UNT (Kazakhstan), mathematics

Kazakhstan's Unified National Testing: equations, inequalities, values of expressions, systems, extrema and integrals.

The tasks are grouped by topic. At the test you pick the answer from options; here you find it yourself, which makes it stick.

Equations OpenClose

Linear, quadratic, fractional, with absolute values, exponential and logarithmic.

How to solve it

Linear equations

In a linear equation the unknown appears only to the first power: $3x + 5 = x - 7$. It has one root, and a few moves find it.

Step by step

Open any brackets.
Move the $x$ terms to the left and the numbers to the right. A term changes its sign as it crosses the «=»: $3x + 5 = x - 7$ becomes $3x - x = -7 - 5$.
Collect like terms: $2x = -12$.
Divide both sides by the number in front of $x$: $x = -6$.
Check by substituting: the left side is $3 \cdot (-6) + 5 = -13$, the right side is $-6 - 7 = -13$. They match.

The number in front of $x$, its coefficient. You may divide by it only when $a \ne 0$.

Whatever is left on the right after moving and collecting terms.

If $x$ cancels out and you get a false statement like $0 = 5$, there are no roots. If you get $0 = 0$, every number is a root.

Common mistakes

Moving a term to the other side and keeping its sign.
Losing a minus when dividing: $-2x = 8$ gives $x = -4$, not $4$.
Opening a bracket with a minus in front and changing only the first sign. Correct: $-(x - 3) = -x + 3$.

Example

Quadratic equations

A quadratic equation looks like $ax^2 + bx + c = 0$ with $a \ne 0$. It has at most two roots, and one number, the discriminant, tells you how many there really are.

Step by step

Move everything to the left so that the right side is zero. Write down $a$, $b$ and $c$ with their signs.
If $c = 0$, take $x$ out of the brackets: $x(ax + b) = 0$, so the roots are $0$ and $-\frac{b}{a}$. If $b = 0$, solve for $x^2$ and take the square root.
Otherwise compute the discriminant $D = b^2 - 4ac$.
If $D > 0$ there are two roots, if $D = 0$ one, if $D < 0$ no real roots.
Check with Vieta's formulas: the roots add up to $-\frac{b}{a}$ and multiply to $\frac{c}{a}$.

The coefficient of $x$ with its sign. In $x^2 - 5x + 6 = 0$ it is $b = -5$, so $-b = 5$.

The discriminant: it decides how many roots there are.

The coefficient of $x^2$.

Example: $x^2 - 5x + 6 = 0$, $D = 25 - 24 = 1$, $x = \frac{5 \pm 1}{2}$, so $x = 2$ and $x = 3$.

Common mistakes

Dropping the sign of $b$: in $x^2 - 5x + 6$ the coefficient is $b = -5$, and $b^2 = 25$.
Computing the discriminant before the right side is zero. Move everything to one side first.
Dividing both sides by $x$ and losing the root $x = 0$. Take $x$ out of the brackets instead.

Example

Equations with fractions

In an equation with fractions the unknown sits in a denominator: $\frac{x^2 - 1}{x - 1} = 0$. You cannot divide by zero, and the whole solution starts from that rule.

Step by step

Find the values of $x$ that make a denominator zero. They are forbidden; the rest of the numbers form the domain of the equation.
Move everything to one side and bring it to a common denominator, so that you have a single fraction.
A fraction is zero when its numerator is zero. Solve «numerator $= 0$».
Throw away the roots that were forbidden in the first step.

The numerator: set it equal to zero.

The denominator: it must not be zero, so such roots are dropped.

Example: in $\frac{x^2 - 1}{x - 1} = 0$ the numerator gives $x = 1$ and $x = -1$, but $x = 1$ makes the denominator zero. Answer: $x = -1$.

Common mistakes

Keeping a root that makes a denominator zero. This is the most common mistake.
Cancelling an expression with $x$ in it and forgetting that it could not be zero.
Multiplying only some of the terms by the missing factor when finding the common denominator.

Example

Equations with absolute values

The absolute value $|a|$ is the distance from $a$ to zero, so it is never negative: $|3| = 3$, $|-3| = 3$. An equation with an absolute value almost always splits into two ordinary equations.

