If the set of all values of a, for which the equation $5 x ^ { 3 } - 15 x - a = 0$ has three distinct real roots, is the interval $( \alpha , \beta )$, then $\beta - 2 \alpha$ is equal to $\_\_\_\_$

Q70. For the function $f ( x ) = ( \cos x ) - x + 1 , x \in \mathbb { R }$, between the following two statements (S1) $f ( x ) = 0$ for only one value of $x$ in $[ 0 , \pi ]$. (S2) $f ( x )$ is decreasing in $\left[ 0 , \frac { \pi } { 2 } \right]$ and increasing in $\left[ \frac { \pi } { 2 } , \pi \right]$.
(1) Both (S1) and (S2) are correct.
(2) Both (S1) and (S2) are incorrect.
(3) Only (S2) is correct.
(4) Only (S1) is correct.

Q71. Let $f ( x ) = 4 \cos ^ { 3 } x + 3 \sqrt { 3 } \cos ^ { 2 } x - 10$. The number of points of local maxima of $f$ in interval $( 0,2 \pi )$ is
(1) 3
(2) 4
(3) 1
(4) 2

Q72. Let the sum of the maximum and the minimum values of the function $f ( x ) = \frac { 2 x ^ { 2 } - 3 x + 8 } { 2 x ^ { 2 } + 3 x + 8 }$ be $\frac { \mathrm { m } } { \mathrm { n } }$, where $\operatorname { gcd } ( \mathrm { m } , \mathrm { n } ) = 1$. Then $\mathrm { m } + \mathrm { n }$ is equal to :
(1) 195
(2) 201
(3) 217
(4) 182
Q73. Let $f : \mathbf { R } \rightarrow \mathbf { R }$ be a function given by $f ( x ) = \left\{ \begin{array} { l l } \frac { 1 - \cos 2 x } { x ^ { 2 } } , & x < 0 \\ \alpha , & x = 0 \\ \frac { \beta \sqrt { 1 - \cos x } } { x } , & x > 0 \end{array} \right.$, where $\alpha , \beta \in \mathbf { R }$. If $f$ is continuous at $x = 0$, then $\alpha ^ { 2 } + \beta ^ { 2 }$ is equal to :
(1) 3
(2) 12
(3) 48
(4) 6

Q72. The number of critical points of the function $f ( x ) = ( x - 2 ) ^ { 2 / 3 } ( 2 x + 1 )$ is
(1) 1
(2) 2
(3) 0
(4) 3

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

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

Q73. If the function $f ( x ) = \left( \frac { 1 } { x } \right) ^ { 2 x } ; x > 0$ attains the maximum value at $x = \frac { 1 } { \mathrm { e } }$ then:
(1) $\mathrm { e } ^ { \pi } < \pi ^ { \mathrm { e } }$
(2) $\mathrm { e } ^ { \pi } > \pi ^ { \mathrm { e } }$
(3) $( 2 e ) ^ { \pi } > \pi ^ { ( 2 e ) }$
(4) $\mathrm { e } ^ { 2 \pi } < ( 2 \pi ) ^ { \mathrm { e } }$

Q73. If the function $f ( x ) = 2 x ^ { 3 } - 9 x ^ { 2 } + 12 \mathrm { a } ^ { 2 } x + 1 , \mathrm { a } > 0$ has a local maximum at $x = \alpha$ and a local minimum at $x = \alpha ^ { 2 }$, then $\alpha$ and $\alpha ^ { 2 }$ are the roots of the equation :
(1) $x ^ { 2 } - 6 x + 8 = 0$
(2) $x ^ { 2 } + 6 x + 8 = 0$
(3) $8 x ^ { 2 } + 6 x - 1 = 0$
(4) $8 x ^ { 2 } - 6 x + 1 = 0$

Q74. Let $f ( x ) = 3 \sqrt { x - 2 } + \sqrt { 4 - x }$ be a real valued function. If $\alpha$ and $\beta$ are respectively the minimum and the maximum values of $f$, then $\alpha ^ { 2 } + 2 \beta ^ { 2 }$ is equal to
(1) 42
(2) 38
(3) 24
(4) 44

Q74. For the function $f ( x ) = \sin x + 3 x - \frac { 2 } { \pi } \left( x ^ { 2 } + x \right)$, where $x \in \left[ 0 , \frac { \pi } { 2 } \right]$, consider the following two statements : (I) f is increasing in ( $0 , \frac { \pi } { 2 }$ ). (II) $f ^ { \prime }$ is decreasing in ( $0 , \frac { \pi } { 2 }$ ).
Between the above two statements,
(1) only (II) is true.
(2) only (I) is true.
(3) neither (I) nor (II) is true.
(4) both (I) and (II) are true

Q74. The interval in which the function $f ( x ) = x ^ { x } , x > 0$, is strictly increasing is
(1) $\left( 0 , \frac { 1 } { e } \right]$
(2) $( 0 , \infty )$
(3) $\left. \left[ \frac { 1 } { e } , \infty \right) \right] _ { V }$
(4) $\left[ \frac { 1 } { e ^ { 2 } } , 1 \right)$

