If y = t - 1/t and x = t + 1/t, then dy/dx is equal to Option 1: (t² - 1)/(t² + 1) Option 2: (t² + 1)/(t² - 1) Option 3: (t - 1)/(t + 1) Option 4: (t + 1)/(t - 1)

Answer: (t² + 1)/(t² - 1). Solution: Given y = t - 1/t and x = t + 1/t, where both x and y depend on parameter t. When both variables are given in terms of a parameter, the derivative is: dy/dx = (dy/dt)/(dx/dt) Differentiating y with respect to t: y = t - 1/t dy/dt = 1 - (-1/t^2) dy/dt = 1 + 1/t^2 Differentiating x with respect to t: x = t + 1/t dx/dt = 1 + (-1/t^2) dx/dt = 1 - 1/t^2 Applying the formula: dy/dx = (dy/dt)/(dx/dt) dy/dx = (1 + 1/t^2)/(1 - 1/t^2) Multiplying numerator and denominator by t^2: dy/dx = (1 + 1/t^2)/(1 - 1/t^2) x t^2/t^2 dy/dx = (t^2 + 1)/(t^2 - 1) Therefore, dy/dx = (t^2 + 1)/(t^2 - 1)

Calculus

Continuity & Differentiability

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If $y = t - \frac{1}{t}$ and $x = t + \frac{1}{t}$ , then $\frac{dy}{dx}$ is equal to

$\frac{t² - 1}{t² + 1}$

$\frac{t² + 1}{t² - 1}$

$\frac{t - 1}{t + 1}$

$\frac{t + 1}{t - 1}$

✅ Correct Option: 2

Given $y = t - \frac{1}{t}$ and $x = t + \frac{1}{t}$ , where both x and y depend on parameter t.

When both variables are given in terms of a parameter, the derivative is:

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

Differentiating y with respect to t:

$y = t - \frac{1}{t}$

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

$\frac{dy}{dt} = 1 + \frac{1}{t^2}$

Differentiating x with respect to t:

$x = t + \frac{1}{t}$

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

$\frac{dx}{dt} = 1 - \frac{1}{t^2}$

Applying the formula:

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

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

Multiplying numerator and denominator by $t^2$ :

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

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

Therefore, $\frac{dy}{dx} = \frac{t^2 + 1}{t^2 - 1}$

2026: 22 May Shift 1

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(A) $f(x) = x|x|$ is differentiable at $x = 0$ .

(B) $f(x) = x|x|$ is continuous at $x = 0$ but not differentiable at $x = 0$ .

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Choose the correct answer from the options given below :

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If y=t−1ty = t - \frac{1}{t}y=t−t1​ and x=t+1tx = t + \frac{1}{t}x=t+t1​, then dydx\frac{dy}{dx}dxdy​ is equal to

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CUET Mathematics Past Year Questions (Topic-Wise)

If $y = t - \frac{1}{t}$ and $x = t + \frac{1}{t}$ , then $\frac{dy}{dx}$ is equal to