If the curve $y = y ( x )$ represented by the solution of the differential equation $\left( 2 x y ^ { 2 } - y \right) d x + x \, d y = 0$, passes through the intersection of the lines $2 x - 3 y = 1$ and $3 x + 2 y = 8$, then $| y ( 1 ) |$ is equal to $\underline{\hspace{1cm}}$.

Let a curve $y = y ( x )$ be given by the solution of the differential equation $\cos \left( \frac { 1 } { 2 } \cos ^ { - 1 } \left( e ^ { - x } \right) \right) dx = \left( \sqrt { e ^ { 2 x } - 1 } \right) dy$. If it intersects $y$-axis at $y = - 1$, and the intersection point of the curve with $x$-axis is $( \alpha , 0 )$, then $e ^ { \alpha }$ is equal to $\underline{\hspace{1cm}}$.

Let $f : R \rightarrow R$ be a continuous function such that $f ( 3 x ) - f ( x ) = x$. If $f ( 8 ) = 7$, then $f ( 14 )$ is equal to:
(1) 4
(2) 10
(3) 11
(4) 16

Let $y = y ( x )$ be the solution curve of the differential equation $\frac { d y } { d x } + \frac { 1 } { x ^ { 2 } - 1 } y = \left( \frac { x - 1 } { x + 1 } \right) ^ { \frac { 1 } { 2 } } , x > 1$ passing through the point $\left( 2 , \sqrt { \frac { 1 } { 3 } } \right)$. Then $\sqrt { 7 } y ( 8 )$ is equal to
(1) $11 + 6 \log _ { e } 3$
(2) $19$
(3) $12 - 2 \log _ { e } 3$
(4) $19 - 6 \log _ { e } 3$

The slope of normal at any point $( x , y ) , x > 0 , y > 0$ on the curve $y = y ( x )$ is given by $\frac { x ^ { 2 } } { x y - x ^ { 2 } y ^ { 2 } - 1 }$. If the curve passes through the point $( 1,1 )$, then $e \cdot y ( e )$ is equal to
(1) $\frac { 1 - \tan ( 1 ) } { 1 + \tan ( 1 ) }$
(2) $\tan ( 1 )$
(3) 1
(4) $\frac { 1 + \tan ( 1 ) } { 1 - \tan ( 1 ) }$

If $y = y ( x )$ is the solution of the differential equation $x \frac { d y } { d x } + 2 y = x e ^ { x } , y ( 1 ) = 0$ then the local maximum value of the function $z ( x ) = x ^ { 2 } y ( x ) - e ^ { x } , x \in R$ is
(1) $1 - e$
(2) 0
(3) $\frac { 1 } { 2 }$
(4) $\frac { 4 } { e } - e$

Let a smooth curve $y = f(x)$ be such that the slope of the tangent at any point $(x, y)$ on it is directly proportional to $\left(\frac{-y}{x}\right)$. If the curve passes through the points $(1, 2)$ and $(8, 1)$, then $\left|y\left(\frac{1}{8}\right)\right|$ is equal to
(1) $2\log_e 2$
(2) 4
(3) 1
(4) $4\log_e 2$

The general solution of the differential equation $(x - y^2)dx + y(5x + y^2)dy = 0$ is
(1) $y ^ { 2 } + x ^ { 4 } = C(y ^ { 2 } + 2x ^ { 3 })$
(2) $y ^ { 2 } + 2x ^ { 4 } = C(y ^ { 2 } + x ^ { 3 })$
(3) $y ^ { 2 } + x ^ { 3 } = C(2y ^ { 2 } + x ^ { 4 })$
(4) $y ^ { 2 } + 2x ^ { 3 } = C(2y ^ { 2 } + x ^ { 4 })$

Let $y = y _ { 1 } ( x )$ and $y = y _ { 2 } ( x )$ be two distinct solutions of the differential equation $\frac { d y } { d x } = x + y$, with $y _ { 1 } ( 0 ) = 0$ and $y _ { 2 } ( 0 ) = 1$ respectively. Then, the number of points of intersection of $y = y _ { 1 } ( x )$ and $y = y _ { 2 } ( x )$ is
(1) 0
(2) 1
(3) 2
(4) 3

The differential equation of the family of circles passing through the points $( 0,2 )$ and $( 0 , - 2 )$ is
(1) $2 x y \frac { d y } { d x } + \left( x ^ { 2 } - y ^ { 2 } + 4 \right) = 0$
(2) $2 x y \frac { d y } { d x } + \left( x ^ { 2 } + y ^ { 2 } - 4 \right) = 0$
(3) $2 x y \frac { d y } { d x } + \left( y ^ { 2 } - x ^ { 2 } + 4 \right) = 0$
(4) $2 x y \frac { d y } { d x } - \left( x ^ { 2 } - y ^ { 2 } + 4 \right) = 0$

If $\frac { d y } { d x } + e ^ { x } \left( x ^ { 2 } - 2 \right) y = \left( x ^ { 2 } - 2 x \right) \left( x ^ { 2 } - 2 \right) e ^ { 2 x }$ and $y ( 0 ) = 0$, then the value of $y ( 2 )$ is
(1) $-1$
(2) 1
(3) 0
(4) $e$

Let the solution curve $y = y ( x )$ of the differential equation $\left( 4 + x ^ { 2 } \right) dy - 2 x \left( x ^ { 2 } + 3 y + 4 \right) dx = 0$ pass through the origin. Then $y ( 2 )$ is equal to $\_\_\_\_$.

Let a curve $y = y ( x )$ pass through the point $( 3,3 )$ and the area of the region under this curve, above the $x$-axis and between the abscissae 3 and $x ( > 3 )$ be $\left( \frac { y } { x } \right) ^ { 3 }$. If this curve also passes through the point $( \alpha , 6 \sqrt { 10 } )$ in the first quadrant, then $\alpha$ is equal to $\_\_\_\_$.

Let $f$ be a differentiable function satisfying $f ( x ) = \frac { 2 } { \sqrt { 3 } } \int _ { 0 } ^ { \sqrt { 3 } } f \left( \frac { \lambda ^ { 2 } x } { 3 } \right) d \lambda , x > 0$ and $f ( 1 ) = \sqrt { 3 }$. If $y = f ( x )$ passes through the point $( \alpha , 6 )$, then $\alpha$ is equal to $\_\_\_\_$ .

Let $S = ( 0 , 2 \pi ) - \left\{ \frac { \pi } { 2 } , \frac { 3 \pi } { 4 } , \frac { 3 \pi } { 2 } , \frac { 7 \pi } { 4 } \right\}$. Let $y = y ( x ) , x \in S$, be the solution curve of the differential equation $\frac { dy } { dx } = \frac { 1 } { 1 + \sin 2 x } , y \left( \frac { \pi } { 4 } \right) = \frac { 1 } { 2 }$. If the sum of abscissas of all the points of intersection of the curve $y = y ( x )$ with the curve $y = \sqrt { 2 } \sin x$ is $\frac { k \pi } { 12 }$, then $k$ is equal to $\_\_\_\_$.

If $\sin\left(\frac{y}{x}\right) = \log_e|x| + \frac{\alpha}{2}$ is the solution of the differential equation $x\cos\left(\frac{y}{x}\right)\frac{dy}{dx} = y\cos\left(\frac{y}{x}\right) + x$ and $y(1) = \frac{\pi}{3}$, then $\alpha^2$ is equal to
(1) 12
(2) 3
(3) 4
(4) 9

Let $f$ be a differentiable function such that $x^2 f(x) - x = 4\int_0^x t\, f(t)\, dt$, $f(1) = \frac{2}{3}$. Then $18\, f(3)$ is equal to
(1) 210
(2) 160
(3) 150
(4) 180

