1. ⇒ (MHT CET 2023 10th May Evening Shift )

If $x = 3 \tan t$ and $y = 3 \sec t$ , then the value of $\frac{d^{2} y}{d x^{2}}$ at $t = \frac{π}{4}$ is

A. $\frac{- 1}{6 \sqrt{2}}$

B. $\frac{1}{6 \sqrt{2}}$

C. $\frac{1}{3 \sqrt{2}}$

D. $\frac{3}{2 \sqrt{2}}$

Correct Option is (B)

$\begin{aligned} x = 3 \tan t \\ ∴ \frac{d x}{dt} = 3 \sec^{2} t \\ y = 3 \sec t \\ ∴ \frac{d y}{dt} = 3 \sec t \tan t \\ Now, \frac{d y}{d x} = \frac{\frac{d y}{dt}}{\frac{d x}{dt}} = \frac{3 sect \tan t}{3 \sec^{2} t} = \sin t \\ ∴ \frac{d^{2} y}{dx} = \frac{d}{dt} (\sin t) \cdot \frac{dt}{d x} \\ = \cos t \times \frac{1}{3 \sec^{2} t} \\ ∴ \frac{d^{2} y}{d x^{2}} = \frac{\cos^{3} t}{3} \\ ∴ {(\frac{d^{2} y}{d x^{2}})}_{(1 - \frac{π}{4})} = \frac{{(\cos \frac{π}{4})}^{3}}{3} = \frac{1}{6 \sqrt{2}} \end{aligned}$

2. ⇒ (MHT CET 2021 20th September Evening Shift )

If $x = a (t + \sin t), y = a (1 - \cos t)$ , then $\frac{d y}{d x} =$

A. $\tan \frac{t}{2}$

B. $- \frac{1}{2} \tan t$

C. $\frac{1}{2} \tan t$

D. $- \tan \frac{t}{2}$

Correct Option is (A)

$\begin{aligned} x = a (t + \sin t) and y = a (1 - \cos t) \\ ∴ \frac{d x}{d t} = a (1 + \cos t) and \frac{d y}{d t} = a (\sin t) \\ ∴ \frac{d y}{d x} = \frac{(\frac{d y}{d t})}{(\frac{d x}{d t})} = \frac{a \sin t}{a (1 + \cos t)} = \frac{2 \sin \frac{t}{2} \cos \frac{t}{2}}{2 \cos^{2} \frac{t}{2}} = \tan \frac{t}{2} \end{aligned}$

3. ⇒ (MHT CET 2021 20th September Morning Shift )

If $x = a \cos θ, y = b \sin θ$ , then ${[\frac{d^{2} y}{d x^{2}}]}_{θ = \frac{π}{4}} =$

A. $2 (\frac{a^{2}}{b})$

B. $\sqrt{2} (\frac{a^{2}}{b})$

C. $- 2 \sqrt{2} (\frac{b}{a^{2}})$

D. $2 \sqrt{2} (\frac{b}{a^{2}})$

Correct Option is (C)

$\begin{aligned} x = a \cos θ, y = b \sin θ \\ ∴ \frac{d x}{d θ} = - a \sin θ, \frac{d y}{d θ} = b \cos θ \\ ∴ \frac{d y}{d x} = \frac{b \cos θ}{- a \sin θ} = (\frac{- b}{a}) \cot θ \\ \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d θ} (\frac{d y}{d x}) \cdot \frac{d θ}{d x} \\ = \frac{d}{d θ} [(\frac{- b}{a}) \cot θ] \times \frac{1}{(\frac{d x}{d θ})} \\ = \frac{(\frac{- b}{a}) (- {cosec}^{2} θ)}{- a \sin θ} = (\frac{- b}{a}) \times \frac{1}{a} \times \frac{1}{\sin^{3} θ} \\ ∴ {(\frac{d^{2} y}{d x^{2}})}_{θ = \frac{π}{4}} = (\frac{- b}{a^{2}}) [\frac{1}{{(\sin \frac{π}{4})}^{3}}] = (\frac{- b}{a^{2}}) (\sqrt{2})^{3} = - 2 \sqrt{2} (\frac{b}{a^{2}}) \end{aligned}$

⇒Derivative