Step by step

If the right side is negative, there are no roots: an absolute value is never negative.
The equation $|f(x)| = c$ with $c > 0$ splits into $f(x) = c$ and $f(x) = -c$. Solve both.
The equation $|f(x)| = |g(x)|$ also gives two: $f(x) = g(x)$ and $f(x) = -g(x)$.
Put the roots of both equations into the answer.

The expression inside the bars.

The number on the right. The rule holds for $c \ge 0$; for a negative $c$ there are no roots.

Example: $|2x - 1| = 5$. The equation $2x - 1 = 5$ gives $x = 3$, and $2x - 1 = -5$ gives $x = -2$. Answer: $-2$ and $3$.

Common mistakes

Solving only $f(x) = c$ and losing the second root.
Looking for roots when the right side is negative, although there are none.
In $|x - 3| = |x + 1|$, forgetting that the two expressions may be opposite: that gives the second equation $x - 3 = -(x + 1)$.

Example

Exponential equations

In an exponential equation the unknown is in the exponent: $2^{x + 1} = 8$. The main trick is to write both sides as powers of the same number.

Step by step

Write both sides as powers of one base: $8 = 2^3$, $\frac{1}{9} = 3^{-2}$, $\sqrt{5} = 5^{1/2}$.
When the bases are equal, the exponents are equal: $2^{x + 1} = 2^3$ means $x + 1 = 3$.
Solve the new equation: here $x = 2$.
When one base will not do, as in $2^x = 5$, the answer is a logarithm: $x = \log_2 5$.

The base: the same positive number, not equal to $1$.

The exponent on the left.

The exponent on the right. The function $a^x$ takes each of its values exactly once, so equal powers mean equal exponents.

Example: $\left(\frac{1}{3}\right)^{x} = 9$. Here $\frac{1}{3} = 3^{-1}$ and $9 = 3^2$, so $3^{-x} = 3^2$ and $x = -2$.

Common mistakes

Equating exponents when the bases differ: $2^x = 3^x$ cannot be solved that way.
Getting the sign of a fraction wrong: $\frac{1}{8} = 2^{-3}$, not $2^3$.
Mixing up the rules: $(a^m)^n = a^{mn}$, while $a^m \cdot a^n = a^{m + n}$.

Example

Logarithmic equations

The logarithm $\log_a b$ answers a question: to what power must $a$ be raised to give $b$? For instance, $\log_2 8 = 3$ because $2^3 = 8$. A logarithmic equation is almost always solved with this definition.

Step by step

Write down the domain: whatever is under a logarithm must be positive, and the base must be positive and not equal to $1$.
If the equation is $\log_a f(x) = c$, the definition gives $f(x) = a^c$.
If it is $\log_a f(x) = \log_a g(x)$, set the arguments equal: $f(x) = g(x)$.
Solve the new equation and check every root against the domain.

The base: $a > 0$ and $a \ne 1$.

The argument, the number under the logarithm. It is always positive.

The exponent to which the base is raised.

Example: $\log_2(x - 1) = 3$, so $x - 1 = 2^3 = 8$ and $x = 9$. Domain check: $9 - 1 = 8 > 0$.

Common mistakes

Skipping the domain check and keeping a root that puts a negative number or zero under the logarithm.
Raising the wrong number to the wrong power: $\log_3 x = 2$ gives $x = 3^2 = 9$, not $x = 2^3$.
Getting lost when the right side is negative: $\log_3 x = -2$ gives $x = 3^{-2} = \frac{1}{9}$. The root exists, it is just a fraction.

Example

Inequalities OpenClose

Quadratic, fractional, exponential and logarithmic.

How to solve it

Quadratic inequalities

A quadratic inequality has a quadratic on the left: $x^2 - 5x + 6 \le 0$. Its graph is a parabola, and the answer is easiest to read straight off the picture: where the parabola is above the axis and where it is below.

Step by step

Move everything to the left so that the right side is zero.
Find the roots of the quadratic by solving $ax^2 + bx + c = 0$. At the roots the parabola crosses the axis.
Sketch the parabola: it opens upwards when $a > 0$ and downwards when $a < 0$.
For $>$ or $\ge$ take the parts where the parabola is above the axis; for $<$ or $\le$, where it is below.
Include the roots only when the inequality is not strict.

The leading coefficient. When it is positive, the parabola opens upwards: it is below the axis between the roots and above it outside. For $a < 0$ it is the other way round.

