Higher Maths · SQA past paper

2025 Paper 2

Calculator · 14 questions
1

Question 1

Altitudes, medians, perpendicular bisectorsIntersection of straight lines
(3, 3, 2)8 Marks
2025 P2 Q1

Triangle ABC has vertices A(9,14),\displaystyle A(-9,-14), B(9,20)\displaystyle B(9,20) and C(21,24).\displaystyle C(21,-24).

Triangle ABC

(a) Find the equation of the altitude through B.
(b) Find the equation of the median through A.
(c) Determine the point of intersection of the altitude through B and the median through A.

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(a) y=3x7\displaystyle y=3x-7
(b) 2y=x19\displaystyle 2y=x-19
(c) (1,10)\displaystyle (-1, -10)
2

Question 2

Completing the square
3 Marks
2025 P2 Q2
Express 2x2+16x+5\displaystyle 2x^{2}+16x+5
in the form p(x+q)2+r\displaystyle p(x+q)^{2}+r
Show answer
2(x+4)227\displaystyle 2(x+4)^{2}-27
3

Question 3

Areas using integration
4 Marks
2025 P2 Q3

The diagram shows the graph of y=x22x+3.\displaystyle y=x^{2}-2x+3.

Graph of y=x^2-2x+3 with shaded area between x=2 and x=4

Calculate the shaded area.

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383\displaystyle \frac{38}{3}
4

Question 4

Inverse functions
3 Marks
2025 P2 Q4

A function, g, is defined by g(x)=(x4)3,\displaystyle g(x)=(x-4)^{3}, where xR.\displaystyle x\in\mathbb{R}.
Find the inverse function, g1(x).\displaystyle g^{-1}(x).

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g1(x)=x3+4\displaystyle g^{-1}(x)=\sqrt[3]{x}+4
5

Question 5

Collinearity (in 3d or 2d)Ratio in which one point divides two others
(3, 1)4 Marks
2025 P2 Q5

(a) Show that the points A(3,2,1),\displaystyle A(-3,2,-1), B(6,1,5)\displaystyle B(6,-1,5) and C(12,3,9)\displaystyle C(12,-3,9) are collinear.
(b) State the ratio in which B divides AC.

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(a) Proof showing BC=23AB\displaystyle \vec{BC} = \frac{2}{3}\vec{AB} (or equivalent) and B is a common point.
(b) 3:2
6

Question 6

Wave function (y = asin x ± bcosx)Trig equation involving compound angle
(4, 3)7 Marks
2025 P2 Q6

(a) Express 5 cos x9 sin x\displaystyle 5~\cos~x-9~\sin~x in the form k cos(x+a)\displaystyle k~\cos(x+a) where k>0\displaystyle k \gt 0 and 0<a<2π.\displaystyle 0 \lt a \lt 2\pi.
(b) Hence solve
5 cos x9 sin x=7\displaystyle 5~\cos~x-9~\sin~x=7
for 0x<2π.\displaystyle 0\le x \lt 2\pi.

Show answer
(a) 106cos(x+1.06)\displaystyle \sqrt{106}\cos(x+1.06)
(b) 4.40, 6.04
7

Question 7

Integrate (definite or indefinite): (px + q)^n
2 Marks
2025 P2 Q7
Find
(3x+2)7dx\displaystyle \int(3x+2)^{7}dx.
Show answer
124(3x+2)8+c\displaystyle \frac{1}{24}(3x+2)^{8}+c
8

Question 8

Vector pathways in geometric diagrams
2 Marks
2025 P2 Q8

E,ABCD is a rectangular-based pyramid as shown.
AD=6i+4j+2k\displaystyle \vec{AD}=6i+4j+2k
DC=2i4j+2k\displaystyle \vec{DC}=2i-4j+2k
DE=4i3j+4k\displaystyle \vec{DE}=-4i-3j+4k

Rectangular-based pyramid

Express BE\displaystyle \vec{BE} in terms of i, j and k.