Q86. Let $f : \mathbb { R } \rightarrow \mathbb { R }$ be a thrice differentiable function such that $f ( 0 ) = 0 , f ( 1 ) = 1 , f ( 2 ) = - 1 , f ( 3 ) = 2$ and $f ( 4 ) = - 2$. Then, the minimum number of zeros of $\left( 3 f ^ { \prime } f ^ { \prime \prime } + f f ^ { \prime \prime \prime } \right) ( x )$ is $\_\_\_\_$

Q87. Let the maximum and minimum values of $\left( \sqrt { 8 x - x ^ { 2 } - 12 } - 4 \right) ^ { 2 } + ( x - 7 ) ^ { 2 } , x \in \mathbf { R }$ be M and m , respectively. Then $\mathrm { M } ^ { 2 } - \mathrm { m } ^ { 2 }$ is equal to $\_\_\_\_$

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 $\_\_\_\_$

Q89. Let the set of all positive values of $\lambda$, for which the point of local minimum of the function $\left( 1 + x \left( \lambda ^ { 2 } - x ^ { 2 } \right) \right)$ satisfies $\frac { x ^ { 2 } + x + 2 } { x ^ { 2 } + 5 x + 6 } < 0$, be $( \alpha , \beta )$. Then $\alpha ^ { 2 } + \beta ^ { 2 }$ is equal to $\_\_\_\_$

Q89. Let the set of all values of $p$, for which $f ( x ) = \left( p ^ { 2 } - 6 p + 8 \right) \left( \sin ^ { 2 } 2 x - \cos ^ { 2 } 2 x \right) + 2 ( 2 - p ) x + 7$ does not have any critical point, be the interval $( a , b )$. Then $16 a b$ is equal to $\_\_\_\_$

Suppose that $x$ and $y$ satisfy
$$3 x + y = 18 , \quad x \geqq 1 , \quad y \geqq 6 .$$
We are to find the maximum value and the minimum value of $x y$.
When we express $x y$ in terms of $x$, we have
$$x y = \mathbf { A B } ( x - \mathbf { C } ) ^ { 2 } + \mathbf { D E } .$$
Also, the range of values which $x$ can take is
$$\mathbf { F } \leqq x \leqq \mathbf { G } .$$
Hence, the value of $x y$ is maximized at $x = \mathbf { H }$ and its value there is $\mathbf { I J }$, and the value of $x y$ is minimized at $x = \mathbf { K }$ and its value there is $\mathbf { L M }$.

Let $a$ be a positive real number. We are to investigate local extrema of the function
$$f(x) = x^2 - 5 + 4a\log(2x + a + 8) \quad \left(-\frac{a}{2} - 4 < x < -2\right).$$
(1) When we differentiate the function $f(x)$ with respect to $x$, we obtain
$$f'(x) = \frac{\mathbf{A}(\mathbf{B}\, x + a)(x + \mathbf{C})}{\mathbf{D}\, x + a + \mathbf{E}}.$$
(2) Since a condition of $a$ is that $a > 0$ and the domain of $f(x)$ is $-\frac{a}{2} - 4 < x < -2$, the range of values of $a$ such that $f(x)$ has both a local maximum and a local minimum is
$$\mathbf{F} < a < \mathbf{G}.$$
In such a case, the sum of the local maximum and the local minimum is
$$\frac{a^2}{\mathbf{H}} + \mathbf{I} + \mathbf{IJ}\, a\log\mathbf{K}\, a.$$

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
$$\sin 2 \theta _ { 0 } = \frac { \sqrt { \mathbf { M } } } { \mathbf { N } } , \quad \cos 2 \theta _ { 0 } = - \frac { \sqrt { \mathbf { O } } } { \mathbf { N } } .$$

We are to find the range of the values of a real number $t$ such that the maximum value of the cubic function $$f(x) = \frac{1}{3}x^3 - \frac{t+2}{2}x^2 + 2tx + \frac{2}{3}$$ over the interval $x \leqq 4$ is greater than 6.
First of all, since the derivative of $f(x)$ is $$f'(x) = (x - \mathbf{A})(x - t),$$ we consider the problem by dividing the range of the values of $t$ as follows:
(i) When $t > \mathbf{A}$, $f(x)$ has a local maximum at $x = \mathbf{A}$ and a local minimum at $x = t$. Since $f(4) = \mathbf{B}$, we only have to find the range of the values of $t$ satisfying $f(\mathbf{A}) > 6$.
(ii) When $t = \mathbf{A}$, the maximum value of $f(x)$ over the interval $x \leqq 4$ is $f(\mathbf{C}) = \mathbf{D}$, and hence the condition is not satisfied.
(iii) When $t < \mathbf{A}$, $f(x)$ has a local maximum at $x = t$ and a local minimum at $x = \mathbf{A}$. Since $f(4) = \mathbf{B}$, we only have to find the range of the values of $t$ satisfying $f(t) > 6$.
Here, we note $$f(t) - 6 = -\frac{1}{6}(t + \mathbf{E})(t - \mathbf{EF})^2.$$
From the above, the range of the values of $t$ is to be determined.

