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. $3t$

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If $y=\tan^{-1}(x+y)$, then $\frac{d^2y}{dx^2}$ is equal to:

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

If $x\sqrt{1+y}+y\sqrt{1+x}=0, x\neq y$, then $\frac{dy}{dx}$ is equal to

For the function $y(x)=\sqrt{x+1}-\sqrt{x-1}$, which of the following are true?

(A) $\left(\frac{dy}{dx}\right)_{at\,x=2}=\frac{1-\sqrt{3}}{2\sqrt{3}}$

(B) $(x^2-1)\frac{d^2y}{dx^2}-\frac{dy}{dx}+4y=0$

(C) $(x^2-1)\left(\frac{dy}{dx}\right)^2-4y^2=0$

(D) If $P(x)\frac{d^2y}{dx^2}+Q(x)\frac{dy}{dx}+\alpha y=0$, then $P(x)-xQ(x)-4\alpha=0$

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Match List I with List II

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If $x=4t^3$ and $y=6t^2$ then $\frac{d^2y}{dx^2}$ at $t=1$ is