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TMUA MioMath Paper 1

20 questions20 marks75Updated July 2026

The TMUA MioMath Paper 1 paper in full: all 20 questions, each with its answer. TMUA is the Test of Mathematics for University Admission. Sit it cold under exam timing, mark it, then work back through anything you missed using the solutions below.

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Question 1

1 mark
Solve the equation

log⁡4(3x−1)2=log⁡4(12x−4)log⁡4(3x−1)\log_4(3x - 1)^2 = \frac{\log_4(12x - 4)}{\log_4(3x - 1)}
  • A.x=35,x=23x = \frac{3}{5}, x = \frac{2}{3}
  • B.x=23,x=53x = \frac{2}{3}, x = \frac{5}{3}
  • C.x=512,x=12x = \frac{5}{12}, x = \frac{1}{2}
  • D.x=53,x=512x = \frac{5}{3}, x = \frac{5}{12}

Answer: D

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Question 2

1 mark
Find the number of real solutions to the equation 8⋅32x+27⋅22x=32⋅6x8 \cdot 3^{2x} + 27 \cdot 2^{2x} = 32 \cdot 6^x.
  • A.1
  • B.2
  • C.4
  • D.6

Answer: B

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Question 3

1 mark
Which of the following is equal to (3+2)6−(3−2)6(\sqrt{3} + \sqrt{2})^6 - (\sqrt{3} - \sqrt{2})^6
  • A.3966396\sqrt{6}
  • B.1926192\sqrt{6}
  • C.3882388\sqrt{2}
  • D.1322132\sqrt{2}
  • E.3846384\sqrt{6}

Answer: A

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Question 4

1 mark
Given that θ2≤2π\theta^2 \le 2\pi and sin⁡(θ2)=12\sin(\theta^2) = \frac{1}{2} find the product of all possible solutions for θ\theta.
  • A.π56\frac{\pi\sqrt{5}}{6}
  • B.5π236\frac{5\pi^2}{36}
  • C.−π6-\frac{\pi}{6}
  • D.7π224\frac{7\pi^2}{24}
  • E.11π41296\frac{11\pi^4}{1296}
  • F.3π216\frac{3\pi^2}{16}

Answer: B

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Question 5

1 mark
Given 11−sin⁡θ=5\frac{1}{1 - \sin\theta} = 5 find

∑r=1∞cos⁡r(θ+π2)\sum_{r=1}^{\infty} \cos^r (\theta + \frac{\pi}{2})
  • A.11
  • B.−13-\frac{1}{3}
  • C.33
  • D.−49-\frac{4}{9}
  • E.−23-\frac{2}{3}

Answer: D

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Question 6

1 mark
A square with a side length of 1 contains an inscribed equilateral triangle. Find the area of the equilateral triangle.
  • A.23−34\frac{2\sqrt{3}-3}{4}
  • B.34\frac{\sqrt{3}}{4}
  • C.32\frac{\sqrt{3}}{2}
  • D.23−32\sqrt{3}-3
  • E.3−34\frac{3-\sqrt{3}}{4}
  • F.1−341 - \frac{\sqrt{3}}{4}

Answer: D

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Question 7

1 mark
Find the number of points (x,y)(x, y) where xx and yy are integers, such that y≥∣x∣y \ge |x| and y+∣x∣≤20y + |x| \le 20
  • A.200
  • B.210
  • C.220
  • D.221
  • E.235
  • F.240

Answer: D

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Question 8

1 mark
Find the number of intersection points between y2−x2=0y^2 - x^2 = 0 and y=x3−x2−8x+14y = x^3 - x^2 - 8x + 14
  • A.1
  • B.2
  • C.4
  • D.6
  • E.8

Answer: C

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Question 9

1 mark
Let R be the set of all points (x,y)(x, y) in the xy-plane such that: {(x+y)2−6(x+y)+8≤0 (x−y)2+4(x−y)+3≤0\begin{cases} (x + y)^2 - 6(x + y) + 8 \le 0 \ (x - y)^2 + 4(x - y) + 3 \le 0 \end{cases} (Note: The boundaries of the region are included in R.) Find the area of R.
  • A.2
  • B.2
  • C.4
  • D.6
  • E.8

Answer: A

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Question 10

1 mark
Given real numbers p,q,rp, q, r such that p+q+r=0p + q + r = 0 and 0<r<p0 < r < p, find the number of solutions of the equation psin⁡2θ+qsin⁡θ+r=0p \sin^2 \theta + q \sin \theta + r = 0 within the interval 0<θ<π0 < \theta < \pi.
  • A.0
  • B.1
  • C.2
  • D.3
  • E.4
  • F.5

Answer: D

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Question 11

1 mark
Given a geometric sequence Sn={an}S_n = \{a_n\} such that: a1+a2+a3+⋯=1a_1 + a_2 + a_3 + \cdots = 1 and a1−a2+a3−a4+a5−⋯=2a_1 - a_2 + a_3 - a_4 + a_5 - \cdots = 2 find the value of the summation: a1+a2−a3+a4+a5−a6+a7+a8−⋯a_1 + a_2 - a_3 + a_4 + a_5 - a_6 + a_7 + a_8 - \cdots
  • A.45\frac{4}{5}
  • B.35\frac{3}{5}
  • C.57\frac{5}{7}
  • D.47\frac{4}{7}
  • E.13\frac{1}{3}

