For each part of the question on Pages 3 and 4, you will be given four possible answers just one of which is correct. Indicate for each part A-J which answer (a), (b), (c), or (d) you think is correct with a tick $( \checkmark )$ in the corresponding column in the table below.
A. The substitution $x = y + t$ transforms the equation $x ^ { 3 } + a x ^ { 2 } + b x + c = 0$ into an equation of the form $y ^ { 3 } + p y + q = 0$ when\ (a) $t = \frac { a } { 3 }$\ (b) $t = - \frac { a } { 3 }$\ (c) $t = a$\ (d) $t = - a$.\
B. The faces of a cube are coloured red or blue. Exactly three are red and three are blue. The number of distinguishable cubes that can be produced (allowing the cube to be turned around) is\ (a) 2\ (b) 4\ (c) 6\ (d) 20 .\
C. The shortest distance from the origin to the line $3 x + 4 y = 25$ is\ (a) 3\ (b) 4\ (c) 5\ (d) 6 .\
D. The numbers 10, 11 and -12 are solutions of the cubic equation\ (a) $x ^ { 3 } - 11 x ^ { 2 } - 122 x + 1320 = 0$\ (b) $x ^ { 3 } - 9 x ^ { 2 } + 122 x - 1320 = 0$\ (c) $x ^ { 3 } - 9 x ^ { 2 } - 142 x + 1320 = 0$\ (d) $x ^ { 3 } + 9 x ^ { 2 } - 58 x - 1320 = 0$.\
E. The maximum gradient of the curve $y = x ^ { 4 } - 4 x ^ { 3 } + 4 x ^ { 2 } + 2$ in the range $0 \leq x \leq 2 \frac { 1 } { 5 }$ occur when\ (a) $x = 0$\ (b) $x = 1 - \frac { 1 } { \sqrt { 3 } }$\ (c) $x = 1 + \frac { 1 } { \sqrt { 3 } }$\ (d) $x = 2 \frac { 1 } { 5 }$.\
F. The expression $x ^ { 2 } y + x y ^ { 2 } + y ^ { 2 } z + y z ^ { 2 } + z ^ { 2 } x + z x ^ { 2 } - x ^ { 3 } - y ^ { 3 } - z ^ { 3 } - 2 x y z$ factorises as\ (a) $( x + y + z ) ( x - y + z ) ( - x + y - z )$\ (b) $( x + y - z ) ( x - y - z ) ( - x + y + z )$\ (c) $( x + y - z ) ( x - y + z ) ( - x + y + z )$\ (d) $( x - y - z ) ( - x - y + z ) ( - x + y - z )$.\
G. The derivative of $x e ^ { - x ^ { 2 } } \cos \left( \frac { 1 } { x } \right)$ is\ (a) $- \frac { 1 } { x } e ^ { - x ^ { 2 } } \sin \left( \frac { 1 } { x } \right) - 2 x ^ { 2 } e ^ { - x ^ { 2 } } \cos \left( \frac { 1 } { x } \right) + e ^ { - x ^ { 2 } } \cos \left( \frac { 1 } { x } \right)$\ (b) $\frac { 1 } { x } e ^ { - x ^ { 2 } } \sin \left( \frac { 1 } { x } \right) - 2 x ^ { 2 } e ^ { - x ^ { 2 } } \cos \left( \frac { 1 } { x } \right) + e ^ { - x ^ { 2 } } \cos \left( \frac { 1 } { x } \right)$\ (c) $\frac { 1 } { x } e ^ { - x ^ { 2 } } \sin \left( \frac { 1 } { x } \right) + 2 x ^ { 2 } e ^ { - x ^ { 2 } } \cos \left( \frac { 1 } { x } \right) + e ^ { - x ^ { 2 } } \cos \left( \frac { 1 } { x } \right)$\ (d) $\frac { 1 } { x } e ^ { - x ^ { 2 } } \cos \left( \frac { 1 } { x } \right) - 2 x ^ { 2 } e ^ { - x ^ { 2 } } \cos \left( \frac { 1 } { x } \right) + e ^ { - x ^ { 2 } } \cos \left( \frac { 1 } { x } \right)$.\
H. You are told that the infinite series $1 + \frac { 1 } { 2 ^ { 2 } } + \frac { 1 } { 3 ^ { 2 } } + \frac { 1 } { 4 ^ { 2 } } + \ldots$ and $1 + \frac { 1 } { 3 ^ { 2 } } + \frac { 1 } { 5 ^ { 2 } } + \frac { 1 } { 7 ^ { 2 } } + \ldots$ have sums $\frac { \pi ^ { 2 } } { 6 }$ and $\frac { \pi ^ { 2 } } { 8 }$ respectively. The infinite series $1 - \frac { 1 } { 2 ^ { 2 } } + \frac { 1 } { 3 ^ { 2 } } - \frac { 1 } { 4 ^ { 2 } } + \ldots + ( - 1 ) ^ { n - 1 } \frac { 1 } { n ^ { 2 } } + \ldots$ has sum equal to\ (a) $\frac { \pi ^ { 2 } } { 9 }$\ (b) $\frac { \pi ^ { 2 } } { 10 }$\ (c) $\frac { \pi ^ { 2 } } { 12 }$\ (d) $\frac { \pi ^ { 2 } } { 16 }$.\
I. A grid of size $3 \mathrm {~cm} \times 5 \mathrm {~cm}$ is drawn, ruled at 1 cm intervals. The number of squares that can be drawn using the grid is\ (a) 15\ (b) 18\ (c) 26\ (d) 37 .\
J. A pack of cards consists of 52 different cards. A malicious dealer changes one of the cards for a second copy of another card in the pack and he then deals the cards to four players, giving thirteen to each. The probability that one player has two identical cards is\ (a) $\frac { 3 } { 13 }$\ (b) $\frac { 12 } { 51 }$\ (c) $\frac { 1 } { 4 }$\ (d) $\frac { 13 } { 51 }$