b}{x^2}y=xa​−x2b​, dydx=0\frac{dy}{dx} = 0dxdy​=0 at x=1x=1x=1 is :

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The condition on a and b, such that for $y = \frac{a}{x} - \frac{b}{x^2}$ , $\frac{dy}{dx} = 0$ at $x=1$ is :