Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a twice differentiable function such that $$\frac{1}{2y} \int_{x-y}^{x+y} f(t)\, dt = f(x), \quad \text{for all } x \in \mathbb{R},\ y > 0$$ Show that there exist $a, b \in \mathbb{R}$ such that $f(x) = ax + b$ for all $x \in \mathbb{R}$.

Suppose $f ( x )$ is a twice differentiable function on $[ a , b ]$ such that $$f ( a ) = 0 = f ( b )$$ and $$x ^ { 2 } \frac { d ^ { 2 } f ( x ) } { d x ^ { 2 } } + 4 x \frac { d f ( x ) } { d x } + 2 f ( x ) > 0 \text { for all } x \in ( a , b )$$ Then,
(A) $f$ is negative for all $x \in ( a , b )$.
(B) $f$ is positive for all $x \in ( a , b )$.
(C) $f ( x ) = 0$ for exactly one $x \in ( a , b )$.
(D) $f ( x ) = 0$ for at least two $x \in ( a , b )$.

Consider a differentiable function $u : [ 0,1 ] \rightarrow \mathbb { R }$. Assume the function $u$ satisfies $$u ( a ) = \frac { 1 } { 2 r } \int _ { a - r } ^ { a + r } u ( x ) d x , \quad \text{for all } a \in ( 0,1 ) \text{ and all } r < \min ( a , 1 - a ).$$ Which of the following four statements must be true?
(A) $u$ attains its maximum but not its minimum on the set $\{ 0,1 \}$.
(B) $u$ attains its minimum but not maximum on the set $\{ 0,1 \}$.
(C) If $u$ attains either its maximum or its minimum on the set $\{ 0,1 \}$, then it must be constant.
(D) $u$ attains both its maximum and its minimum on the set $\{ 0,1 \}$.

If $f : [ 0 , \infty ) \rightarrow \mathbb { R }$ is a continuous function such that $$f ( x ) + \ln 2 \int _ { 0 } ^ { x } f ( t ) d t = 1 , x \geq 0$$ then for all $x \geq 0$,
(A) $f ( x ) = e ^ { x } \ln 2$.
(B) $f ( x ) = e ^ { - x } \ln 2$.
(C) $f ( x ) = 2 ^ { x }$.
(D) $f ( x ) = \left( \frac { 1 } { 2 } \right) ^ { x }$.

Consider the function:
$$f(x) = \begin{cases} -1 + \arctan x & x < 0 \\ ax + b & x \geq 0 \end{cases}$$
Determine for which values of the real parameters $a, b$ the function is differentiable. Establish whether there exists an interval of $\mathbb{R}$ in which the function $f$ satisfies the hypotheses of Rolle's theorem. Justify your answer.

19. A solution of the differential equation $( d x / d y ) 2 - x ( d x / d y ) + y = 0$ is :

Trans Web Educational Services Pvt. Ltd
B - 147,1st Floor, Sec-6, NOIDA, UP-201301

Website:www.askiitians.com \href{mailto:Email.info@askiitians.com}{Email.info@askiitians.com} Tel:0120-4616500 Ext - 204

III askllTians ||

... Powered By IITians
(A) $y = 2$
(B) $y = 2 x$
(B) $y = 2 x - 4$
(D) $y = 2 \times 2 - 4$

32. The differential equation representing the family of curves $y 2 = 2 c ( x + \sqrt { } c )$, where $c$ is a positive parameter, is of :

III askllTians ||

... Powered By IITians
(A) order 1
(B) order 2
(C) degree 3
(D) 10

6. Let $\mathrm { f } ( \mathrm { x } ) , \mathrm { x } > 0$, be a nornegative continuous function, and let $\mathrm { F } ( \mathrm { x } ) = \int 0 \mathrm { x } \mathrm { f } ( \mathrm { t } ) \mathrm { dt } , \mathrm { x } > 0$. If for some $\mathrm { c } > 0 , \mathrm { f } ( \mathrm { x } ) < \mathrm { cF } ( \mathrm { x } )$ for all $\mathrm { x } > 0$, then show that $\mathrm { f } ( \mathrm { x } ) = 0$ for all $\mathrm { x } > 0$.

13. A hemispherical tank of radius 2 metres is initially full of water and has an outlet of 12 cm 2 cross sectional are at the bottom. The outlet is opened at some instant.
The flow through the outlet is according to the law $\mathrm { v } ( \mathrm { t } ) = 0.6 \sqrt { } ( 2 \mathrm { gh } ( \mathrm { t } ) )$, where $\mathrm { v } ( \mathrm { t } )$ and $\mathrm { h } ( \mathrm { t } )$ are respectively the velocity of the flow through the outlet and the height of water level above the outlet at time $t$, and $g$ is the acceleration due to gravity. Find the time it takes to empty the tank. (Hint : Form a differential equation by relating the decrease of water level to the outflow).