The roots of the quadratic, $x_1 < x_2$.

Example: $x^2 - 5x + 6 \le 0$. The roots are $2$ and $3$, the parabola opens upwards, and we need the points below the axis together with the roots: $[2, 3]$.

Common mistakes

Forgetting that a negative $a$ turns the parabola upside down, and taking the wrong part.
Mixing up «between the roots» and «outside». A quick check helps: substitute one number from your answer.
Not knowing what to do when $D < 0$. Then the quadratic has the same sign everywhere, and the answer is either every number or the empty set.

Example

Inequalities with fractions

In an inequality with fractions the unknown is in a denominator: $\frac{x - 1}{x + 2} \le 0$. You may not multiply by the denominator: its sign is unknown, and the inequality could flip. The method of intervals works instead.

Step by step

Move everything to one side and bring it to a single fraction.
Find the zeros of the numerator and of the denominator and mark them on the number line. Zeros of the denominator are always excluded (open circles): you cannot divide by zero.
These points cut the line into intervals. On each interval the fraction keeps its sign; find it by substituting any number from the interval.
Take the intervals with the sign you need. Include the zeros of the numerator only for a non-strict inequality.

The zero of the numerator: here the fraction equals zero. It is included for a non-strict sign.

The zero of the denominator: here the fraction is not defined. It is never included.

Example: $\frac{x - 1}{x + 2} \le 0$. Mark $-2$ (open) and $1$ (filled). Right of $1$ the fraction is positive, between $-2$ and $1$ negative, left of $-2$ positive. Answer: $(-2, 1]$.

Common mistakes

Multiplying both sides by the denominator without knowing its sign.
Including a zero of the denominator, where the fraction does not exist.
Alternating the signs without checking a single point. The sign need not change, for example at a factor that is squared.

Example

Exponential inequalities

An exponential inequality such as $2^{3x + 1} > \frac{1}{4}$ starts like an equation: write both sides as powers of one base. Then look at the base itself, whether it is greater or less than one.

Step by step

Write both sides as powers of one base: $\frac{1}{4} = 2^{-2}$.
If the base is greater than $1$, compare the exponents with the same sign: $2^{3x + 1} > 2^{-2}$ means $3x + 1 > -2$.
If the base is between $0$ and $1$, the sign flips when you pass to the exponents.
Solve the inequality for the exponents and write the answer as an interval.

The base. For $a > 1$ the function $a^x$ increases, so a larger power has a larger exponent. For $0 < a < 1$ it decreases, and everything is reversed.

Example: $\left(\frac{1}{3}\right)^{x - 1} \ge \frac{1}{9}$. Here $\frac{1}{9} = \left(\frac{1}{3}\right)^2$, the base is less than $1$, the sign flips: $x - 1 \le 2$, so $x \le 3$.

Common mistakes

Not flipping the sign when the base is less than one.
Comparing the exponents of powers with different bases.
Looking for a solution of something like $2^x > -5$, which holds for every $x$: a power of a positive number is always positive.

Example

Logarithmic inequalities

A logarithmic inequality, for example $\log_2(x - 3) < 2$, is like an exponential one: the base decides again. One more condition comes in, and without it the answer is wrong: the domain.

Step by step

Write down the domain: everything under a logarithm must be positive.
Write the right side as a logarithm with the same base: $2 = \log_2 4$.
If the base is greater than $1$, the sign stays when you pass to the arguments; if it is between $0$ and $1$, it flips.
Solve the inequality for the arguments and intersect the result with the domain.

The base. For $a > 1$ the logarithm increases and the sign stays; for $0 < a < 1$ it decreases and the sign flips. The conditions $> 0$ are the domain.

Example: $\log_2(x - 3) < 2$. Domain: $x > 3$. Then $2 = \log_2 4$, the base is greater than $1$: $x - 3 < 4$, $x < 7$. With the domain: $(3, 7)$.

Common mistakes

Forgetting the domain, so that the answer contains numbers where the logarithm does not exist.
Not flipping the sign for a base less than one.
Writing $\log_2 x < 3 \iff x < 8$ and losing the left end: in fact $0 < x < 8$.

Example

Values of expressions OpenClose

Logarithms, powers, trigonometry.

