IGCSE Mathematics Mock – Set 2 Paper 2-Updated 2026
IGCSE Mathematics Mock – Set 2 Paper 2 – Updated 2026
Preparing for your CIE IGCSE Mathematics exam can be daunting, but with the right approach, you can achieve your goals with CIE iGCSE Mathematics Mock test.
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iGCSE Practice Questions, Past Papers, Flashcards and notes available for iGCSE Students at IITian Academy.
Question
The graph of $y=$f$(x)$ is drawn on the grid
(a) Draw the tangent to the graph at the point $x=3.$
(b) Use your tangent to find an estimate for the gradient of the curve at the point $x=3.$
▶️Answer/Explanation
(a) tangent ruled at $x = 3$
(b) $4.8$ to $5.8$
(a)
(b)
$
\text{Gradient} = \frac{\Delta y}{\Delta x} = \frac{y_2 – y_1}{x_2 – x_1}
$
From the diagram
Point 1: \((3, 5)\)
Point 2: \((2, 0)\)
$
\text{Gradient} = \frac{5 – 0}{3 – 2} = \frac{1}{1} = 1
$
So, the gradient of the line is 1
Question
(a) For each sketch, put a ring around the correct type of function shown.
$\mathbf{( i) }$ On the grid, sketch the curve $y= \sin x$ for $0^\circ\leqslant x\leqslant360^\circ.$
(ii) Solve the equation $\sin x+ 0. 4= 0$ for $0^{\circ}\leqslant x\leqslant360^{\circ}$
▶️Answer/Explanation
(a)(i) cubic
(a)(ii) reciprocal
(b)(i) 
(b)(ii) $203.6$ and $336.4$
(a)
Graph (i):
The curve has multiple turning points, which is a cubic function.
Graph (ii):
The curve has asymptotes (it gets close to the axes but never touches them), characteristic of a reciprocal function (e.g., \( y = \frac{1}{x} \)).
(b) (i)
$
\sin(x) + 0.4 = 0
$
$
\sin(x) = -0.4
$
$
x_{\text{ref}} = \sin^{-1}(0.4)
$
$
x_{\text{ref}} = 23.58^\circ
$
Sine value is negative, the angle will be in the 3rd and 4th quadrants
In the 3rd quadrant
$
x = 180^\circ + 23.58^\circ = 203.58^\circ
$
In the 4th quadrant
$
x = 360^\circ – 23.58^\circ = 336.42^\circ
$
Question
$O$ is the origin and $OPQR$ is a parallelogram
$M$ is the midpoint of $PQ$ and $\hat{N}$ divides $QR$ in the ratio 2:1.
$\overrightarrow{OP}=$a and $\overrightarrow OR=\mathbf{b}.$
$( \mathbf{a} )$ Find $\overrightarrow{MN}.$
Give your answer in terms of a and/or b and in its simplest form.
$(\mathbf{b})$ The lines $MN$ and $OR$ are extended to meet at S.
Find the position vector of $S.$
Give your answer in terms of a and/or b and in its simplest form.
▶️Answer/Explanation
(a) $\frac{1}{2}\mathbf{b}-\frac{2}{3}\mathbf{a}$
(b) $\frac{5}{4}\mathbf{b}$
(a)
\( \overrightarrow{OP} = \mathbf{a} \)
\( \overrightarrow{OR} = \mathbf{b} \)
Since \( OPQR \) is a parallelogram
\( \overrightarrow{OQ} = \mathbf{a} + \mathbf{b} \)
$
\overrightarrow{M} = \frac{1}{2}(\overrightarrow{P} + \overrightarrow{Q})
$
$
= \frac{1}{2}(\mathbf{a} + (\mathbf{a} + \mathbf{b}))
$
$
= \frac{1}{2}(2\mathbf{a} + \mathbf{b}) = \mathbf{a} + \frac{1}{2}\mathbf{b}
$
N divide QR in 2:1.
$
\overrightarrow{N} = \frac{1(\overrightarrow{Q}) + 2(\overrightarrow{R})}{1 + 2}
$
$
= \frac{1(\mathbf{a} + \mathbf{b}) + 2(\mathbf{b})}{3}
$
$
= \frac{\mathbf{a} + 3\mathbf{b}}{3}
$
$
\overrightarrow{MN} = \overrightarrow{N} – \overrightarrow{M}
$
$
= \frac{\mathbf{a} + 3\mathbf{b}}{3} – \left(\mathbf{a} + \frac{1}{2}\mathbf{b}\right)
$
$
= -\frac{2}{3}\mathbf{a} + \frac{1}{2}\mathbf{b}
$
(b)
The equation for line \( MN \) is:
$
\overrightarrow{S} = \overrightarrow{M} + k(\overrightarrow{MN})
$
$
\overrightarrow{S} = \left( \mathbf{a} + \frac{1}{2}\mathbf{b} \right) + k\left( -\frac{2}{3}\mathbf{a} + \frac{1}{2}\mathbf{b} \right)
$
$
= \mathbf{a} + \frac{1}{2}\mathbf{b} – \frac{2k}{3}\mathbf{a} + \frac{k}{2}\mathbf{b}
$
this equal to the equation of line \( OR \) also
$
\overrightarrow{S} = t\mathbf{b}
$
Match coefficients of \( \mathbf{a} \) and \( \mathbf{b} \)
\( \mathbf{a} \)
$
1 – \frac{2k}{3} = 0 \implies k = \frac{3}{2}
$
\( \mathbf{b} \)
\( k = \frac{3}{2} \)
$
\frac{1}{2} + \frac{3}{4} = t \implies t = \frac{5}{4}
$
$
\overrightarrow{S} = \frac{5}{4}\mathbf{b}
$