If x = a/(1 + t) and y = a/((1 + t)^2) where a > 0, then d^2y/dx^2 at t = 1 is Option 1: 1/a Option 2: 2/a Option 3: 4/a Option 4: Not defined

Answer: 2/a. Solution: Given x = a/(1 + t) and y = a/((1 + t)^2) where a > 0. To find d^2y/dx^2 at t = 1, both x and y depend on parameter t. For parametric equations: dy/dx = (dy/dt)/(dx/dt) Finding dx/dt: x = a(1 + t)^(-1) dx/dt = a * (-1)(1 + t)^(-2) dx/dt = -a/((1 + t)^2) Finding dy/dt: y = a(1 + t)^(-2) dy/dt = a * (-2)(1 + t)^(-3) dy/dt = -2a/((1 + t)^3) Finding dy/dx: dy/dx = (-2a/((1 + t)^3))/(-a/((1 + t)^2)) dy/dx = 2a/((1 + t)^3) x ((1 + t)^2)/a dy/dx = 2/(1 + t) For the second derivative: d^2y/dx^2 = (d/dt(dy/dx))/(dx/dt) Finding d/dt(dy/dx): dy/dx = 2(1 + t)^(-1) d/dt(dy/dx) = 2 * (-1)(1 + t)^(-2) d/dt(dy/dx) = -2/((1 + t)^2) Finding d^2y/dx^2: d^2y/dx^2 = (-2/((1 + t)^2))/(-a/((1 + t)^2)) d^2y/dx^2 = 2/((1 + t)^2) x ((1 + t)^2)/a d^2y/dx^2 = 2/a The (1 + t)^2 terms cancel, so the result is independent of t. Therefore, d^2y/dx^2 at t = 1 is 2/a.

Calculus

Continuity & Differentiability

Medium

applied

If $x = \frac{a}{1 + t}$ and $y = \frac{a}{(1 + t)^2}$ where $a > 0$ , then $\frac{d^2y}{dx^2}$ at $t = 1$ is

$\frac{1}{a}$

$\frac{2}{a}$

$\frac{4}{a}$

Not defined

✅ Correct Option: 2

Given $x = \frac{a}{1 + t}$ and $y = \frac{a}{(1 + t)^2}$ where $a > 0$ .

To find $\frac{d^2y}{dx^2}$ at $t = 1$ , both $x$ and $y$ depend on parameter $t$ .

For parametric equations:

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$

Finding $\frac{dx}{dt}$ :

$x = a(1 + t)^{-1}$

$\frac{dx}{dt} = a \cdot (-1)(1 + t)^{-2}$

$\frac{dx}{dt} = -\frac{a}{(1 + t)^2}$

Finding $\frac{dy}{dt}$ :

$y = a(1 + t)^{-2}$

$\frac{dy}{dt} = a \cdot (-2)(1 + t)^{-3}$

$\frac{dy}{dt} = -\frac{2a}{(1 + t)^3}$

Finding $\frac{dy}{dx}$ :

$\frac{dy}{dx} = \frac{-\frac{2a}{(1 + t)^3}}{-\frac{a}{(1 + t)^2}}$

$\frac{dy}{dx} = \frac{2a}{(1 + t)^3} \times \frac{(1 + t)^2}{a}$

$\frac{dy}{dx} = \frac{2}{1 + t}$

For the second derivative:

$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$

Finding $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ :

$\frac{dy}{dx} = 2(1 + t)^{-1}$

$\frac{d}{dt}\left(\frac{dy}{dx}\right) = 2 \cdot (-1)(1 + t)^{-2}$

$\frac{d}{dt}\left(\frac{dy}{dx}\right) = -\frac{2}{(1 + t)^2}$

Finding $\frac{d^2y}{dx^2}$ :

$\frac{d^2y}{dx^2} = \frac{-\frac{2}{(1 + t)^2}}{-\frac{a}{(1 + t)^2}}$

$\frac{d^2y}{dx^2} = \frac{2}{(1 + t)^2} \times \frac{(1 + t)^2}{a}$

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

The $(1 + t)^2$ terms cancel, so the result is independent of $t$ .

Therefore, $\frac{d^2y}{dx^2}$ at $t = 1$ is $\frac{2}{a}$ .

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:

2024: 16 May Shift 1

If $f(x)$ , defined by $f(x)=\left\{\begin{array}{lll}k x+1 & \text { if } & x \leq \pi \\ \cos x & \text { if } & x>\pi\end{array}\right.$ is continuous at $x=\pi$ , then the value of $k$ is :

2025: 22 May Shift 1

Let $[x]$ denote the greatest integer $\leq t$ and $a \mathbb{Z} = [ax: x \in \mathbb{Z}, a \in \mathbb{R}]$ (where $\mathbb{Z}$ is set of integer and $\mathbb{R}$ is set of real number). The set of points of discontinuity of the function $f(x) = [2x]$ is given by

2026: 29 May Shift 2

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

2026: 18 May Shift 2

If the function $f(x)=\begin{cases} x^2, & x\le c \\ ax+b, & x>c \end{cases}$ is differentiable at $x=c$ , then the relation between $b$ and $c$ is:

2025: 14 May Shift 2

Match List-I with List-II. Here [x] denotes the greatest integer function

$\begin{array}{|l|l|} \hline \rule{0pt}{2.8ex}\text{List-I} & \text{List-II} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(A) } f(x) = [x] & \text{(I) is continuous everywhere but not differentiable at } x=-1 \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(B) } f(x) = |x-1| & \text{(II) is continuous everywhere except at all integral values} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(C) } f(x) = e^{|x|} & \text{(III) is continuous everywhere but not differentiable at } x=1 \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(D) } f(x) = |x+1| & \text{(IV) is continuous everywhere but not differentiable at } x=0 \\[1.2ex] \hline \end{array}$

Choose the correct answer from the options given below:

If x=a1+tx = \frac{a}{1 + t}x=1+ta​ and y=a(1+t)2y = \frac{a}{(1 + t)^2}y=(1+t)2a​ where a>0a > 0a>0 , then d2ydx2\frac{d^2y}{dx^2}dx2d2y​ at t=1t = 1t=1 is

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If $x = \frac{a}{1 + t}$ and $y = \frac{a}{(1 + t)^2}$ where $a > 0$ , then $\frac{d^2y}{dx^2}$ at $t = 1$ is