10 Hardest TMUA Past Paper Questions

Ready for TMUA 2026? Try this 10-question past-paper challenge

Try this as a 38-minute challenge without a calculator, or work at your own pace. Select an option and use Check answer for immediate feedback. Open Show solution beneath any question when you want the explanation.

Our mentors selected ten demanding questions from the harder end of real TMUA-style past-paper practice. The set concentrates on algebraic control, unfamiliar reasoning and the small logical details that often separate a correct method from a lost mark. It is a stretch set—not a prediction that every question in the real exam will be this difficult.

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Why Try This Test?

Face questions that mirror the difficulty of the hardest problems on the TMUA.
Build familiarity with the exam’s unique style and tricky logic-based structures.
Review the method behind each answer and revisit the questions you found difficult.

10 Hardest TMUA Past Paper Questions

Question 1

The function $F(x)$ , where $x$ is a non-negative real number, is the result of subtracting the integer part of $x$ from $x$ . For example, $F(3)=0$ , $F(5.43)=0.43$ , $F(\pi)=0.14159265\ldots$ . Find an expression for:

$\int_{0}^{\sqrt{k}} F\!\left(x^{2}\right)\, dx$

where $k$ is a positive integer.

\tfrac{1}{3} k^{3/2} - F\!\left( \tfrac{1}{3} k^{3/2} \right)

\tfrac{1}{3} k^{3/2} - \sum_{n=0}^{k} n\!\left( \sqrt{n+1} - \sqrt{n} \right)

\tfrac{1}{3} k^{3/2} + \sum_{n=1}^{k-1} n\!\left( \sqrt{n+1} - \sqrt{n} \right)

\tfrac{1}{3} k^{3/2} - \sum_{n=1}^{k-1} n\!\left( \sqrt{n+1} + \sqrt{n} \right)

\tfrac{1}{3} k^{3/2} + \left( \sum_{n=0}^{k-1} \sqrt{n} \right) + (k-1)\sqrt{k}

\tfrac{1}{3} k^{3/2} + \left( \sum_{n=0}^{k-1} \sqrt{n} \right) + (1-k)\sqrt{k}

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Write $F(x^2)=x^2-\lfloor x^2\rfloor$ . On $[\sqrt n,\sqrt{n+1})$ , the integer part of $x^2$ is $n$ . Hence

I=\frac13k^{3/2}-\sum_{n=0}^{k-1}n(\sqrt{n+1}-\sqrt n).

Shift the index in the first half of the sum. The terms telescope:

\sum_{n=0}^{k-1}n(\sqrt{n+1}-\sqrt n)=(k-1)\sqrt k-\sum_{n=1}^{k-1}\sqrt n.

Therefore

I=\frac13k^{3/2}+\sum_{n=0}^{k-1}\sqrt n+(1-k)\sqrt k.

Answer F.

Question 2

The function is defined by

$f(x) = 1! \cdot x^{2! \cdot x^{3! \cdot x^{4! \cdot x^{5! \cdot x^{6! \cdot x}}}}}$

Which option is closest to $f\!\left(\tfrac12\right)$ ?

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Evaluate the power tower from the top down. At $x=\tfrac12$ , define the successive exponents by

\begin{aligned} a&=6!\cdot\tfrac12=360,\\ b&=5!\cdot2^{-a}=120\cdot2^{-360}\approx0,\\ c&=4!\cdot2^{-b}\approx24,\\ d&=3!\cdot2^{-c}\approx6\cdot2^{-24}\approx0,\\ e&=2!\cdot2^{-d}\approx2. \end{aligned}

So $f(\tfrac12)=2^{-e}\approx2^{-2}=\tfrac14$ . The tiny positive corrections do not change the nearest option.

Answer C. Work from the top; simply discarding the higher exponents would not justify the result.

Question 3

The function $f(x)$ has the property that $f(x)=f(6-x)$ for all real $x$ . Given that:

\begin{aligned} &\left(\int_2^3 f(x)\,dx\right)^2+\left(\int_4^6 f(x)\,dx\right)^2\\[4pt] &\quad+\left(\int_2^4 f(x)\,dx\right)\left(\int_0^2 f(x)\,dx\right)\\[4pt] &\quad-\int_6^3 3f(x)\,dx=-2, \end{aligned}

find the sum of the possible values of $\int_0^3 f(x)\,dx$ .

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Let $A=\int_2^3 f(x)\,dx$ and $B=\int_0^2 f(x)\,dx$ . Reflection about $x=3$ gives

\int_4^6 f=B,\qquad\int_2^4 f=2A,\qquad\int_6^3 3f=-3(A+B).

The given equation becomes $A^2+B^2+2AB+3(A+B)=-2$ . Setting $w=A+B=\int_0^3 f(x)\,dx$ ,

w^2+3w+2=0\quad\Longrightarrow\quad(w+1)(w+2)=0.

The two possible values are $-1$ and $-2$ , whose sum is $-3$ .

Answer C.

Question 4

In the range $0 \leq x < 2\pi$ the equation

$\cos(\sin x) = \tfrac{1}{2}$

has

Ano solutionsBone solutionCtwo solutionsDthree solutions

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For every real $x$ , $-1\leq\sin x\leq1$ . Cosine is even and decreases on $[0,1]$ , so

\cos(\sin x)\geq\cos1>\cos\!\left(\frac\pi3\right)=\frac12,

because $1<\pi/3$ . The equation therefore has no solutions.

Answer A.

Question 5

The turning point of the parabola

$y = x^{2} - 2ax + 1$

is closest to the origin when

a = 0

a = \pm 1

a = \pm \tfrac{1}{\sqrt{2}} \text{ or } a = 0

a = \pm \tfrac{1}{\sqrt{2}}

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Complete the square: $y=(x-a)^2+1-a^2$ . The turning point is $(a,1-a^2)$ , so its squared distance from the origin is

D^2=a^2+(1-a^2)^2=\left(a^2-\frac12\right)^2+\frac34.

