If the value of the integral $\int _ { - \frac { \pi } { 2 } } ^ { \frac { \pi } { 2 } } \left( \frac { x ^ { 2 } \cos x } { 1 + \pi ^ { x } } + \frac { 1 + \sin ^ { 2 } x } { 1 + e ^ { ( \sin x ) ^ { 2023 } } } \right) d x = \frac { \pi } { 4 } ( \pi + a ) - 2$, then the value of $a$ is
(1) 3
(2) $- \frac { 3 } { 2 }$
(3) 2
(4) $\frac { 3 } { 2 }$

Let $\mathrm { fx } = \int _ { 0 } ^ { \mathrm { x } } \mathrm { gt } \log _ { \mathrm { e } } \frac { 1 - \mathrm { t } } { 1 + \mathrm { t } } \mathrm { dt }$, where g is a continuous odd function. If $\int _ { - \frac { \pi } { 2 } } ^ { \frac { \pi } { 2 } } \mathrm { fx } + \frac { \mathrm { x } ^ { 2 } \cos \mathrm { x } } { 1 + \mathrm { e } ^ { \mathrm { x } } } \mathrm { dx } = \frac { \pi ^ { 2 } } { \alpha } - \alpha$, then $\alpha$ is equal to $\_\_\_\_$.

Let for $f ( x ) = 7 \tan ^ { 8 } x + 7 \tan ^ { 6 } x - 3 \tan ^ { 4 } x - 3 \tan ^ { 2 } x , \quad \mathrm { I } _ { 1 } = \int _ { 0 } ^ { \pi / 4 } f ( x ) \mathrm { d } x$ and $\mathrm { I } _ { 2 } = \int _ { 0 } ^ { \pi / 4 } x f ( x ) \mathrm { d } x$. Then $7 \mathrm { I } _ { 1 } + 12 \mathrm { I } _ { 2 }$ is equal to:
(1) 2
(2) 1
(3) $2 \pi$
(4) $\pi$

If $\mathrm { I } = \int _ { 0 } ^ { \frac { \pi } { 2 } } \frac { \sin ^ { \frac { 3 } { 2 } } x } { \sin ^ { \frac { 3 } { 2 } } x + \cos ^ { \frac { 3 } { 2 } } x } \mathrm {~d} x$, then $\int _ { 0 } ^ { 2\mathrm{I} } \frac { x \sin x \cos x } { \sin ^ { 4 } x + \cos ^ { 4 } x } \mathrm {~d} x$ equals :
(1) $\frac { \pi ^ { 2 } } { 12 }$
(2) $\frac { \pi ^ { 2 } } { 4 }$
(3) $\frac { \pi ^ { 2 } } { 16 }$
(4) $\frac { \pi ^ { 2 } } { 8 }$

Let $f(x) = \frac{2^{x+2} + 16}{2^{2x+1} + 2^{x+4} + 32}$. Then the value of $8\left(f\left(\frac{1}{15}\right) + f\left(\frac{2}{15}\right) + \ldots + f\left(\frac{59}{15}\right)\right)$ is equal to
(1) 92
(2) 118
(3) 102
(4) 108

The value of $\int _ { e ^ { 2 } } ^ { e ^ { 4 } } \frac { 1 } { x } \left( \frac { e ^ { \left( \left( \log _ { e } x \right) ^ { 2 } + 1 \right) ^ { - 1 } } } { e ^ { \left( \left( \log _ { e } x \right) ^ { 2 } + 1 \right) ^ { - 1 } } + e ^ { \left( \left( 6 - \log _ { e } x \right) ^ { 2 } + 1 \right) ^ { - 1 } } } \right) d x$ is
(1) 2
(2) $\log _ { e } 2$
(3) 1
(4) $e ^ { 2 }$

If $\int _ { - \frac { \pi } { 2 } } ^ { \frac { \pi } { 2 } } \frac { 96 x ^ { 2 } \cos ^ { 2 } x } { \left( 1 + e ^ { x } \right) } \mathrm { d } x = \pi \left( \alpha \pi ^ { 2 } + \beta \right) , \alpha , \beta \in \mathbb { Z }$, then $( \alpha + \beta ) ^ { 2 }$ equals
(1) 64
(2) 196
(3) 144
(4) 100

Q87. If $\int _ { 0 } ^ { \frac { \pi } { 4 } } \frac { \sin ^ { 2 } x } { 1 + \sin x \cos x } \mathrm {~d} x = \frac { 1 } { \mathrm { a } } \log _ { \mathrm { e } } \left( \frac { \mathrm { a } } { 3 } \right) + \frac { \pi } { \mathrm { b } \sqrt { 3 } }$, where $\mathrm { a } , \mathrm { b } \in \mathbf { N }$, then $\mathrm { a } + \mathrm { b }$ is equal to $\_\_\_\_$

Q75. If the value of the integral $\int _ { - 1 } ^ { 1 } \frac { \cos \alpha x } { 1 + 3 ^ { x } } d x$ is $\frac { 2 } { \pi }$. Then, a value of $\alpha$ is
(1) $\frac { \pi } { 3 }$
(2) $\frac { \pi } { 6 }$
(3) $\frac { \pi } { 4 }$
(4) $\frac { \pi } { 2 }$

Q75. The value of $\int _ { - \pi } ^ { \pi } \frac { 2 y ( 1 + \sin y ) } { 1 + \cos ^ { 2 } y } d y$ is :
(1) $2 \pi ^ { 2 }$
(2) $\frac { \pi ^ { 2 } } { 2 }$
(3) $\frac { \pi } { 2 }$
(4) $\pi ^ { 2 }$