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 $\left( \pm \dfrac { 1 } { \mathbf { X } } , \dfrac { \mathbf { Y } } { \mathbf { Z } } \right)$.

Q2 Consider the quadratic function
$$f ( x ) = \frac { 3 } { 4 } x ^ { 2 } - 3 x + 4$$
Let $a$ and $b$ be real numbers satisfying $0 < a < b$ and $2 < b$. We are to find the values of $a$ and $b$ such that the range of the values of the function $y = f ( x )$ on $a \leqq x \leqq b$ is $a \leqq y \leqq b$.
Since the equation of the axis of symmetry of the graph of $y = f ( x )$ is $x = \mathbf { M }$, we divide the problem into two cases as follows:
(i) $\mathbf{M} \leqq a$;
(ii) $0 < a < \mathbf{M}$.
In the case of (i), since the values of $f ( x )$ increase with $x$ on $a \leqq x \leqq b$, the equations $f ( a ) = a$ and $f ( b ) = b$ have to be satisfied. By solving these, we obtain $a = \frac { \mathbf { N } } { \mathbf { O } }$ and $b = \mathbf { P }$. However, this $a$ does not satisfy (i).
In the case of (ii), since the minimum value of $f ( x )$ on $a \leqq x \leqq b$ is $\mathbf { Q }$, we have
$$a = \mathbf { R } .$$
This satisfies (ii). Then since $f ( a ) = \frac { \mathbf { S } } { \mathbf { T } } < b$, we have $f ( b ) = b$. Hence, we obtain
$$b = \mathbf { U } .$$

Given the function
$$f ( x ) = x ^ { 3 } - 3 a x ^ { 2 } - 3 ( 2 a + 1 ) x + a + 2 ,$$
answer the following questions.
(1) For $\mathbf { G }$ $\sim$ $\mathbf { K }$, choose the correct answers from among (0) $\sim$ (5) below, and for the other $\square$, enter the correct numbers.
Since
$$f ^ { \prime } ( x ) = \mathbf { A } ( x - \mathbf { B } a - \mathbf { C } ) ( x + \mathbf { D } ) ,$$
we see that
(i) when $a > \mathbf { EF }$, $f ( x )$ is $\mathbf { G }$ at $x = - \square \mathbf { D }$ and is $\square$ H at $x =$ $\square$ B $a +$ $\square$ C;
(ii) when $a =$ $\square$ EF, $f ( x )$ is always $\square$ I;
(iii) when $a < \mathbf{EF}$, $f ( x )$ is $\square$ J at $x = -$ $\square$ D and is $\square$ K at $x =$ $\square$ B $a +$ $\square$ C. (0) locally maximized
(1) locally minimized
(2) increasing
(3) decreasing
(4) maximized
(5) minimized
(2) When we express the minimum value $m$ of $f ( x )$ over the range $- 1 \leqq x \leqq 1$ in terms of $a$, we have that
(i) when $a \geqq \mathbf { L }$, $m = \mathbf { MN } a$;
(ii) when $\mathbf { OP } \leqq a < \mathbf { L }$, $m = \mathbf { QR } \left( a ^ { 3 } + \mathbf { S } a ^ { 2 } + \mathbf { T } a \right)$;
(iii) when $a < \mathbf{OP}$, $m = \mathbf { U } a + \mathbf { V }$.
(3) The value of $m$ in (2) is maximized at $a = \frac { - \mathbf { W } + \sqrt { \mathbf { X } } } { \square \mathbf { Y } }$.

3. For APPLICANTS IN $\left\{ \begin{array} { l } \text { MATHEMATICS } \\ \text { MATHEMATICS \& STATISTICS } \\ \text { MATHEMATICS \& PHILOSOPHY } \\ \text { MATHEMATICS \& COMPUTER SCIENCE } \end{array} \right\}$ ONLY.

Computer Science applicants should turn to page 14. In this question we shall consider the function $f ( x )$ defined by
$$f ( x ) = x ^ { 2 } - 2 p x + 3$$
where $p$ is a constant.
(i) Show that the function $f ( x )$ has one stationary value in the range $0 < x < 1$ if $0 < p < 1$, and no stationary values in that range otherwise.
In the remainder of the question we shall be interested in the smallest value attained by $f ( x )$ in the range $0 \leqslant x \leqslant 1$. Of course, this value, which we shall call $m$, will depend on $p$.
(ii) Show that if $p \geqslant 1$ then $m = 4 - 2 p$.
(iii) What is the value of $m$ if $p \leqslant 0$ ?
(iv) Obtain a formula for $m$ in terms of $p$, valid for $0 < p < 1$.
(v)Using the axes opposite, sketch the graph of $m$ as a function of $p$ in the range $- 2 \leqslant p \leqslant 2$. [Figure]