If e^y(1+x)=1, then d^2y/dx^2 is: Option 1: (dy/dx)^2 Option 2: dy/dx Option 3: -dy/dx Option 4: e^y(dy/dx)

Answer: (dy/dx)^2. Solution: From e^y(1+x)=1, taking logs gives y = -(1+x). Then dy/dx = (-1)/(1+x) and d^2y/dx^2 = 1/((1+x)^2). Since 1/((1+x)^2) = ((-1)/(1+x))^2, we get d^2y/dx^2 = (dy/dx)^2.

Calculus

Continuity & Differentiability

Medium

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If $e^y(1+x)=1$ , then $\dfrac{d^2y}{dx^2}$ is:

$\left(\dfrac{dy}{dx}\right)^2$

$\dfrac{dy}{dx}$

$-\dfrac{dy}{dx}$

$e^y\left(\dfrac{dy}{dx}\right)$

✅ Correct Option: 1

From $e^y(1+x)=1$ , taking logs gives $y = -\log(1+x)$ .

Then $\dfrac{dy}{dx} = \dfrac{-1}{1+x}$ and $\dfrac{d^2y}{dx^2} = \dfrac{1}{(1+x)^2}$ .

Since $\dfrac{1}{(1+x)^2} = \left(\dfrac{-1}{1+x}\right)^2$ , we get $\dfrac{d^2y}{dx^2} = \left(\dfrac{dy}{dx}\right)^2$ .

2025: 15 May Shift 2

If $x = a\cos\alpha + b\sin\alpha$ and $y = a\sin\alpha - b\cos\alpha$ , then $\left(x\frac{dy}{dx} - y^2\frac{d^2y}{dx^2}\right)$ is equal to:

2026: 30 May Shift 2

If $y=e^{2\log_e t}$ and $x=\log_e e^{t^2}$ , then the value of $\frac{d^2y}{dx^2}$ is

2026: 15 May Shift 1

Match the LIST-I with LIST-II

LIST-I (Function in parametric form)	LIST-II ( $\dfrac{dy}{dx}=$ )
A. $x=t, y=t$	I. 1
B. $x=t^2, y=2t$	II. $\dfrac{1}{t}$
C. $x=3t, y=t^2$	III. $\dfrac{2t}{3}$
D. $x=2t, y=3t^2$	IV. $3t$

Choose the correct answer from the options given below:

2026: 12 May Shift 2

Function $f(x)=[x]$ , greatest integer function is

(A) Continuous at $x=2.5$

(B) Continuous at $x=1$

(D) Differentiable at $x=-1.2$

Choose the correct answer from the options given below:

2025: 21 May Shift 2

$f(x) = \begin{cases} mx + 1,\ x \geq \pi/2 \\sin x + n, x \leq \pi/2, & \end{cases}$

is continuous at $x = \pi/2$ , where $m \in \mathbb{Z}$ (set of integers), then $\sin 2n =$

2023: 15 June Shift 2

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 :

If ey(1+x)=1e^y(1+x)=1ey(1+x)=1, then d2ydx2\dfrac{d^2y}{dx^2}dx2d2y​ is:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

If $e^y(1+x)=1$ , then $\dfrac{d^2y}{dx^2}$ is: