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Advanced Higher Maths · Qualifications Scotland past paper

2026 Paper 1

Non-Calculator · 7 questions
1

Question 1

Differentiation
(2, 3)5 Marks
2026 P1 Q1

Differentiate:

(a)y=3x4sec⁡2x\displaystyle y=3x^{4}\sec 2x

(b)f(x)=e5x2x+1\displaystyle f(x)=\frac{e^{5x}}{2x+1}, simplifying your answer.

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(a) dydx=12x3sec⁡2x+6x4sec⁡2xtan⁡2x=6x3sec⁡2x(2+xtan⁡2x)\displaystyle \frac{dy}{dx}=12x^{3}\sec 2x+6x^{4}\sec 2x\tan 2x=6x^{3}\sec 2x\left(2+x\tan 2x\right)
(b) f′(x)=(2x+1)5e5x−2e5x(2x+1)2=e5x(10x+3)(2x+1)2\displaystyle f'(x)=\frac{(2x+1)5e^{5x}-2e^{5x}}{(2x+1)^{2}}=\frac{e^{5x}(10x+3)}{(2x+1)^{2}}
2

Question 2

Systems of Equations
4 Marks
2026 P1 Q2

A system of equations is given by

x+y−z=9\displaystyle x+y-z=9
2x−y+3z=−2\displaystyle 2x-y+3z=-2
3x+2y−2z=21\displaystyle 3x+2y-2z=21

Use Gaussian elimination to solve this system of equations.

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Reducing the augmented matrix to upper triangular form gives 2z=−2\displaystyle 2z=-2, so by back substitution
x=3, y=5, z=−1\displaystyle x=3,\ y=5,\ z=-1
3

Question 3

Complex Numbers
(2, 2)4 Marks
2026 P1 Q3

A complex number is defined by z=3+i.\displaystyle z=\sqrt{3}+i.

(a)Express z\displaystyle z in polar form.

(b)Use de Moivre's theorem to show that z3\displaystyle z^{3} is purely imaginary.

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(a) ∣z∣=3+1=2\displaystyle |z|=\sqrt{3+1}=2 and arg⁡z=π6\displaystyle \arg z=\frac{\pi}{6}, so z=2(cos⁡π6+isin⁡π6)\displaystyle z=2\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)
(b) z3=23(cos⁡π2+isin⁡π2)=8i.\displaystyle z^{3}=2^{3}\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)=8i. The real part is zero, so z3\displaystyle z^{3} is purely imaginary.
4

Question 4

Differential Equations
5 Marks
2026 P1 Q4

Find the particular solution of the differential equation

2d2ydx2−3dydx+y=0\displaystyle 2\frac{d^{2}y}{dx^{2}}-3\frac{dy}{dx}+y=0

given that y=2\displaystyle y=2 and dydx=−1\displaystyle \frac{dy}{dx}=-1 when x=0.\displaystyle x=0.

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Auxiliary equation 2m2−3m+1=0⇒(2m−1)(m−1)=0\displaystyle 2m^{2}-3m+1=0\Rightarrow(2m-1)(m-1)=0, so m=12\displaystyle m=\frac{1}{2} and m=1.\displaystyle m=1.
General solution y=Ae12x+Bex\displaystyle y=Ae^{\frac{1}{2}x}+Be^{x}, giving A+B=2\displaystyle A+B=2 and 12A+B=−1.\displaystyle \frac{1}{2}A+B=-1.
A=6, B=−4\displaystyle A=6,\ B=-4, so y=6e12x−4ex\displaystyle y=6e^{\frac{1}{2}x}-4e^{x}
5

Question 5

Matrices
(1, 1, 2)4 Marks
2026 P1 Q5

Matrix A\displaystyle A is defined by A=(35−2x).\displaystyle A=\begin{pmatrix}3&5\\-2&x\end{pmatrix}.