Answer: C

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Question 12

1 mark
Given the equations of two curves: y=x3−x2−7x+19y = x^3 - x^2 - 7x + 19 and y=x+7y = x + 7 How many regions do these two curves divide the plane into?
  • A.2
  • B.3
  • C.4
  • D.5
  • E.6
  • F.7

Answer: D

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Question 13

1 mark
Given the following equations: y=ax2+by = ax^2 + b and x2+y2=4x^2 + y^2 = 4. Let P and Q be the intersection points of the circle and the quadratic function. If ab=−52ab = -\frac{5}{2}, find the value of a+ba + b.
  • A.53\frac{5}{3}
  • B.16\frac{1}{6}
  • C.±16\pm \frac{1}{6}
  • D.−32-\frac{3}{2}
  • E.92\frac{9}{2}
  • F.±32\pm \frac{3}{2}

Answer: F

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Question 14

1 mark
Circle C1C_1 has the equation (x−32)2+(y−32)2=18(x - 3\sqrt{2})^2 + (y - 3\sqrt{2})^2 = 18. A new circle, C2C_2, is drawn so that it passes through the point (0,32)(0, 3\sqrt{2}) and is internally tangent to C1C_1. The difference between the area of C1C_1 and C2C_2 is 32π32\pi. Which of the following is an equation for C2C_2?
  • A.(x−52)2+(y−32)2=25(x - 5\sqrt{2})^2 + (y - 3\sqrt{2})^2 = 25
  • B.(x+52)2+y2=50(x + 5\sqrt{2})^2 + y^2 = 50
  • C.(x+42)2+(y−32)2=32(x + 4\sqrt{2})^2 + (y - 3\sqrt{2})^2 = 32
  • D.x2+(y−32)2=18x^2 + (y - 3\sqrt{2})^2 = 18
  • E.(x−52)2+(y−32)2=50(x - 5\sqrt{2})^2 + (y - 3\sqrt{2})^2 = 50

Answer: E

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Question 15

1 mark
Point S lies on the line y=12xy = \frac{1}{2}x with OS=215OS = 2\sqrt{15}. Point T lies on the x-axis, with ST=4ST = 4, and ∠OTS>∠OST\angle OTS > \angle OST. Find the area of △OST\triangle OST.
  • A.6−36 - \sqrt{3}
  • B.12+2312 + 2\sqrt{3}
  • C.12
  • D.83−38\sqrt{3} - 3
  • E.12−2312 - 2\sqrt{3}

Answer: E

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Question 16

1 mark
The curve C is defined as y=x33+x22+176y = \frac{x^3}{3} + \frac{x^2}{2} + \frac{17}{6}. The tangent and the normal to C at the point where x=−1x = -1 form a triangle with the y-axis. Find the area of the triangle.
  • A.1
  • B.56\frac{5}{6}
  • C.54\frac{5}{4}
  • D.23\frac{2}{3}
  • E.32\frac{3}{2}
  • F.34\frac{3}{4}
  • G.43\frac{4}{3}

Answer: C

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Question 17

1 mark
ABCD is a cyclic quadrilateral. The lengths of its sides are AB=xAB = x, BC=2x+1BC = 2x + 1, CD=3x+8CD = 3x + 8, and DA=4x−1DA = 4x - 1. Angle DAB is a right angle. Find the value of x.
  • A.8
  • B.12
  • C.16
  • D.18
  • E.22
  • F.32

Answer: C

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Question 18

1 mark
Let O be the center of a circle, and M be a point outside the circle. Two tangents are drawn from M to the circle O touching the circle at points A and B. The line segment AB is connected, and MO intersects AB at point P. It is given that OP=yOP = y, OM=10yOM = 10y and AB=2y+6AB = 2y + 6. Find the area of the SOBMAS_{OBMA}.
  • A.270
  • B.120
  • C.60
  • D.135
  • E.1352\frac{135}{2}

Answer: E

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Question 19

1 mark
Find the area enclosed by the curves y=p∣x∣−py = p|x| - p and y=−1p∣x∣+py = -\frac{1}{p}|x| + p when p is a positive constant.
  • A.4p3p2+1\frac{4p^3}{p^2+1}
  • B.2p3p2+1\frac{2p^3}{p^2+1}
  • C.8p3p2+1\frac{8p^3}{p^2+1}
  • D.4p3p−1\frac{4p^3}{p-1}

Answer: A

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Question 20

1 mark
Let f(x)f(x) be the greatest integer less than or equal to xx eg. f(4)=4,f(4.9)=4f(4) = 4, f(4.9) = 4. Evaluate ∫050⌊x⌋dx\int_{0}^{50} \lfloor \sqrt{x} \rfloor dx.
  • A.195
  • B.203
  • C.210
  • D.217

Answer: C

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TMUA MioMath Paper 1: Questions & Worked Solutions | tmua.fyi