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5j+4k\displaystyle 5j+4k
9

Question 9

Find a specific term of a recurrence relationLimits of recurrence relations
(2, 1)3 Marks
2025 P2 Q9

A sequence satisfies the recurrence relation un+1=mun+4\displaystyle u_{n+1}=mu_{n}+4, where m is a constant.
(a) The sequence approaches a limit of 10 as n.\displaystyle n\rightarrow\infty. Determine the value of m.
(b) Given that u1=19\displaystyle u_{1}=19, calculate the value of u0.\displaystyle u_{0}.

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(a) m=35\displaystyle m=\frac{3}{5} (or 0.6)
(b) u0=25\displaystyle u_{0}=25
10

Question 10

Optimisation
(3, 6)9 Marks
2025 P2 Q10

A hotel owner is designing signs showing the room numbers. Each sign is a rectangle with a right-angled triangle above it. The length and breadth of the rectangle are 5x centimetres and y centimetres respectively. The shorter sides of the triangle are 3x centimetres and 4x centimetres.

Room signSign dimensions

The area of the sign is 150 square centimetres.
(a) Show that the perimeter, P cm, of the sign is given by P=9.6x+60x.\displaystyle P=9.6x+\frac{60}{x}.
Each sign will be lit using a lighting strip placed around its perimeter.
The hotel owner requires the perimeter, P, of the sign to be as small as possible.
(b) Find the minimum value of P.

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(a) Proof
(b) 48 cm
11

Question 11

Solving a trigonometric equation using formula for sin(2x)
4 Marks
2025 P2 Q11
Solve
3 sin 2x+4 cos x=0\displaystyle 3~\sin~2x^{\circ}+4~\cos~x^{\circ}=0
for 0x<360\displaystyle 0\le x \lt 360.
Show answer
x=90,221.8,270,318.2\displaystyle x=90, 221.8, 270, 318.2
12

Question 12

Composite functionsDifferentiate or evaluate derivative: composite function
(2, 2)4 Marks
2025 P2 Q12

Functions f and g are defined on the set of real numbers by:
f(x)=x5+3\displaystyle f(x)=x^{5}+3
g(x)=1x3.\displaystyle g(x)=1-x^{3}.
(a) Find an expression for h(x)\displaystyle h(x), where h(x)=f(g(x)).\displaystyle h(x)=f(g(x)).
(b) Find h(x)\displaystyle h^{\prime}(x)

Show answer
(a) h(x)=(1x3)5+3\displaystyle h(x)=(1-x^{3})^{5}+3
(b) h(x)=15x2(1x3)4\displaystyle h^{\prime}(x)=-15x^{2}(1-x^{3})^{4}
13

Question 13

Solving equations where the unknown is in the exponent
(1, 4)5 Marks
2025 P2 Q13

A radioactive substance, which has been collected, decays over time.
The mass of the radioactive substance remaining is modelled by
M=150e0.0054t\displaystyle M=150e^{-0.0054t}
where M is the mass, in micrograms, t years after the radioactive substance was collected.
(a) Determine the initial mass of the radioactive substance.
(b) Calculate the time taken for the mass of the radioactive substance to decay to 120 micrograms.

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(a) 150 micrograms
(b) 41.3 years
14

Question 14

Circle equation from radius/centre or vice versaIntersections of two circles
(2, 2, 3)7 Marks
2025 P2 Q14

Circle C1\displaystyle C_{1} has equation (x+5)2+(y6)2=9.\displaystyle (x+5)^{2}+(y-6)^{2}=9.
(a) State the centre and radius of C1.\displaystyle C_{1}.
Circle C2\displaystyle C_{2} has equation x2+y214x+6y+54=0.\displaystyle x^{2}+y^{2}-14x+6y+54=0.
(b) State the centre and radius of C2.\displaystyle C_{2}.
Circles C1\displaystyle C_{1} C2\displaystyle C_{2} and C3\displaystyle C_{3} are touching as shown in the diagram. The centre of circle C3\displaystyle C_{3} lies on the line joining the centres of C1\displaystyle C_{1} and C2.\displaystyle C_{2}.

Circles C1, C2 and C3

(c) Determine the equation of C3.\displaystyle C_{3}.

Show answer
(a) Centre (-5, 6), Radius 3
(b) Centre (7, -3), Radius 2
(c) (x+1)2+(y3)2=64\displaystyle (x+1)^{2}+(y-3)^{2}=64