How to solve it

Expressions with logarithms

No calculator is needed here: an expression like $\log_5 65 - \log_5 13$ folds into a single number by the rules of logarithms. The job is to see which rule applies.

Step by step

Look at the bases. If they are equal, a sum of logarithms becomes the logarithm of a product, and a difference the logarithm of a quotient: $\log_5 65 - \log_5 13 = \log_5 5 = 1$.
A number in front of a logarithm can move inside as an exponent: $2\log_3 5 = \log_3 25$.
An expression of the form $a^{\log_a b}$ equals $b$.
What usually remains at the end is $\log_a a^k = k$.

The first argument.

The second argument. Both logarithms have the same base.

Example: $\log_6 4 + \log_6 9 = \log_6 36 = 2$.

The same base below and in the logarithm.

The number you get. By definition $\log_a b$ is the exponent that turns $a$ into $b$.

Example: $3^{\log_3 7 + 1} = 3^{\log_3 7} \cdot 3 = 21$.

Common mistakes

Adding logarithms with different bases, although the rule works only for equal ones.
Splitting the logarithm of a sum: $\log_a(x + y)$ is not $\log_a x + \log_a y$.
Confusing a difference of logarithms with their quotient: $\log_a x - \log_a y = \log_a \frac{x}{y}$, not $\frac{\log_a x}{\log_a y}$.

Example

Expressions with powers

An expression like $\frac{2^5 \cdot 4^3}{8^3}$ looks heavy, but once every number is written as a power of two it folds into a single power. No big multiplications needed.

Step by step

Bring all the powers to one base: $4 = 2^2$, $8 = 2^3$, $\frac{1}{2} = 2^{-1}$.
When multiplying powers add the exponents, when dividing subtract them, when raising a power to a power multiply them.
Work out the final exponent and compute the power.

The exponent of the first power.

The exponent of the second power. The rules hold only when the bases are equal.

Example: $\frac{2^5 \cdot 4^3}{8^3} = \frac{2^5 \cdot 2^6}{2^9} = 2^{5 + 6 - 9} = 2^2 = 4$.

Common mistakes

Multiplying the exponents when multiplying powers: $2^3 \cdot 2^4 = 2^7$, not $2^{12}$.
Merging powers with different bases: $2^3 \cdot 3^2$ is not a single power.
Confusing a negative exponent with a negative number: $2^{-3} = \frac{1}{8}$, not $-8$.

Example

Trigonometric expressions

You need a value like $10\cos 420^\circ$. The angle is large, but sine and cosine repeat every full turn, $360^\circ$, and the reduction formulas bring any angle down to an acute one from the table.

Step by step

Remove full turns: $\cos 420^\circ = \cos(420^\circ - 360^\circ) = \cos 60^\circ$.
For a negative angle use symmetry: $\cos(-\alpha) = \cos \alpha$, $\sin(-\alpha) = -\sin \alpha$.
Bring the angle down to an acute one with the reduction formulas: $\sin(180^\circ - \alpha) = \sin \alpha$, $\cos(180^\circ + \alpha) = -\cos \alpha$.
Take the value from the table: $\sin 30^\circ = \frac{1}{2}$, $\cos 45^\circ = \frac{\sqrt{2}}{2}$, $\sin 60^\circ = \frac{\sqrt{3}}{2}$.

With $180^\circ$ the function stays the same. The sign is that of the original function in the quadrant where the angle lies.

With $90^\circ$ and $270^\circ$ sine turns into cosine and back. The sign is found the same way, from the quadrant.

Example: $\sin 150^\circ = \sin(180^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2}$.

Common mistakes

Forgetting the sign: cosine is negative in the second and third quadrants, sine in the third and fourth.
Swapping sine and cosine where it is not needed, with $180^\circ$ and $360^\circ$.
Losing the minus of a negative angle in a sine: $\sin(-30^\circ) = -\frac{1}{2}$.

Example

Systems of equations OpenClose

Linear and nonlinear systems in two unknowns.

How to solve it

Systems of linear equations

A system of two linear equations, such as $x + 2y = 7$ and $3x - y = 7$, puts two conditions on the same pair of numbers $(x, y)$. You need the pair that satisfies both. The usual tools are substitution and adding the equations.