A right circular cone with radius R and height H contains a liquid which evaporates at a rate proportional to its surface area in contact with air (proportionality constant $= \mathrm { k } > 0$ ). Find the time after which the cone is empty.

11. A curve passes through $( 2,0 )$ and the slope of tangent at point $P ( x , y )$ equals $\frac { ( x + 1 ) ^ { 2 } + y - 3 } { ( x + 1 ) }$. Find the equation of the curve and area enclosed by the curve and the $x$-axis in the fourth quadrant.
Sol. $\frac { d y } { d x } = \frac { ( x + 1 ) ^ { 2 } + y - 3 } { x + 1 }$ or, $\frac { \mathrm { dy } } { \mathrm { dx } } = ( \mathrm { x } + 1 ) + \frac { \mathrm { y } - 3 } { \mathrm { x } + 1 }$ Putting $\mathrm { x } + 1 = \mathrm { X } , \mathrm { y } - 3 = \mathrm { Y }$ $\frac { d Y } { d X } = X + \frac { Y } { X }$ $\frac { d Y } { d X } - \frac { Y } { X } = X$ [Figure] I. $F = \frac { 1 } { X } \Rightarrow \frac { 1 } { X } \cdot Y = X + c$ $\frac { \mathrm { y } - 3 } { \mathrm { x } + 1 } = ( \mathrm { x } + 1 ) + \mathrm { c }$. It passes through $( 2,0 ) \Rightarrow \mathrm { c } = - 4$. So, $\mathrm { y } - 3 = ( \mathrm { x } + 1 ) ^ { 2 } - 4 ( \mathrm { x } + 1 )$ $\Rightarrow \mathrm { y } = \mathrm { x } ^ { 2 } - 2 \mathrm { x }$. ⇒ Required area $= \left| \int _ { 0 } ^ { 2 } \left( x ^ { 2 } - 2 x \right) d x \right| = \left| \left[ \frac { x ^ { 3 } } { 3 } - x ^ { 2 } \right] _ { 0 } ^ { 2 } \right| = \frac { 4 } { 3 }$ sq. units.

11. A curve passes through $( 2,0 )$ and the slope of tangent at point $P ( x , y )$ equals $\frac { ( x + 1 ) ^ { 2 } + y - 3 } { ( x + 1 ) }$. Find the equation of the curve and area enclosed by the curve and the $x$-axis in the fourth quadrant.
Sol. $\frac { d y } { d x } = \frac { ( x + 1 ) ^ { 2 } + y - 3 } { x + 1 }$ or, $\frac { \mathrm { dy } } { \mathrm { dx } } = ( \mathrm { x } + 1 ) + \frac { \mathrm { y } - 3 } { \mathrm { x } + 1 }$ Putting $\mathrm { x } + 1 = \mathrm { X } , \mathrm { y } - 3 = \mathrm { Y }$ $\frac { d Y } { d X } = X + \frac { Y } { X }$ $\frac { d Y } { d X } - \frac { Y } { X } = X$ [Figure] I. $F = \frac { 1 } { X } \Rightarrow \frac { 1 } { X } \cdot Y = X + c$ $\frac { \mathrm { y } - 3 } { \mathrm { x } + 1 } = ( \mathrm { x } + 1 ) + \mathrm { c }$. It passes through $( 2,0 ) \Rightarrow \mathrm { c } = - 4$. So, $\mathrm { y } - 3 = ( \mathrm { x } + 1 ) ^ { 2 } - 4 ( \mathrm { x } + 1 )$ $\Rightarrow \mathrm { y } = \mathrm { x } ^ { 2 } - 2 \mathrm { x }$. ⇒ Required area $= \left| \int _ { 0 } ^ { 2 } \left( x ^ { 2 } - 2 x \right) d x \right| = \left| \left[ \frac { x ^ { 3 } } { 3 } - x ^ { 2 } \right] _ { 0 } ^ { 2 } \right| = \frac { 4 } { 3 }$ sq. units.

12. If length of tangent at any point on the curve $y = f ( x )$ intercepted between the point and the $x$-axis is of length 1 . Find the equation of the curve.

