If x = e^t and y = e^2t then d^2y/dx^2 = Option 1: 2x Option 2: 2y Option 3: 2 Option 4: 1/2x

Answer: 2. Solution: Given x = e^t and y = e^2t. Since x = e^t, we have: t = (x) Substituting into the expression for y: y = e^2t y = e^(2(x)) y = e^((x^2)) y = x^2 Taking the first derivative with respect to x: dy/dx = 2x Taking the second derivative with respect to x: d^2y/dx^2 = d/dx(2x) d^2y/dx^2 = 2 Therefore, d^2y/dx^2 = 2

Calculus

Continuity & Differentiability

Medium

common

If $x = e^t$ and $y = e^{2t}$ then $\frac{d^2y}{dx^2} =$

$2x$

$2y$

$2$

$\frac{1}{2x}$

✅ Correct Option: 3

Given $x = e^t$ and $y = e^{2t}$ .

Since $x = e^t$ , we have:

$t = \ln(x)$

Substituting into the expression for $y$ :

$y = e^{2t}$

$y = e^{2\ln(x)}$

$y = e^{\ln(x^2)}$

$y = x^2$

Taking the first derivative with respect to $x$ :

$\frac{dy}{dx} = 2x$

Taking the second derivative with respect to $x$ :

$\frac{d^2y}{dx^2} = \frac{d}{dx}(2x)$

$\frac{d^2y}{dx^2} = 2$

Therefore, $\frac{d^2y}{dx^2} = 2$

2026: 19 May Shift 2

Observe the graph of a function $y = f(x)$ and identify the correct statements (s):

A. Function is continuous but not differentiable at $x = a$

B. Function is not defined at $x = a$

C. Function is defined at $x = a$ but not differentiable.

D. Given that at $x = a, y = b$ .

Choose the correct answer from the options given below:

2026: 29 May Shift 1

Match List-I with List-II

List-I (Function $y=$ )	List-II ( $\dfrac{dy}{dx}$ at $x=0$ )
(A) $x+e^x$	(I) 1
(B) $\tan^{-1}x-\dfrac{x}{2}$	(II) $\dfrac{1}{2}$
(C) $\tan^{-1}x$	(III) 0
(D) $x+\cos^{-1}x$	(IV) 2

Choose the correct answer from the options given below:

2026: 31 May Shift 2

If the function $f(x) = \dfrac{(4^x-1)^3}{\sin\dfrac{x}{4}\cdot\log\left(1+\dfrac{x^2}{3}\right)}$ is continuous at $x = 0$ , then the value of $f(0)$ is: (Take $\log_e x = \log x$ )

2025: 29 May Shift 2

If $xy = e^{(x-y)}$ , then $\frac{dy}{dx}$ 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:

2026: 19 May Shift 2

The derivative of $|x|$ with respect to $x$ , where $x \in R \sim \{0\}$ is

If x=etx = e^tx=et and y=e2ty = e^{2t}y=e2t then d2ydx2=\frac{d^2y}{dx^2} =dx2d2y​=

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If $x = e^t$ and $y = e^{2t}$ then $\frac{d^2y}{dx^2} =$