If x^2 - y^2 = t - 1/t, and x^4 + y^4 = t^2 + 1/t^2, then which of the following is correct? Option 1: xy^3dy/dx + 1 = 0 Option 2: x^3dy/dx + y = 1 Option 3: x^2ydy/dx - 1 = 0 Option 4: x^3ydy/dx + 1 = 0

Answer: x^3ydy/dx + 1 = 0. Solution: Given: x^2 - y^2 = t - 1/t s (i) x^4 + y^4 = t^2 + 1/t^2 s (ii) (x^2 - y^2)^2 = (t - 1/t)^2 x^4 - 2x^2y^2 + y^4 = t^2 - 2 + 1/t^2 s (iii) From (ii), replace x^4 + y^4 with t^2 + 1/t^2 in (iii): t^2 + 1/t^2 - 2x^2y^2 = t^2 - 2 + 1/t^2 -2x^2y^2 = -2 x^2y^2 = 1 s (iv) Differentiating x^2y^2 = 1 with respect to x using the product rule: 2xy^2 + x^2 * 2ydy/dx = 0 xy^2 + x^2ydy/dx = 0 s (v) From (iv), y^2 = 1/x^2 Substituting into (v): x 1/x^2 + x^2ydy/dx = 0 1/x + x^2ydy/dx = 0 1 + x^3ydy/dx = 0 x^3ydy/dx + 1 = 0

Calculus

Continuity & Differentiability

Hard

applied

If $x^2 - y^2 = t - \frac{1}{t}$ , and $x^4 + y^4 = t^2 + \frac{1}{t^2}$ , then which of the following is correct?

$xy^3\frac{dy}{dx} + 1 = 0$

$x^3\frac{dy}{dx} + y = 1$

$x^2y\frac{dy}{dx} - 1 = 0$

$x^3y\frac{dy}{dx} + 1 = 0$

✅ Correct Option: 4

Given:

$x^2 - y^2 = t - \dfrac{1}{t} \quad \cdots (i)$

$x^4 + y^4 = t^2 + \dfrac{1}{t^2} \quad \cdots (ii)$

$(x^2 - y^2)^2 = \left(t - \dfrac{1}{t}\right)^2$

$x^4 - 2x^2y^2 + y^4 = t^2 - 2 + \dfrac{1}{t^2} \quad \cdots (iii)$

From (ii), replace $x^4 + y^4$ with $t^2 + \dfrac{1}{t^2}$ in (iii):

$t^2 + \dfrac{1}{t^2} - 2x^2y^2 = t^2 - 2 + \dfrac{1}{t^2}$

$-2x^2y^2 = -2$

$x^2y^2 = 1 \quad \cdots (iv)$

Differentiating $x^2y^2 = 1$ with respect to $x$ using the product rule:

$2xy^2 + x^2 \cdot 2y\dfrac{dy}{dx} = 0$

$xy^2 + x^2y\dfrac{dy}{dx} = 0 \quad \cdots (v)$

From (iv), $y^2 = \dfrac{1}{x^2}$

Substituting into (v):

$x \cdot \dfrac{1}{x^2} + x^2y\dfrac{dy}{dx} = 0$

$\dfrac{1}{x} + x^2y\dfrac{dy}{dx} = 0$

$1 + x^3y\dfrac{dy}{dx} = 0$

$x^3y\dfrac{dy}{dx} + 1 = 0$

2026: 18 May Shift 2

For the function $(ex)^y=e^x$ ;

Match the LIST-I with LIST-II

LIST-I	LIST-II
A. $y$ is equal to	I. $\dfrac{1-\log_e x}{x(1+\log_e x)^3}$
B. value of $\dfrac{dy}{dx}$	II. $\dfrac{x}{1+\log_e x}$
C. value of $\dfrac{d^2y}{dx^2}$	III. $\dfrac{\log_e x}{(1+\log_e x)^2}$
D. $x^y$ is equal to	IV. $e^{x-y}$

Choose the correct answer from the options given below:

2025: 30 May Shift 1

Match List-I with List-II

List-I	List-II
Function f(x)	Points of Non-Differentiability
---	---
(A) $f(x) = \vert x\vert + 1$	(I) Not differentiable at $x = 3$ only
(B) $f(x) = \vert x - 3\vert$	(II) Not differentiable at $x = -3$ only
(C) $f(x) = \vert x + 3\vert$	(III) Not differentiable at $x = 3, -3$ only
(D) $f(x) = \vert x^2 - 9\vert$	(IV) Not differentiable at $x = 0$ only

Choose the correct answer from the options given below:

2025: 30 May Shift 2

The value of derivative of the function $\cot^{-1}\{(\cos 2x)^{1/2}\}$ at $x = \frac{\pi}{6}$ is

2025: 22 May Shift 2

Match List-I with List-II

Where $\mathbb{R}$ is set of real numbers

List-I	List-II
(A) $\sin x$ is continuous on:	(I) $\mathbb{R} - \{0\}$
(B) $\tan x$ is continuous on:	(II) $\mathbb{R}$
(C) $\cot x$ is continuous on:	(III) $\mathbb{R} - \{n\pi: n \in \mathbb{Z}\}$
(D) $x^{-n}, n \in \mathbb{N}$ is continuous on:	(IV) $\mathbb{R} - \left\{(2n + 1)\frac{\pi}{2}: n \in \mathbb{Z}\right\}$

Choose the correct answer from the options given below:

2026: 29 May Shift 1

For the function $f(x)=|\cos x|$ , which of the following statements are correct?

(A) $f(x)$ is continuous for all $x \in \mathbb{R}$

(B) $f(x)$ is differentiable for all $x \in \mathbb{R}$

(D) $f(x)$ is not differentiable at $x=n\pi, n \in \mathbb{Z}$

Choose the correct answer from the options given below:

2025: 15 May Shift 1

If the function defined by $f(x) = \begin{cases} \ kx^2 + 1, & \text{if } x \le 1 \\ 2 , & \text{if } x > 1 \end{cases}$ is continuous at $x = 1$ , then k is equal to

If x2−y2=t−1tx^2 - y^2 = t - \frac{1}{t}x2−y2=t−t1​, and x4+y4=t2+1t2x^4 + y^4 = t^2 + \frac{1}{t^2}x4+y4=t2+t21​, then which of the following is correct?

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If $x^2 - y^2 = t - \frac{1}{t}$ , and $x^4 + y^4 = t^2 + \frac{1}{t^2}$ , then which of the following is correct?