2015-II-01
Ans: D
$\begin{array}{cl}
& (x+1)(x^2 + x +1) \\
=
… Read $\begin{array}{cl}
& (x+1)(x^2 + x +1) \\
=
HKDSE-MATH - 2015 - Paper 2
2015-II-02
Ans: D
$\begin{array}{cl}
& \dfrac{(3y^6)^4}{3y^2
… Read $\begin{array}{cl}
& \dfrac{(3y^6)^4}{3y^2
2015-II-03
Ans: A
$\left\{ \begin{array}{ll}
p+3q = 4 & \ldots \u
… Read $\left\{ \begin{array}{ll}
p+3q = 4 & \ldots \u
2015-II-04
Ans: D
$\begin{array}{cl}
& 0.002\ 345\ 678\ 9 \\
= �
… Read $\begin{array}{cl}
& 0.002\ 345\ 678\ 9 \\
= �
2015-II-05
Ans: B
$\begin{array}{rcl}
x^2 + mx + n & \equiv &
… Read $\begin{array}{rcl}
x^2 + mx + n & \equiv &
2015-II-06
Ans: A
$\begin{array}{rcl}
18 + 7x & & 4 \\
7x
… Read $\begin{array}{rcl}
18 + 7x & & 4 \\
7x
2015-II-07
Ans: A
Since $\beta$ is a root of the equation $4x^2 -5x ̵
… Read Since $\beta$ is a root of the equation $4x^2 -5x ̵
2015-II-08
Ans: D
Since the graph opens upwards, then $a>0$.
… Read Since the graph opens upwards, then $a>0$.
Note th
2015-II-09
Ans: B
Let $\$x$ be the price of the souvenir. Then the new pri
… Read Let $\$x$ be the price of the souvenir. Then the new pri
2015-II-10
Ans: D
The required amount
… Read The required amount
$\begin{array}{cl}
= & 50\
2015-II-11
Ans: C
$\begin{array}{ccccccccccc}
a & : &
… Read $\begin{array}{ccccccccccc}
a & : &
2015-II-12
Ans: D
Let $z = kx^3y^2$, where $k$ is a non-zero constant. Su
… Read Let $z = kx^3y^2$, where $k$ is a non-zero constant. Su
2015-II-13
Ans: B
Note that $T(1)=5$ and $T(n+1) = T(n) + 4$ for any posit
… Read Note that $T(1)=5$ and $T(n+1) = T(n) + 4$ for any posit
2015-II-14
Ans: C
Note that the greatest possible weight of the bag of wh
… Read Note that the greatest possible weight of the bag of wh
2015-II-15
Ans: C
By applying the Pythagorus Theorem to $\Delta CDN$, w
… Read By applying the Pythagorus Theorem to $\Delta CDN$, w
2015-II-16
Ans: B
… Read Let $x\text{ cm}$ be the radius of the upper base of the
2015-II-17
Ans: D
Note that $ABCD$ is a parallelogram, then $AD//BC$. H
… Read Note that $ABCD$ is a parallelogram, then $AD//BC$. H
2015-II-18
Ans: A
Consider $\Delta ABC$, we have
… Read Consider $\Delta ABC$, we have
$\begin{array}{rcl}
2015-II-19
Ans: C
$\begin{array}{cl}
& \dfrac{\cos 180^\circ}
… Read $\begin{array}{cl}
& \dfrac{\cos 180^\circ}
2015-II-20
Ans: C
… Read Join $AC$.
$\begin{array}{ll}
BC = CD & \text{(