(a)State an expression for the determinant of A\displaystyle A in terms of x.\displaystyle x.

Matrix A\displaystyle A is multiplied by matrix B\displaystyle B such that det⁡AB=12x+40.\displaystyle \det AB=12x+40.

(b)State the determinant of B.\displaystyle B.

The inverse of matrix B\displaystyle B is B−1=(1−1−5432).\displaystyle B^{-1}=\begin{pmatrix}1&-1\\[7pt]-\dfrac{5}{4}&\dfrac{3}{2}\end{pmatrix}.

(c)Find matrix B.\displaystyle B.

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(a) det⁡A=3x+10\displaystyle \det A=3x+10
(b) det⁡AB=det⁡Adet⁡B\displaystyle \det AB=\det A\det B, so (3x+10)det⁡B=4(3x+10)\displaystyle (3x+10)\det B=4(3x+10) and det⁡B=4\displaystyle \det B=4
(c) det⁡B−1=32−54=14\displaystyle \det B^{-1}=\frac{3}{2}-\frac{5}{4}=\frac{1}{4}, so B=4(321541)=(6454)\displaystyle B=4\begin{pmatrix}\dfrac{3}{2}&1\\[7pt]\dfrac{5}{4}&1\end{pmatrix}=\begin{pmatrix}6&4\\5&4\end{pmatrix}
6

Question 6

Integration
(3, 4)7 Marks
2026 P1 Q6

(a)Use the substitution u=x−1\displaystyle u=x-1 to find ∫x(x−1)4 dx.\displaystyle \int x(x-1)^{4}\,dx.

(b)Hence find the exact volume of the solid formed by rotating the curve with equation y=2x(x−1)2\displaystyle y=2\sqrt{x}(x-1)^{2} about the x\displaystyle x axis through 2π\displaystyle 2\pi radians, from x=0\displaystyle x=0 to x=1.\displaystyle x=1.

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(a) With u=x−1\displaystyle u=x-1, ∫(u+1)u4 du=u66+u55+C=(x−1)66+(x−1)55+C\displaystyle \int(u+1)u^{4}\,du=\frac{u^{6}}{6}+\frac{u^{5}}{5}+C=\frac{(x-1)^{6}}{6}+\frac{(x-1)^{5}}{5}+C
(b) V=π∫014x(x−1)4 dx=4π[(x−1)66+(x−1)55]01=4π(0−(16−15))=2π15\displaystyle V=\pi\int_{0}^{1}4x(x-1)^{4}\,dx=4\pi\left[\frac{(x-1)^{6}}{6}+\frac{(x-1)^{5}}{5}\right]_{0}^{1}=4\pi\left(0-\left(\frac{1}{6}-\frac{1}{5}\right)\right)=\frac{2\pi}{15} cubic units
7

Question 7

Complex Numbers
(1, 5)6 Marks
2026 P1 Q7

The complex number z=2+i\displaystyle z=2+i is a root of the polynomial equation

z4−2z3−z2+2z+10=0.\displaystyle z^{4}-2z^{3}-z^{2}+2z+10=0.

(a)State a second root of the equation.

(b)Find the remaining roots.

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(a) The coefficients are real, so roots occur in conjugate pairs: z=2−i\displaystyle z=2-i
(b) (z−(2+i))(z−(2−i))=z2−4z+5\displaystyle (z-(2+i))(z-(2-i))=z^{2}-4z+5, and z4−2z3−z2+2z+10=(z2−4z+5)(z2+2z+2).\displaystyle z^{4}-2z^{3}-z^{2}+2z+10=(z^{2}-4z+5)(z^{2}+2z+2).
Solving z2+2z+2=0\displaystyle z^{2}+2z+2=0 gives z=−2±−42=−1±i\displaystyle z=\frac{-2\pm\sqrt{-4}}{2}=-1\pm i, so the remaining roots are −1+i\displaystyle -1+i and −1−i\displaystyle -1-i