13. If $x d y = y ( d x + y d y ) , y ( 1 ) = 1$ and $y ( x ) > 0$. Then $y ( - 3 ) = :$
(a) 3
(b) 2
(c) 1
(d) 0

7. If $f ^ { \prime } ( x ) = - f ( x )$ and $g ( x ) = f ^ { \prime } ( x )$ and $F ( x ) = \left( f \left( \frac { x } { 2 } \right) \right) ^ { 2 } + \left( g \left( \frac { x } { 2 } \right) \right) ^ { 2 }$ and given that $F ( 5 ) = 5$, then $F ( 10 )$ is equal to
(A) 5
(B) 10
(C) 0
(D) 15

Sol. (A)

$\mathrm { f } ^ { \prime } ( \mathrm { x } ) = - \mathrm { f } ( \mathrm { x } )$ and $\mathrm { f } ^ { \prime } ( \mathrm { x } ) = \mathrm { g } ( \mathrm { x } )$ $\Rightarrow \mathrm { f } ^ { \prime \prime } ( \mathrm { x } ) \cdot \mathrm { f } ^ { \prime } ( \mathrm { x } ) + \mathrm { f } ( \mathrm { x } ) \cdot \mathrm { f } ^ { \prime } ( \mathrm { x } ) = 0$ $\Rightarrow \mathrm { f } ( \mathrm { x } ) ^ { 2 } + \left( \mathrm { f } ^ { \prime } ( \mathrm { x } ) \right) ^ { 2 } = \mathrm { c } \Rightarrow \left( \mathrm { f } ( \mathrm { x } ) ^ { 2 } + ( \mathrm { g } ( \mathrm { x } ) ) ^ { 2 } = \mathrm { c } \right.$ $\Rightarrow \mathrm { F } ( \mathrm { x } ) = \mathrm { c } \Rightarrow \mathrm { F } ( 10 ) = 5$.

15. A tangent drawn to the curve $\mathrm { y } = \mathrm { f } ( \mathrm { x } )$ at $\mathrm { P } ( \mathrm { x } , \mathrm { y } )$ cuts the x -axis and y -axis at A and B respectively such that $\mathrm { BP } : \mathrm { AP } = 3 : 1$, given that $\mathrm { f } ( 1 ) = 1$, then
(A) equation of curve is $x \frac { d y } { d x } - 3 y = 0$
(B) normal at (1,1) is $x + 3 y = 4$
(C) curve passes through ( 2 , 1/8)
(D) equation of curve is $x \frac { d y } { d x } + 3 y = 0$

Sol. (C), (D)

Equation of the tangent is
$$\mathrm { Y } - \mathrm { y } = \frac { \mathrm { dy } } { \mathrm { dx } } ( \mathrm { X } - \mathrm { x } )$$
Given $\frac { \mathrm { BP } } { \mathrm { AP } } = \frac { 3 } { 1 }$ so that
$$\begin{aligned} & \Rightarrow \quad \frac { d x } { x } = - \frac { d y } { 3 y } \Rightarrow x \frac { d y } { d x } + 3 y = 0 \\ & \Rightarrow \quad \ln x = - \frac { 1 } { 3 } \ln y - \ln c \Rightarrow \ln x ^ { 3 } = - ( \ln c y ) \\ & \Rightarrow \quad \frac { 1 } { x ^ { 3 } } = c y . \text { Given } f ( 1 ) = 1 \Rightarrow c = 1 \\ & \therefore y = \frac { 1 } { x ^ { 3 } } . \end{aligned}$$
[Figure]

49. The differential equation $\frac { d y } { d x } = \frac { \sqrt { 1 - y ^ { 2 } } } { y }$ determines a family of circles with
(A) variable radii and a fixed centre at $( 0,1 )$
(B) variable radii and a fixed centre at $( 0 , - 1 )$
(C) fixed radius 1 and variable centres along the $x$-axis
(D) fixed radius 1 and variable centres along the $y$-axis Answer O O O O
(A)
(B)
(C)
(D)

Let a solution $y = y ( x )$ of the differential equation
$$x \sqrt { x ^ { 2 } - 1 } d y - y \sqrt { y ^ { 2 } - 1 } d x = 0$$
satisfy $y ( 2 ) = \frac { 2 } { \sqrt { 3 } }$. STATEMENT-1 : $y ( x ) = \sec \left( \sec ^ { - 1 } x - \frac { \pi } { 6 } \right)$ and STATEMENT-2 : $y ( x )$ is given by
$$\frac { 1 } { y } = \frac { 2 \sqrt { 3 } } { x } - \sqrt { 1 - \frac { 1 } { x ^ { 2 } } }$$
(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
(B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
(C) STATEMENT-1 is True, STATEMENT-2 is False
(D) STATEMENT-1 is False, STATEMENT-2 is True

