The solution curve of the differential equation, $\left( 1 + e ^ { - x } \right) \left( 1 + y ^ { 2 } \right) \frac { d y } { d x } = y ^ { 2 }$ which passes through the point $( 0,1 )$, is
(1) $y ^ { 2 } + 1 = y \left( \log _ { e } \left( \frac { 1 + e ^ { - x } } { 2 } \right) + 2 \right)$
(2) $y ^ { 2 } + 1 = y \left( \log _ { e } \left( \frac { 1 + e ^ { x } } { 2 } \right) + 2 \right)$
(3) $y ^ { 2 } = 1 + y \log _ { e } \left( \frac { 1 + e ^ { x } } { 2 } \right)$
(4) $y ^ { 2 } = 1 + y \log _ { e } \left( \frac { 1 + e ^ { - x } } { 2 } \right)$

If $\frac { d y } { d x } = \frac { x y } { x ^ { 2 } + y ^ { 2 } } ; y ( 1 ) = 1$; then a value of $x$ satisfying $y(x) = e$ is:

If $y = y ( x )$ is the solution of the differential equation $\frac { 5 + e ^ { x } } { 2 + y } \cdot \frac { d y } { d x } + e ^ { x } = 0$ satisfying $y ( 0 ) = 1$ then value of $y \left( \log _ { e } 13 \right)$ is
(1) 1
(2) - 1
(3) 0
(4) 2

Let $f ( x ) = \int _ { 0 } ^ { x } e ^ { t } f ( t ) d t + e ^ { x }$ be a differentiable function for all $x \in R$. Then $f ( x )$ equals:
(1) $e ^ { \left( e ^ { x } - 1 \right) }$
(2) $e ^ { e ^ { x } } - 1$
(3) $2 e ^ { e ^ { x } } - 1$

If $y = y ( x )$ is the solution of the differential equation, $\frac { d y } { d x } + 2 y \tan x = \sin x , y \left( \frac { \pi } { 3 } \right) = 0$, then the maximum value of the function $y ( x )$ over $R$ is equal to :
(1) 8
(2) $\frac { 1 } { 2 }$
(3) $- \frac { 15 } { 4 }$
(4) $\frac { 1 } { 8 }$

If a curve passes through the origin and the slope of the tangent to it at any point $( x , y )$ is $\frac { x ^ { 2 } - 4 x + y + 8 } { x - 2 }$, then this curve also passes through the point:
(1) $( 5,4 )$
(2) $( 4,4 )$
(3) $( 4,5 )$
(4) $( 5,5 )$

Let $y = y ( x )$ satisfies the equation $\frac { d y } { d x } - | A | = 0$, for all $x > 0$, where $A = \left[ \begin{array} { c c c } y & \sin x & 1 \\ 0 & - 1 & 1 \\ 2 & 0 & \frac { 1 } { x } \end{array} \right]$. If $y ( \pi ) = \pi + 2$, then the value of $y \left( \frac { \pi } { 2 } \right)$ is:
(1) $\frac { \pi } { 2 } + \frac { 4 } { \pi }$
(2) $\frac { \pi } { 2 } - \frac { 1 } { \pi }$
(3) $\frac { 3 \pi } { 2 } - \frac { 1 } { \pi }$
(4) $\frac { \pi } { 2 } - \frac { 4 } { \pi }$

If the solution curve of the differential equation $\left( 2 x - 10 y ^ { 3 } \right) d y + y d x = 0$, passes through the points $( 0,1 )$ and $( 2 , \beta )$, then $\beta$ is a root of the equation? (1) $y ^ { 5 } - 2 y - 2 = 0$ (2) $y ^ { 5 } - y ^ { 2 } - 1 = 0$ (3) $2 y ^ { 5 } - y ^ { 2 } - 2 = 0$ (4) $2 y ^ { 5 } - 2 y - 1 = 0$

A differential equation representing the family of parabolas with axis parallel to $y$-axis and whose length of latus rectum is the distance of the point $( 2 , - 3 )$ from the line $3 x + 4 y = 5$, is given by: (1) $11 \frac { d ^ { 2 } x } { d y ^ { 2 } } = 10$ (2) $11 \frac { d ^ { 2 } y } { d x ^ { 2 } } = 10$ (3) $10 \frac { d ^ { 2 } y } { d x ^ { 2 } } = 11$ (4) $10 \frac { d ^ { 2 } x } { d y ^ { 2 } } = 11$

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 $\underline{\hspace{1cm}}$.

If $x \phi ( x ) = \int _ { 5 } ^ { x } \left( 3 t ^ { 2 } - 2 \phi ^ { \prime } ( t ) \right) d t , x > - 2 , \phi ( 0 ) = 4$, then $\phi ( 2 )$ is

Let the curve $y = y ( x )$ be the solution of the differential equation, $\frac { d y } { d x } = 2 ( x + 1 )$. If the numerical value of area bounded by the curve $y = y ( x )$ and $x$-axis is $\frac { 4 \sqrt { 8 } } { 3 }$, then the value of $y ( 1 )$ is equal to $\_\_\_\_$.

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 $f : \mathbb{R} \rightarrow \mathbb{R}$ be defined as $$f(x) = \begin{cases} \left[e^x\right], & x < 0 \\ ae^x + [x-1], & 0 \leq x < 1 \\ b + [\sin(\pi x)], & 1 \leq x < 2 \\ \left[e^{-x}\right] - c, & x \geq 2 \end{cases}$$ where $a, b, c \in \mathbb{R}$ and $[t]$ denotes greatest integer less than or equal to $t$. Then, which of the following statements is true?
(1) There exists $a, b, c \in \mathbb{R}$ such that $f$ is continuous
(2) If $f$ is discontinuous at exactly one point, then $a + b + c = 1$
(3) If $f$ is discontinuous at exactly one point, then $a + b + c \neq 1$
(4) $f$ is discontinuous at at least two points, for any values of $a, b, c \in \mathbb{R}$

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) $y ^ { 2 } + 2x ^ { 3 } = C(2y ^ { 2 } + x ^ { 4 })$

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) $2 x y \frac { d y } { d x } - \left( x ^ { 2 } - y ^ { 2 } + 4 \right) = 0$

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 $y = y ( x )$ be the solution of the differential equation $\left( 1 - x ^ { 2 } \right) d y = \left( x y + \left( x ^ { 3 } + 2 \right) \sqrt { 1 - x ^ { 2 } } \right) d x , - 1 < x < 1$ and $y ( 0 ) = 0$. If $\int _ { - \frac { 1 } { 2 } } ^ { \frac { 1 } { 2 } } \sqrt { 1 - x ^ { 2 } } y ( x ) d x = k$ then $k ^ { - 1 }$ 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 $y = y ( x )$ be the solution curve of the differential equation $\sin \left( 2 x ^ { 2 } \right) \log _ { e } \left( \tan x ^ { 2 } \right) d y + \left( 4 x y - 4 \sqrt { 2 } x \sin \left( x ^ { 2 } - \frac { \pi } { 4 } \right) \right) d x = 0,0 < x < \sqrt { \frac { \pi } { 2 } }$, which passes through the point $\left( \sqrt { \frac { \pi } { 6 } } , 1 \right)$. Then $\left| y \left( \sqrt { \frac { \pi } { 3 } } \right) \right|$ 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 $\_\_\_\_$.