Australian Mathematics Competition

Conducted By: Australian Maths Trust
For: Primary and secondary school students
Divisions: Primary, Junior, Intermediate, and Senior
Question Format: Multiple choice and integer answers
Duration: 60 to 75 minutes
Marking: Points for correct answers; guessing is penalised on some questions
Frequency: Held once a year

The Australian Mathematics Competition (AMC) is an annual problem-solving event for primary and secondary school students. It is organised by the Australian Maths Trust and is one of the largest competitions of its kind in the world, with participants from Australia and many other countries. The AMC is designed to be accessible and enjoyable, encouraging students to develop their mathematical thinking rather than testing curriculum knowledge directly.

The competition consists of a single paper with questions arranged in order of increasing difficulty. The first 25 questions are multiple choice, beginning with straightforward problems and becoming progressively more complex. The final five questions are designed to be the most challenging and require integer answers from 1 to 999.

In practice, the difficulty of the AMC comes from its emphasis on creative problem-solving under time pressure. The later questions require students to apply familiar mathematical concepts in novel and ingenious ways. The marking scheme, which penalises incorrect answers on the last five questions, also adds a strategic element, forcing students to balance speed with accuracy and confidence.

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From a mathematics paper rather than Australian Mathematics Competition, worked the way every question on this site is. Australian Mathematics Competition questions are being added.

JEE Advanced 2020

Let the functions $f:\mathbb{R} \to \mathbb{R}$ and $g:\mathbb{R} \to \mathbb{R}$ be defined by $f(x) = e^{x-1} - e^{-|x-1|}$ and $g(x) = \frac{1}{2}(e^{x-1} + e^{1-x})$ . Then the area of the region in the first quadrant bounded by the curves $y = f(x)$ , $y = g(x)$ and $x = 0$ is

Applications of Integrals Let the functions

f:\mathbb{R} \to \mathbb{R}

and

g:\mathbb{R} \to \mathbb{R}

be defined by

f(x) = e^{x-1} - e^{-|x-1|}

and

g(x) = \frac{1}{2}(e^{x-1} + e^{1-x})

. Then the area of the region in the first quadrant bounded by the curves

y = f(x)

y = g(x)

and

x = 0

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$f(x) = e^{x-1} - e^{-|x-1|},\qquad g(x) = \frac{1}{2}\left(e^{x-1} + e^{1-x}\right)$ $\text{Region in first quadrant bounded by } y=f(x),\ y=g(x),\ x=0$ $\text{To find: area of the bounded region}$ $\text{Plan: write } f \text{ piecewise; note } g(x) = \cosh(x-1)$ $\text{find intersection of } f, g; \text{ split area into two integrals}$ $f(x) = \begin{cases} 0, & x \leq 1 \\ e^{x-1} - e^{1-x} = 2\sinh(x-1), & x \geq 1 \end{cases}$ $g(x) = \dfrac{e^{x-1}+e^{1-x}}{2} = \cosh(x-1)$ $u = x-1:\qquad f = 2\sinh u,\qquad g = \cosh u$ $2\sinh u = \cosh u \Rightarrow e^{2u} = 3 \Rightarrow u = \tfrac{1}{2}\ln 3$ $x^{*} = 1 + \tfrac{1}{2}\ln 3$ $A = \int_{0}^{1} g(x)\,dx + \int_{1}^{x^{*}} \bigl(g(x)-f(x)\bigr)\,dx$ $\int_{0}^{1} g(x)\,dx = \bigl[\sinh(x-1)\bigr]_{0}^{1} = \sinh 1 = \dfrac{1}{2}(e-e^{-1})$ $g-f = \cosh u - 2\sinh u = -\tfrac{1}{2}e^{u} + \tfrac{3}{2}e^{-u}$ $\int_{0}^{\frac{1}{2}\ln 3}\!\left(-\tfrac{1}{2}e^{u}+\tfrac{3}{2}e^{-u}\right)du = [-\tfrac{1}{2}e^{u}-\tfrac{3}{2}e^{-u}]_{0}^{\frac{1}{2}\ln 3} = -\sqrt{3}-(-2) = 2-\sqrt{3}$ $A = (2-\sqrt{3}) + \dfrac{1}{2}(e-e^{-1})$ $\boxed{\text{option (1)}}$

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