The maximum of the areas of the isosceles triangles with base on the positive $x$-axis and which lie below the curve $y = e^{-x}$ is:
(A) $1/e$
(B) 1
(C) $1/2$
(D) $e$

The minimum area of the triangle formed by any tangent to the ellipse $\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1$ and the coordinate axes is
(A) $a b$
(B) $\frac { a ^ { 2 } + b ^ { 2 } } { 2 }$
(C) $\frac { ( a + b ) ^ { 2 } } { 2 }$
(D) $\frac { a ^ { 2 } + a b + b ^ { 2 } } { 3 }$

A truck is to be driven 300 kilometres (kms.) on a highway at a constant speed of $x$ kms. per hour. Speed rules of the highway require that $30 \leq x \leq 60$. The fuel costs ten rupees per litre and is consumed at the rate $2 + \left( x ^ { 2 } / 600 \right)$ litres per hour. The wages of the driver are 200 rupees per hour. The most economical speed (in kms. per hour) to drive the truck is
(A) 30
(B) 60
(C) $30 \sqrt { 3.3 }$
(D) $20 \sqrt { 33 }$

The maximum of the areas of the isosceles triangles with base on the positive $x$-axis and which lie below the curve $y = e ^ { - x }$ is:
(A) $1 / e$
(B) 1
(C) $1 / 2$
(D) $e$

A circular lawn of diameter 20 meters on a horizontal plane is to be illuminated by a light-source placed vertically above the centre of the lawn. It is known that the illuminance at a point $P$ on the lawn is given by the formula $I = \frac{C \sin\theta}{d^2}$ for some constant $C$, where $d$ is the distance of $P$ from the light-source and $\theta$ is the angle between the line joining the centre of the lawn to $P$ and the line joining the light-source to $P$. Then the maximum possible illuminance at a point on the circumference of the lawn is
(A) $\frac{C}{75\sqrt{3}}$
(B) $\frac{C}{100\sqrt{3}}$
(C) $\frac{C}{150\sqrt{3}}$
(D) $\frac{C}{250\sqrt{3}}$.

Let $A B C D$ be a rectangle with its shorter side $a > 0$ units and perimeter $2 s$ units. Let $P Q R S$ be any rectangle such that vertices $A , B , C$ and $D$ respectively lie on the lines $P Q , Q R , R S$ and $S P$. Then the maximum area of such a rectangle $P Q R S$ in square units is given by
(A) $s ^ { 2 }$
(B) $2 a ( s - a )$
(C) $\frac { s ^ { 2 } } { 2 }$
(D) $\frac { 5 } { 2 } a ( s - a )$.

Consider the following subsets of the plane: $$C_{1} = \left\{(x, y) : x > 0,\ y = \frac{1}{x}\right\}$$ and $$C_{2} = \left\{(x, y) : x < 0,\ y = -1 + \frac{1}{x}\right\}$$ Given any two points $P = (x, y)$ and $Q = (u, v)$ of the plane, their distance $d(P, Q)$ is defined by $$d(P, Q) = \sqrt{(x - u)^{2} + (y - v)^{2}}$$ Show that there exists a unique choice of points $P_{0} \in C_{1}$ and $Q_{0} \in C_{2}$ such that $$d(P_{0}, Q_{0}) \leq d(P, Q) \quad \text{for all } P \in C_{1} \text{ and } Q \in C_{2}.$$

Let $S$ be the set consisting of all those real numbers that can be written as $p - 2 a$ where $p$ and $a$ are the perimeter and area of a right-angled triangle having base length 1 . Then $S$ is
(A) $( 2 , \infty )$
(B) $( 1 , \infty )$
(C) $( 0 , \infty )$
(D) the real line $\mathbb { R }$.

Prove that the largest pentagon (in terms of area) that can be inscribed in a circle of radius 1 is regular (i.e., has equal sides).

A rectangular sheet of fixed perimeter with sides having their lengths in the ratio $8 : 15$ is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is 100, the resulting box has maximum volume. Then the lengths of the sides of the rectangular sheet are
(A) 24
(B) 32
(C) 45
(D) 60

A line $L : y = m x + 3$ meets $y$-axis at $E ( 0,3 )$ and the arc of the parabola $y ^ { 2 } = 16 x$, $0 \leq y \leq 6$ at the point $F \left( x _ { 0 } , y _ { 0 } \right)$. The tangent to the parabola at $F \left( x _ { 0 } , y _ { 0 } \right)$ intersects the $y$-axis at $G \left( 0 , y _ { 1 } \right)$. The slope $m$ of the line $L$ is chosen such that the area of the triangle $E F G$ has a local maximum.
Match List I with List II and select the correct answer using the code given below the lists:
List I

• [P.] $m =$
• [Q.] Maximum area of $\triangle EFG$ is
• [R.] $y_0 =$
• [S.] $y_1 =$

List II

1. $\frac{1}{2}$
2. $4$
3. $2$
4. $1$

Codes:

