List IList IIP. Let y(x)=cos left(3 cos ^{-1} | vedclass

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(C) $(P)$ Given $y(x) = \cos(3 \cos^{-1} x) = 4x^3 - 3x$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = 12x^2 - 3$ and $\frac{d^2y}{dx^2

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$4 \quad 3 \quad 1 \quad 2$

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(C) $(P)$ Given $y(x) = \cos(3 \cos^{-1} x) = 4x^3 - 3x$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = 12x^2 - 3$ and $\frac{d^2y}{dx^2} = 24x$. Substituting these into the expression: $(x^2-1)(24x) + x(12x^2-3) = 24x^3 - 24x + 12x^3 - 3x = 36x^3 - 27x = 9(4x^3 - 3x) = 9y$. Thus,$\frac{1}{y} \{9y\} = 9$. So $P 	o 4$. $(Q)$ The magnitude of the cross product is $\sum |\vec{a}_k| |\vec{a}_{k+1}| \sin(\frac{2\pi}{n}) = (n-1) \lambda^2 \sin(\frac{2\pi}{n})$. The magnitude of the dot product is $\sum |\vec{a}_k| |\vec{a}_{k+1}| \cos(\frac{2\pi}{n}) = (n-1) \lambda^2 \cos(\frac{2\pi}{n})$. Equating them gives $	an(\frac{2\pi}{n}) = 1$,so $\frac{2\pi}{n} = \frac{\pi}{4}$,which implies $n = 8$. So $Q 	o 3$. $(R)$ The normal to $\frac{x^2}{6} + \frac{y^2}{3} = 1$ at $(x_1, y_1)$ is $\frac{6x}{x_1} - \frac{3y}{y_1} = 3$. Given $P(h, 1)$ lies on the ellipse,$\frac{h^2}{6} + \frac{1}{3} = 1 \implies h^2 = 4 \implies h = 2$ (for positive $h$). The slope of the normal is $\frac{6/x_1}{3/y_1} = \frac{2y_1}{x_1}$. Since it is perpendicular to $x+y=8$ (slope $-1$),the normal slope is $1$. Thus $2y_1 = x_1$. Substituting into the ellipse equation: $\frac{(2y_1)^2}{6} + \frac{y_1^2}{3} = 1 \implies \frac{4y_1^2}{6} + \frac{2y_1^2}{6} = 1 \implies y_1^2 = 1 \implies y_1 = 1$. Then $x_1 = 2$. The normal equation is $\frac{6x}{2} - \frac{3y}{1} = 3 \implies 3x - 3y = 3 \implies x - y = 1$. Since $P(h, 1)$ is on the normal,$h - 1 = 1 \implies h = 2$. So $R 	o 2$. $(S)$ Using $	an^{-1} A + 	an^{-1} B = 	an^{-1}(\frac{A+B}{1-AB})$,we get $\frac{\frac{1}{2x+1} + \frac{1}{4x+1}}{1 - \frac{1}{(2x+1)(4x+1)}} = \frac{2}{x^2} \implies \frac{6x+2}{8x^2+6x} = \frac{2}{x^2} \implies \frac{3x+1}{4x^2+3x} = \frac{2}{x^2} \implies 3x^3 + x^2 = 8x^2 + 6x \implies 3x^3 - 7x^2 - 6x = 0$. Since $x > 0$,$3x^2 - 7x - 6 = 0 \implies (3x+2)(x-3) = 0$. Only $x=3$ is a positive solution. Thus,there is $1$ positive solution. So $S 	o 1$.

List $I$	List $II$
$P.$ Let $y(x)=\cos \left(3 \cos ^{-1} x\right), x \in[-1,1], x \neq \pm \frac{\sqrt{3}}{2}$. Then $\frac{1}{y(x)}\left\{\left(x^2-1\right) \frac{d^2 y(x)}{d x^2}+x \frac{d y(x)}{d x}\right\}$ equals	$1. \ 1$
$Q.$ Let $A_1, A_2, \ldots, A_n(n>2)$ be the vertices of a regular polygon of $n$ sides with its centre at the origin. Let $\vec{a}_k$ be the position vector of the point $A_k, k=1,2, \ldots, n$. If $\left\|\sum_{k=1}^{n-1}\left(\vec{a}_k \times \vec{a}_{k+1}\right)\right\|=\left\|\sum_{k=1}^{n-1}\left(\vec{a}_k \cdot \vec{a}_{k+1}\right)\right\|$,then the minimum value of $n$ is	$2. \ 2$
$R.$ If the normal from the point $P(h, 1)$ on the ellipse $\frac{x^2}{6}+\frac{y^2}{3}=1$ is perpendicular to the line $x+y=8$,then the value of $h$ is	$3. \ 8$
$S.$ Number of positive solutions satisfying the equation $\tan ^{-1}\left(\frac{1}{2 x+1}\right)+\tan ^{-1}\left(\frac{1}{4 x+1}\right)=\tan ^{-1}\left(\frac{2}{x^2}\right)$ is	$4. \ 9$

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