The slope of tangent at any point $(x, y)$ on a curve $y = y(x)$ is $\frac{x^2 + y^2}{2xy}$, $x > 0$. If $y(2) = 0$, then a value of $y(8)$ is
(1) $-4\sqrt{2}$
(2) $2\sqrt{3}$
(3) $-2\sqrt{3}$
(4) $4\sqrt{3}$

The area enclosed by the closed curve $C$ given by the differential equation $\frac{dy}{dx} + \frac{x + a}{y - 2} = 0$, $y(1) = 0$ is $4\pi$. Let $P$ and $Q$ be the points of intersection of the curve $C$ and the $y$-axis. If normals at $P$ and $Q$ on the curve $C$ intersect $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $RS$ is
(1) $2\sqrt{3}$
(2) $\frac{2\sqrt{3}}{3}$
(3) 2
(4) $\frac{4\sqrt{3}}{3}$

Consider a function $\mathrm { f } : \mathbb { N } \rightarrow \mathbb { R }$, satisfying $f ( 1 ) + 2 f ( 2 ) + 3 f ( 3 ) + \ldots + x f ( x ) = x ( x + 1 ) f ( x ) ; x \geq 2$ with $f ( 1 ) = 1$. Then $\frac { 1 } { f ( 2022 ) } + \frac { 1 } { f ( 2028 ) }$ is equal to (1) 8200 (2) 8000 (3) 8400 (4) 8100

Let $y = y ( x ) , y > 0$, be a solution curve of the differential equation $\left( 1 + x ^ { 2 } \right) d y = y ( x - y ) d x$. If $y ( 0 ) = 1$ and $y ( 2 \sqrt { 2 } ) = \beta$, then
(1) $e ^ { 3 \beta - 1 } = e ( 3 + 2 \sqrt { 2 } )$
(2) $e ^ { 3 \beta - 1 } = e ( 5 + \sqrt { 2 } )$
(3) $e ^ { \beta - 1 } = e ^ { - 2 } ( 3 + 2 \sqrt { 2 } )$
(4) $e ^ { \beta - 1 } = e ^ { - 2 } ( 5 + \sqrt { 2 } )$

Let $\mathrm { y } = \mathrm { f } ( \mathrm { x } )$ be the solution of the differential equation $y ( x + 1 ) d x - x ^ { 2 } d y = 0 , y ( 1 ) = e$. Then $\lim _ { x \rightarrow 0 ^ { + } } f ( x )$ is equal to
(1) 0
(2) $\frac { 1 } { e }$
(3) $e ^ { 2 }$
(4) $\frac { 1 } { e ^ { 2 } }$

The solution of the differential equation $\frac{dy}{dx} = -\left(\frac{x^{2} + 3y^{2}}{3x^{2} + y^{2}}\right)$, $y(1) = 0$ is
(1) $\log_{e}|x + y| - \frac{xy}{(x+y)^{2}} = 0$
(2) $\log_{e}|x + y| + \frac{xy}{(x+y)^{2}} = 0$
(3) $\log_{e}|x + y| + \frac{2xy}{(x+y)^{2}} = 0$
(4) $\log_{e}|x + y| - \frac{2xy}{(x+y)^{2}} = 0$

If the solution curve of the differential equation $\left( y - 2 \log _ { e } x \right) d x + \left( x \log _ { e } x ^ { 2 } \right) d y = 0 , x > 1$ passes through the points $\left( e , \frac { 4 } { 3 } \right)$ and $\left( e ^ { 4 } , \alpha \right)$, then $\alpha$ is equal to $\_\_\_\_$

Let $\int _ { 0 } ^ { x } \sqrt { 1 - \left( y ^ { \prime } ( t ) \right) ^ { 2 } } d t = \int _ { 0 } ^ { x } y ( t ) d t , 0 \leq x \leq 3 , y \geq 0 , y ( 0 ) = 0$. Then at $x = 2 , y ^ { \prime \prime } + y + 1$ is equal to
(1) 1
(2) 2
(3) $\sqrt { 2 }$
(4) $1 / 2$