	P	Q	R	S
(A)	4	1	2	3
(B)	3	4	1	2
(C)	1	3	2	4
(D)	1	3	4	2

Consider the hyperbola $H : x ^ { 2 } - y ^ { 2 } = 1$ and a circle $S$ with center $N \left( x _ { 2 } , 0 \right)$. Suppose that $H$ and $S$ touch each other at a point $P \left( x _ { 1 } , y _ { 1 } \right)$ with $x _ { 1 } > 1$ and $y _ { 1 } > 0$. The common tangent to $H$ and $S$ at $P$ intersects the $x$-axis at point $M$. If ( $l , m$ ) is the centroid of the triangle $\triangle P M N$, then the correct expression(s) is(are)
(A) $\frac { d l } { d x _ { 1 } } = 1 - \frac { 1 } { 3 x _ { 1 } ^ { 2 } }$ for $x _ { 1 } > 1$
(B) $\frac { d m } { d x _ { 1 } } = \frac { x _ { 1 } } { 3 \left( \sqrt { x _ { 1 } ^ { 2 } - 1 } \right) }$ for $x _ { 1 } > 1$
(C) $\frac { d l } { d x _ { 1 } } = 1 + \frac { 1 } { 3 x _ { 1 } ^ { 2 } }$ for $x _ { 1 } > 1$
(D) $\frac { d m } { d y _ { 1 } } = \frac { 1 } { 3 }$ for $y _ { 1 } > 0$

A wire of length 2 units is cut into two parts which are bent respectively to form a square of side $= x$ units and a circle of radius $= r$ units. If the sum of the areas of the square and the circle so formed is minimum, then:
(1) $2x = (\pi + 4)r$
(2) $(4 - \pi)x = \pi r$
(3) $x = 2r$
(4) $2x = r$

A wire of length 2 units is cut into two parts which are bent respectively to form a square of side $= x$ units and a circle of radius $= r$ units. If the sum of the areas of the square and the circle so formed is minimum, then: (1) $2x = (\pi + 4)r$ (2) $(4-\pi)x = \pi r$ (3) $x = 2r$ (4) $2x = r$

Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is:
(1) 12.5
(2) 10
(3) 25
(4) 30

Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is:
(1) 12.5
(2) 10
(3) 25
(4) 30

A man in a car at location Q on a straight highway is moving with speed v . He decides to reach a point $P$ in a field at a distance $d$ from highway (point $M$) as shown in the figure. Speed of the car in the field is half to that on the highway. What should be the distance RM, so that the time taken to reach $P$ is minimum? [Figure]
(1) $\frac { \mathrm { d } } { \sqrt { 3 } }$
(2) $\frac { d } { 2 }$
(3) $\frac { d } { \sqrt { 2 } }$
(4) d

If a right circular cone having maximum volume, is inscribed in a sphere of radius 3 cm , then the curved surface area ( $\mathrm { in } \mathrm { cm } ^ { 2 }$ ) of this cone is
(1) $8 \sqrt { 3 } \pi$
(2) $6 \sqrt { 2 } \pi$
(3) $6 \sqrt { 3 } \pi$
(4) $8 \sqrt { 2 } \pi$

The shortest distance between the point $\left( \frac { 3 } { 2 } , 0 \right)$ and the curve $y = \sqrt { x } , ( x > 0 )$, is
(1) $\frac { \sqrt { 3 } } { 2 }$
(2) $\frac { 5 } { 4 }$
(3) $\frac { 3 } { 2 }$
(4) $\frac { \sqrt { 5 } } { 2 }$

The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is:
(1) $\sqrt { 3 }$
(2) $\frac { 2 } { 3 } \sqrt { 3 }$
(3) $\sqrt { 6 }$
(4) $2 \sqrt { 3 }$

The shortest distance between the line $y = x$ and the curve $y^2 = x - 2$ is
(1) $\frac{7}{4\sqrt{2}}$
(2) $\frac{7}{8}$
(3) $\frac{11}{4\sqrt{2}}$
(4) 2

The area (in sq. units) of the largest rectangle $ABCD$ whose vertices $A$ and $B$ lie on the $x$-axis and vertices $C$ and $D$ lie on the parabola, $y = x ^ { 2 } - 1$ below the $x$-axis, is:
(1) $\frac { 2 } { 3 \sqrt { 3 } }$
(2) $\frac { 1 } { 3 \sqrt { 3 } }$
(3) $\frac { 4 } { 3 }$
(4) $\frac { 4 } { 3 \sqrt { 3 } }$

If $P$ is a point on the parabola $y = x ^ { 2 } + 4$ which is closest to the straight line $y = 4 x - 1$, then the coordinates of $P$ are:
(1) $( - 2,8 )$
(2) $( 1,5 )$
(3) $( 2,8 )$
(4) $( 3,13 )$

The triangle of maximum area that can be inscribed in a given circle of radius ' $r$ ' is :
(1) An equilateral triangle having each of its side of length $\sqrt { 3 } r$.
(2) An isosceles triangle with base equal to $2 r$.
(3) An equilateral triangle of height $\frac { 2 r } { 3 }$.
(4) A right angle triangle having two of its sides of length $2 r$ and $r$.

If the point on the curve $y ^ { 2 } = 6 x$, nearest to the point $\left( 3 , \frac { 3 } { 2 } \right)$ is $( \alpha , \beta )$, then $2 ( \alpha + \beta )$ is equal to $\underline{\hspace{1cm}}$.