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

Moment of inertia of a cylinder of mass m, length L and radius R about an axis passing through its centre and perpendicular to the axis of the cylinder is $I = M \left( \frac { R ^ { 2 } } { 4 } + \frac { L ^ { 2 } } { 12 } \right)$. If such a cylinder is to be made for a given mass of a material, the ratio $\frac { L } { R }$ for it to have minimum possible $I$ is:
(1) $\frac { 2 } { 3 }$
(2) $\frac { 3 } { 2 }$
(3) $\sqrt { \frac { 3 } { 2 } }$
(4) $\sqrt { \frac { 2 } { 3 } }$

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

A box open from top is made from a rectangular sheet of dimension $a \times b$ by cutting squares each of side $x$ from each of the four corners and folding up the flaps. If the volume of the box is maximum, then $x$ is equal to: (1) $\frac { a + b + \sqrt { a ^ { 2 } + b ^ { 2 } - a b } } { 6 }$ (2) $\frac { a + b - \sqrt { a ^ { 2 } + b ^ { 2 } - a b } } { 12 }$ (3) $\frac { a + b - \sqrt { a ^ { 2 } + b ^ { 2 } - a b } } { 6 }$ (4) $\frac { a + b - \sqrt { a ^ { 2 } + b ^ { 2 } + a b } } { 6 }$

Let $x , y > 0$. If $x ^ { 3 } y ^ { 2 } = 2 ^ { 15 }$, then the least value of $3 x + 2 y$ is
(1) 30
(2) 32
(3) 36
(4) 40

Consider a cuboid of sides $2 x , 4 x$ and $5 x$ and a closed hemisphere of radius $r$. If the sum of their surface areas is constant $k$, then the ratio $x : r$, for which the sum of their volumes is maximum, is
(1) $2 : 5$
(2) $19 : 45$
(3) $3 : 8$
(4) $19 : 15$

A wire of length 22 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into an equilateral triangle. Then, the length of the side of the equilateral triangle, so that the combined area of the square and the equilateral triangle is minimum, is equal to (if the full question text was truncated, this is the standard formulation of this problem).

Let $S$ be the set of all $a \in N$ such that the area of the triangle formed by the tangent at the point $P(b, c)$, $b, c \in N$, on the parabola $y^2 = 2ax$ and the lines $x = b$, $y = 0$ is 16 unit$^2$, then $\sum_{a \in S} a$ is equal to $\_\_\_\_$.

A wire of length 20 m is to be cut into two pieces. A piece of length $\ell_1$ is bent to make a square of area $A_1$ and the other piece of length $\ell_2$ is made into a circle of area $A_2$. If $2A_1 + 3A_2$ is minimum then $\pi\ell_1 : \ell_2$ is equal to:
(1) 6:1
(2) $3:1$
(3) $1:6$
(4) $4:1$

A square piece of tin of side 30 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. If the volume of the box is maximum, then its surface area (in $\mathrm{cm}^2$) is equal to
(1) 800
(2) 675
(3) 1025
(4) 900

The maximum area of a triangle whose one vertex is at $( 0,0 )$ and the other two vertices lie on the curve $y = - 2 x ^ { 2 } + 54$ at points $( x , y )$ and $( - x , y )$ where $\mathrm { y } > 0$ is :
(1) 88
(2) 122
(3) 92
(4) 108

Let A be the region enclosed by the parabola $y ^ { 2 } = 2 x$ and the line $x = 24$. Then the maximum area of the rectangle inscribed in the region A is $\_\_\_\_$

Consider the region $R = \left\{(x, y) : x \leq y \leq 9 - \frac{11}{3}x^2,\, x \geq 0\right\}$. The area of the largest rectangle of sides parallel to the coordinate axes and inscribed in $R$, is:
(1) $\frac{730}{119}$
(2) $\frac{625}{111}$
(3) $\frac{821}{123}$
(4) $\frac{567}{121}$

Q73. Let a rectangle $A B C D$ of sides 2 and 4 be inscribed in another rectangle $P Q R S$ such that the vertices of the rectangle $A B C D$ lie on the sides of the rectangle $P Q R S$. Let $a$ and $b$ be the sides of the rectangle $P Q R S$ when its area is maximum. Then $( a + b ) ^ { 2 }$ is equal to :
(1) 72
(2) 60
(3) 64
(4) 80

Q87. Let A be the region enclosed by the parabola $y ^ { 2 } = 2 x$ and the line $x = 24$. Then the maximum area of the rectangle inscribed in the region A is $\_\_\_\_$

Q72. A variable line $L$ passes through the point $( 3,5 )$ and intersects the positive coordinate axes at the points A and B . The minimum area of the triangle OAB , where O is the origin, is :
(1) 30
(2) 25
(3) 40
(4) 35

Given real numbers $x$ and $y$ that satisfy
$$\frac { x ^ { 2 } } { 2 } + \frac { y ^ { 2 } } { 4 } = 1 , \quad x \geqq 0 , \quad y \geqq 0$$
we are to find the maximum value of
$$P = x ^ { 2 } + x y + y ^ { 2 } .$$
Let $x$ and $y$ satisfy the conditions. When we set $x = \sqrt { 2 } \cos \theta \left( 0 \leqq \theta \leqq \frac { \pi } { 2 } \right)$, we have
$$y = \mathbf { A } \sin \theta .$$
Thus $P$ can be represented as
$$\begin{aligned} P & = \sqrt { \mathbf { B } } \sin 2 \theta - \cos 2 \theta + \mathbf { C } \\ & = \sqrt { \mathbf { D } } \sin ( 2 \theta - \alpha ) + \mathbf { E } \end{aligned}$$
where
$$\sin \alpha = \frac { \sqrt { \mathbf { F } } } { \mathbf { G } } , \quad \cos \alpha = \frac { \sqrt { \mathbf { H } } } { \mathbf { I } } \quad \left( 0 < \alpha < \frac { \pi } { 2 } \right) .$$
Hence the maximum value of $P$ is $\sqrt { \square \mathbf { J } } + \mathbf { K }$. Let us denote the $\theta$ at which the value of $P$ is maximized by $\theta _ { 0 }$. Then we have
$$2 \theta _ { 0 } = \alpha + \frac { \pi } { \square } ,$$
and hence
$$\sin 2 \theta _ { 0 } = \frac { \sqrt { \mathbf { M } } } { \mathbf { N } } , \quad \cos 2 \theta _ { 0 } = - \frac { \sqrt { \mathbf { O } } } { \mathbf { N } } .$$

Consider all segments PQ of length 2 such that the end points P and Q are on the parabola $y = x ^ { 2 }$. Denote the mid-point of the segment PQ by M. Among all M, we are to find the coordinates of the ones nearest to the $x$-axis.
Let us denote the coordinates of the end points of segment PQ by $\mathrm{ P }\left( p , p ^ { 2 } \right)$ and $\mathrm{ Q }\left( q , q ^ { 2 } \right)$. Then the $y$-coordinate $m$ of M is
$$m = \frac { p ^ { 2 } + q ^ { 2 } } { \mathbf { M } } .$$
Next, since $\mathrm{ PQ } = 2$, then
$$( p - q ) ^ { 2 } + \left( p ^ { 2 } - q ^ { 2 } \right) ^ { 2 } = \mathbf { N }$$
by the Pythagorean theorem.
Now, when we set $t = p q$, we obtain from (1) and (2) the quadratic equation in $m$
$$\mathbf { O } m ^ { 2 } + m - \mathbf { P } t ^ { 2 } - t - \mathbf { Q } = 0 .$$
When we solve this for $m$, noting that $m > 0$, we have
$$m = - \frac { 1 } { \mathbf { R } } + \sqrt { \left( t + \frac { 1 } { \mathbf { S } } \right) ^ { 2 } + \mathbf{T} } .$$
This shows that $m$ is minimized when $t = - \dfrac { 1 } { \mathbf{U} }$. In this case, $p q = - \dfrac { 1 } { \mathbf{U} }$ and $p ^ { 2 } + q ^ { 2 } = \dfrac { \mathbf { V } } { \mathbf { V } }$, and so we have $p + q = \pm \mathbf { W }$.
Thus the coordinates of the M nearest to the $x$-axis are $\left( \pm \dfrac { 1 } { \mathbf { X } } , \dfrac { \mathbf { Y } } { \mathbf { Z } } \right)$.

Consider a rhombus ABCD with sides of length $a$, where $a$ is a constant. Let $r$ be the radius of the circle O inscribed in the rhombus ABCD, and $\mathrm{K}, \mathrm{L}$, $\mathrm{M}, \mathrm{N}$ be the points of tangency of the circle O and the rhombus. Let $S$ denote the area of the part of the rhombus outside circle O.
We are to find the range of the values of $r$, and the maximum value of $S$.
(1) For each of $\mathbf{A}$ $\sim$ $\mathbf{C}$ below, choose the correct answer from among (0) $\sim$ (9).
Let $\angle \mathrm{ABO} = \theta$. We have $\mathrm{OB} = \mathbf{A}$, and hence $\mathrm{OK} = \mathbf{B}$. Hence, since $( \cos \theta - \sin \theta ) ^ { 2 } \geqq 0$, the range of the values taken by $r$ is
$$0 < r \leqq \mathbf { C } .$$
(0) $a$ (1) $\frac { a } { 2 }$ (2) $\frac { a } { 3 }$ (3) $a \sin \theta$ (4) $a \cos \theta$ (5) $a \tan \theta$ (6) $a \sin ^ { 2 } \theta$ (7) $a \cos ^ { 2 } \theta$ (8) $a \sin \theta \cos \theta$ (9) $a \tan ^ { 2 } \theta$
(2) For each of $\mathbf { D } \sim$ $\mathbf{F}$ below, choose the correct answer from among (0) $\sim$ (9).
When the area $S$ is expressed in terms of $r$, we have
$$S = \mathbf { D } .$$
Here, we observe that when $r = \mathbf { E }$, the value of $\mathbf{D}$ is maximized, and this value for $r$ satisfies (1). Thus, at $r = \mathbf { E }$, $S$ takes the maximum value $\mathbf { F }$.
(0) $2 a r - \pi r ^ { 2 }$ (1) $a r - \frac { \pi } { 2 } r ^ { 2 }$ (2) $\frac { a } { 2 } r - \pi r ^ { 2 }$ (3) $\frac { a r - \pi r ^ { 2 } } { 2 }$ (4) $\frac { 2 a } { \pi }$ (5) $\frac { a } { \pi }$ (6) $\frac { a } { 2 \pi }$ (7) $\frac { 4 a ^ { 2 } } { \pi }$ (8) $\frac { a ^ { 2 } } { \pi }$ (9) $\frac { a ^ { 2 } } { 4 \pi }$

4. For APPLICANTS IN $\left\{ \begin{array} { l } \text { MATHEMATICS } \\ \text { MATHEMATICS \& STATISTICS } \\ \text { MATHEMATICS \& PHILOSOPHY } \end{array} \right\}$ ONLY. Mathematics \& Computer Science and Computer Science applicants should turn to page 14.
As shown in the diagram below: $C$ is the parabola with equation $y = x ^ { 2 } ; P$ is the point $( 0,1 ) ; Q$ is the point ( $a , a ^ { 2 }$ ) on $C ; L$ is the normal to $C$ which passes through $Q$. [Figure]
(i) Find the equation of $L$.
(ii) For what values of $a$ does $L$ pass through $P$ ?
(iii) Determine $| Q P | ^ { 2 }$ as a function of $a$, where $| Q P |$ denotes the distance from $P$ to $Q$.
(iv) Find the values of $a$ for which $| Q P |$ is smallest.
(v) Find a point $R$, in the $x y$-plane but not on $C$, such that $| R Q |$ is smallest for a unique value of $a$. Briefly justify your answer.

a) (1.5 points) In a laboratory experiment, 5 measurements of the same object have been made, which gave the following results: $\mathrm { m } _ { 1 } = 0.92 ; \mathrm { m } _ { 2 } = 0.94 ; \mathrm { m } _ { 3 } = 0.89 ; \mathrm { m } _ { 4 } = 0.90$; $\mathrm { m } _ { 5 } = 0.91$.
The result will be taken as the value of x such that the sum of the squares of the errors is minimized. That is, the value for which the function
$E ( x ) = \left( x - m _ { 1 } \right) ^ { 2 } + \left( x - m _ { 2 } \right) ^ { 2 } + \left( x - m _ { 3 } \right) ^ { 2 } + \left( x - m _ { 4 } \right) ^ { 2 } + \left( x - m _ { 5 } \right) ^ { 2 }$ reaches its minimum.
Calculate this value x .
b) (1 point) Apply the integration by parts method to calculate the integral $\int _ { 1 } ^ { 2 } x ^ { 2 } \ln ( x ) d x$, where ln denotes the natural logarithm.

A right circular cylinder is contained within a sphere of radius 5 cm in such a way that the whole of the circumferences of both ends of the cylinder are in contact with the sphere.
The diagram shows a planar cross section through the centre of the sphere and cylinder.
Find, in cubic centimetres, the maximum possible volume of the cylinder.
A $250 \pi$ B $500 \pi$ C $1000 \pi$ D $\frac { 250 \sqrt { 3 } } { 3 } \pi$ E $\frac { 500 \sqrt { 3 } } { 9 } \pi$ F $\frac { 1000 \sqrt { 3 } } { 9 } \pi$