This is smallest when $a^2=\tfrac12$ , giving $a=\pm\tfrac1{\sqrt2}$ .

Answer D.

Question 6

The equation

$\left(x^{2} + 1\right)^{10} = 2x - x^{2} - 2$

Ahas

x = 2

as a solutionBhas no real solutionsChas an odd number of real solutionsDhas twenty real solutions

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Compare the signs of the two sides:

(x^2+1)^{10}\geq1,\qquad2x-x^2-2=-(x-1)^2-1\leq-1.

A positive quantity cannot equal a negative quantity, so there are no real solutions.

Answer B.

Question 7

A triangle $ABC$ is to be drawn with $AB = 10 \,\text{cm}$ , $BC = 7 \,\text{cm}$ and the angle at $A$ equal to $\theta$ , where $\theta$ is a certain specified angle.

Of the two possible triangles that could be drawn, the larger triangle has three times the area of the smaller one.

What is the value of $\cos \theta$ ?

\tfrac{5}{7}

\tfrac{151}{200}

\tfrac{2\sqrt{2}}{5}

\tfrac{\sqrt{17}}{5}

\tfrac{\sqrt{51}}{8}

\tfrac{\sqrt{34}}{8}

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Let the two possible lengths of $AC$ be $u$ and $v$ , with $v>u$ . Their areas are $5u\sin\theta$ and $5v\sin\theta$ , so $v=3u$ .

By the cosine rule, both lengths solve

t^2-20t\cos\theta+51=0.

The product of the roots is $uv=51$ . Thus $3u^2=51$ , so $u=\sqrt{17}$ and $v=3\sqrt{17}$ . Using their sum,

20\cos\theta=u+v=4\sqrt{17}\quad\Longrightarrow\quad\cos\theta=\frac{\sqrt{17}}5.

Answer D.

Question 8

The graph of $y = \tan(\cos(\sin(x)))$ has a least positive period of $P$ and a maximum value of $M$ . Which row is correct?

P = 1, \; M < 1

P = 1, \; M > 1

P = \tfrac{\pi}{2}, \; M < 1

P = \tfrac{\pi}{2}, \; M > 1

P = \pi, \; M < 1

P = \pi, \; M > 1

P = 2\pi, \; M < 1

P = 2\pi, \; M > 1

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Since $\sin(x+\pi)=-\sin x$ and cosine is even,

\cos(\sin(x+\pi))=\cos(\sin x).

So the whole function repeats after $\pi$ . Its maxima occur exactly when $\sin x=0$ , at consecutive multiples of $\pi$ ; no smaller positive period is possible.

Also $\cos(\sin x)\in[\cos1,1]$ , where tangent is increasing. Hence

M=\tan1>\tan\!\left(\frac\pi4\right)=1.

Thus $P=\pi$ and $M>1$ .

Answer F.

Question 9

Find the fraction of the interval $0 \leq \theta \leq 2\pi$ for which the inequality

$\left( \cos\theta + \sin\theta \right) \left( \tfrac{\sqrt{3}}{2} + \cos\tfrac{1}{2}\theta \right) \left( \cos^{2}\theta - 1 \right) \leq 0$

is satisfied.

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The final factor is $\cos^2\theta-1=-\sin^2\theta\leq0$ . Away from its zeros, the other two factors must have the same sign.

Now $\cos\theta+\sin\theta=\sqrt2\sin(\theta+\pi/4)$ , with sign changes at $3\pi/4$ and $7\pi/4$ . The middle factor changes from positive to negative at $\theta=5\pi/3$ .

The intervals satisfying the inequality are

\left[0,\frac{3\pi}4\right]\quad\text{and}\quad\left[\frac{5\pi}3,\frac{7\pi}4\right].

The isolated points $\pi$ and $2\pi$ also satisfy it, but add no interval length. The required fraction is

\frac{\frac{3\pi}4+\left(\frac{7\pi}4-\frac{5\pi}3\right)}{2\pi}=\frac5{12}.

Answer B.

Question 10

Let $\lfloor x \rfloor$ denote the largest whole number that is less than or equal to $x$ . For example, $\lfloor -\pi \rfloor = -4$ .

A function $f(x)$ is defined as follows: if $0 < x < 2$ then

$f(x) = \left(\tfrac{3}{4}\right)^{\lfloor \log_{2}(x) \rfloor}$

and $f(x) = 0$ otherwise. Note that, for example, $f\!\left(\tfrac{1}{2}\right) = \tfrac{4}{3}$ . The value of $\int_{0}^{2} f(x)\, dx$ is

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On $[2^k,2^{k+1})$ , for each integer $k\leq0$ , the floor of $\log_2x$ is $k$ . Each rectangular strip has area

\left(\frac34\right)^k(2^{k+1}-2^k)=\left(\frac32\right)^k.

Adding the strips from $0$ to $2$ gives a convergent geometric series:

\int_0^2 f(x)\,dx=1+\frac23+\left(\frac23\right)^2+\cdots=\frac1{1-\frac23}=3.

Answer C.

These questions represent the harder end of TMUA practice. A real paper contains a range of difficulty, so use this mini-test to sharpen selection, accuracy and recovery under pressure—not as a forecast that an entire paper will feel this hard.

For each mistake, identify whether the issue was mathematical knowledge, interpretation, algebra, or time management. That short diagnosis turns one difficult question into a reusable exam strategy.

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Ready for TMUA 2026? Try this 10-question past-paper challenge

Why Try This Test?

10 Hardest TMUA Past Paper Questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

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