The value of $\int_{-\pi/6}^{\pi/6} \left(\frac{\pi + 4x^{11}}{1 - \sin(|x| + \frac{\pi}{6})}\right)dx$ is equal to
(A) $8\pi$ (B) $7\pi$ (C) $5\pi$ (D) $4\pi$

Find $\int _ { \frac { \pi } { 24 } } ^ { \frac { 5 \pi } { 24 } } \frac { 1 + ( \tan 2 \mathrm { x } ) ^ { 1 / 3 } } { 1 + ( \tan 2 x ) ^ { 1 / 3 } } \mathrm { dx }$\ (A) $\frac { \pi } { 24 }$\ (B) $\frac { \pi } { 12 }$\ (C) $\frac { \pi } { 48 }$\ (D) $\frac { \pi } { 6 }$

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 and Computer Science \& Philosophy applicants should turn to page 14.
In this question we fix a real number $\alpha$ which will be the same throughout. We say that a function $f$ is bilateral if
$$f ( x ) = f ( 2 \alpha - x )$$
for all $x$.
(i) Show that if $f ( x ) = ( x - \alpha ) ^ { 2 }$ for all $x$ then the function $f$ is bilateral.
(ii) On the other hand show that if $f ( x ) = x - \alpha$ for all $x$ then the function $f$ is not bilateral.
(iii) Show that if $n$ is a non-negative integer and $a$ and $b$ are any real numbers then
$$\int _ { a } ^ { b } x ^ { n } \mathrm {~d} x = - \int _ { b } ^ { a } x ^ { n } \mathrm {~d} x$$
(iv) Hence show that if $f$ is a polynomial (and $a$ and $b$ are any reals) then
$$\int _ { a } ^ { b } f ( x ) \mathrm { d } x = - \int _ { b } ^ { a } f ( x ) \mathrm { d } x$$
(v) Suppose that $f$ is any bilateral function. By considering the area under the graph of $y = f ( x )$ explain why for any $t \geqslant \alpha$ we have
$$\int _ { \alpha } ^ { t } f ( x ) \mathrm { d } x = \int _ { 2 \alpha - t } ^ { \alpha } f ( x ) \mathrm { d } x$$
If $f$ is a function then we write $G$ for the function defined by
$$G ( t ) = \int _ { \alpha } ^ { t } f ( x ) \mathrm { d } x$$
for all $t$.
(vi) Suppose now that $f$ is any bilateral polynomial. Show that
$$G ( t ) = - G ( 2 \alpha - t )$$
for all $t$.
(vii) Suppose $f$ is a bilateral polynomial such that $G$ is also bilateral. Show that $G ( x ) = 0$ for all $x$.
If you require additional space please use the pages at the end of the booklet

The polynomial function $f ( x )$ is such that $f ( x ) > 0$ for all values of $x$.
Given $\int _ { 2 } ^ { 4 } f ( x ) d x = A$, which one of the following statements must be correct?
A $\int _ { 0 } ^ { 2 } [ f ( x + 2 ) + 1 ] d x = A + 1$
B $\quad \int _ { 0 } ^ { 2 } [ f ( x + 2 ) + 1 ] d x = A + 2$
C $\int _ { 2 } ^ { 4 } [ f ( x + 2 ) + 1 ] d x = A + 1$
D $\int _ { 2 } ^ { 4 } [ f ( x + 2 ) + 1 ] d x = A + 2$
E $\quad \int _ { 4 } ^ { 6 } [ f ( x + 2 ) + 1 ] d x = A + 1$
F $\quad \int _ { 4 } ^ { 6 } [ f ( x + 2 ) + 1 ] d x = A + 2$

$f ( x )$ is a function defined for all real values of $x$.
Which one of the following is a sufficient condition for $\int _ { 1 } ^ { 3 } f ( x ) d x = 0$ ?
A $f ( 2 ) = 0$
B $f ( 1 ) = f ( 3 ) = 0$
C $f ( - x ) = - f ( x )$ for all $x$
D $f ( x + 2 ) = - f ( 2 - x )$ for all $x$
E $\quad f ( x - 2 ) = - f ( 2 - x )$ for all $x$

The function $\mathrm { f } ( x )$ is defined for all real values of $x$. Which of the following conditions on $\mathrm { f } ( x )$ is/are necessary to ensure that
$$\int _ { - 5 } ^ { 0 } \mathrm { f } ( x ) \mathrm { d } x = \int _ { 0 } ^ { 5 } \mathrm { f } ( x ) \mathrm { d } x$$
Condition I: $\quad \mathrm { f } ( x ) = \mathrm { f } ( - x ) $ for $- 5 \leq x \leq 5$ Condition II: $\mathrm { f } ( x ) = c$ for $- 5 \leq x \leq 5$, where $c$ is a constant Condition III: $\mathrm { f } ( x ) = - \mathrm { f } ( - x ) $ for $- 5 \leq x \leq 5$
A none of them
B I only
C II only
D III only
E I and II only F I and III only G II and III only H I, II and III

For the function $f$ whose graph is given above in the rectangular coordinate plane
$$\begin{aligned} & \int_{a}^{c} |f(x)|\, dx = 20 \\ & \int_{a}^{c} f(x)\, dx = 8 \end{aligned}$$
the equalities are satisfied.
What is the value of $$\int_{a/2}^{b/2} f(2x)\, dx$$ ?
A) $-3$ B) $-4$ C) $-5$ D) $-6$ E) $-7$

For a continuous function $f$ defined on the set of real numbers and the function $g(x) = 2x + 2$ defined as,
$$\begin{aligned} & \int_{-1}^{1} f(g(x))\, dx = 18 \\ & \int_{2}^{4} g(f(x))\, dx = 18 \end{aligned}$$
are satisfied. Accordingly, what is the value of the integral $\int_{0}^{2} f(x)\, dx$?
A) 20 B) 23 C) 26 D) 29 E) 32