Let $f:\mathbf{R}\rightarrow\mathbf{R}$ be a continuous function which satisfies $$f(x)=\int_{0}^{x}f(t)\,dt.$$ Then the value of $f(\ln5)$ is

Match the statements/expressions in Column I with the open intervals in Column II.
Column I
(A) Interval contained in the domain of definition of non-zero solutions of the differential equation $( x - 3 ) ^ { 2 } y ^ { \prime } + y = 0$
(B) Interval containing the value of the integral $$\int _ { 1 } ^ { 5 } ( x - 1 ) ( x - 2 ) ( x - 3 ) ( x - 4 ) ( x - 5 ) d x$$ (C) Interval in which at least one of the points of local maximum of $\cos ^ { 2 } x + \sin x$ lies
(D) Interval in which $\tan ^ { - 1 } ( \sin x + \cos x )$ is increasing
Column II
(p) $\left( - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right)$
(q) $\left( 0 , \frac { \pi } { 2 } \right)$
(r) $\left( \frac { \pi } { 8 } , \frac { 5 \pi } { 4 } \right)$
(s) $\left( 0 , \frac { \pi } { 8 } \right)$
(t) $( - \pi , \pi )$

Let f be a real-valued differentiable function on $\mathbf { R }$ (the set of all real numbers) such that $f ( 1 ) = 1$. If the $y$-intercept of the tangent at any point $P ( x , y )$ on the curve $y = f ( x )$ is equal to the cube of the abscissa of $P$, then the value of $f ( - 3 )$ is equal to

A curve passes through the point $\left( 1 , \frac { \pi } { 6 } \right)$. Let the slope of the curve at each point $( x , y )$ be $\frac { y } { x } + \sec \left( \frac { y } { x } \right) , x > 0$. Then the equation of the curve is
(A) $\quad \sin \left( \frac { y } { x } \right) = \log x + \frac { 1 } { 2 }$
(B) $\quad \operatorname { cosec } \left( \frac { y } { x } \right) = \log x + 2$
(C) $\quad \sec \left( \frac { 2 y } { x } \right) = \log x + 2$
(D) $\quad \cos \left( \frac { 2 y } { x } \right) = \log x + \frac { 1 } { 2 }$

Consider the family of all circles whose centers lie on the straight line $y = x$. If this family of circles is represented by the differential equation $P y ^ { \prime \prime } + Q y ^ { \prime } + 1 = 0$, where $P , Q$ are functions of $x , y$ and $y ^ { \prime }$ (here $y ^ { \prime } = \frac { d y } { d x } , y ^ { \prime \prime } = \frac { d ^ { 2 } y } { d x ^ { 2 } }$), then which of the following statements is (are) true?
(A) $P = y + x$
(B) $P = y - x$
(C) $P + Q = 1 - x + y + y ^ { \prime } + \left( y ^ { \prime } \right) ^ { 2 }$
(D) $P - Q = x + y - y ^ { \prime } - \left( y ^ { \prime } \right) ^ { 2 }$

If $y = y ( x )$ satisfies the differential equation
$$8 \sqrt { x } ( \sqrt { 9 + \sqrt { x } } ) d y = ( \sqrt { 4 + \sqrt { 9 + \sqrt { x } } } ) ^ { - 1 } d x , \quad x > 0$$
and $y ( 0 ) = \sqrt { 7 }$, then $y ( 256 ) =$
[A] 3
[B] 9
[C] 16
[D] 80

Let $\Gamma$ denote a curve $y = y ( x )$ which is in the first quadrant and let the point $( 1,0 )$ lie on it. Let the tangent to $\Gamma$ at a point $P$ intersect the $y$-axis at $Y _ { P }$. If $P Y _ { P }$ has length 1 for each point $P$ on $\Gamma$, then which of the following options is/are correct?
(A) $y = \log _ { e } \left( \frac { 1 + \sqrt { 1 - x ^ { 2 } } } { x } \right) - \sqrt { 1 - x ^ { 2 } }$
(B) $\quad x y ^ { \prime } + \sqrt { 1 - x ^ { 2 } } = 0$
(C) $y = - \log _ { e } \left( \frac { 1 + \sqrt { 1 - x ^ { 2 } } } { x } \right) + \sqrt { 1 - x ^ { 2 } }$
(D) $x y ^ { \prime } - \sqrt { 1 - x ^ { 2 } } = 0$