Step by step

Find an equation where one variable is easy to express (coefficient $1$ or $-1$) and express it: $x + 2y = 7$ gives $x = 7 - 2y$.
Substitute it into the other equation: $3(7 - 2y) - y = 7$. One unknown is left.
Solve it: $21 - 7y = 7$, $y = 2$. Then find the other variable: $x = 7 - 4 = 3$.
If expressing is awkward, add the equations after multiplying them so that the coefficients of one variable become opposite.
Check the pair in both equations and write the answer: $(3, 2)$.

The coefficients of $x$ are opposite, so adding the equations removes $x$.

Then $y = 2$, and either equation gives $x = \frac{1}{2}$. Check: $2 \cdot \frac{1}{2} + 3 \cdot 2 = 7$.

Common mistakes

Multiplying only one side of an equation by a number; both sides must be multiplied.
Substituting the found number and not checking the pair in the other equation.
Writing the pair in the wrong order: $x$ comes first, then $y$.

Example

Nonlinear systems

In a nonlinear system one equation is linear and the other is not: $x + y = 1$, $xy = -12$. There are usually two solutions, each a pair of numbers. Substitution works: express a variable from the linear equation and put it into the other.

Step by step

Express one variable from the linear equation: $y = 1 - x$.
Substitute it into the other equation: $x(1 - x) = -12$.
Solve the quadratic equation you get: $x^2 - x - 12 = 0$, with roots $4$ and $-3$.
For each root find the other variable: $y = 1 - 4 = -3$ and $y = 1 - (-3) = 4$. That gives two pairs.
Check both pairs and write them in the answer: $(4, -3)$ and $(-3, 4)$.

The sum of the unknowns.

Their product. This is Vieta's theorem read backwards.

Example: $x + y = 1$, $xy = -12$ give $t^2 - t - 12 = 0$ with roots $4$ and $-3$.

Common mistakes

Finding only $x$ and forgetting to compute $y$.
Writing one pair, although the quadratic has two roots and so the system has two solutions.
Swapping $x$ and $y$ in a pair.

Example

The greatest and least value OpenClose

The derivative helps find where a function is largest and smallest.

How to solve it

The greatest and least value on a segment

You need the greatest or least value of a function on a segment, for example $y = x^3 - 3x - 7$ on $[0, 2]$. A function reaches its extreme values either at the ends of the segment or where its derivative is zero. So a few points are enough to check.

Step by step

Find the derivative. The rule $(x^n)' = nx^{n - 1}$ is usually enough: $(x^3 - 3x - 7)' = 3x^2 - 3$.
Solve $y' = 0$: these are the critical points, $x = \pm 1$.
Keep only those inside the segment: here $x = 1$.
Compute the function at these points and at the ends: $y(0) = -7$, $y(1) = -9$, $y(2) = -5$.
The largest of these numbers is the greatest value, the smallest is the least: here $-5$ and $-9$.

The ends of the segment: they are always on the list of candidates.

The critical points inside the segment, where $f'(x) = 0$.

The least value is found the same way: take the smallest of the same numbers.

Common mistakes

Forgetting the values at the ends, although the answer is often there.
Taking a critical point that lies outside the segment.
Answering with the point $x$ instead of the value of the function. The question asks for $y$.

Example

Antiderivatives and integrals OpenClose

Find an antiderivative and compute a definite integral.

How to solve it

Antiderivatives

An antiderivative is a function $F$ whose derivative is the given one: $F' = f$. Finding it means reading the table of derivatives backwards. The answer can always be checked: differentiate it and compare with what was under the integral.

Step by step

The table: $\int x^n\,dx = \frac{x^{n + 1}}{n + 1} + C$ for $n \ne -1$, $\int \frac{dx}{x} = \ln|x| + C$, $\int \cos x\,dx = \sin x + C$, $\int e^x\,dx = e^x + C$.
When the function is applied to $kx + b$, divide the result by $k$: $\int \cos 3x\,dx = \frac{\sin 3x}{3} + C$.
When the derivative of an inner function stands next to it, substitute $t = g(x)$: $\int 2x e^{x^2}\,dx = e^{x^2} + C$.
A polynomial times $e^x$, $\sin x$ or $\cos x$ is integrated by parts.
Check the answer by differentiating it.

What gets simpler when differentiated: a polynomial or a logarithm.

What is easy to integrate: $e^x$, $\sin x$, $\cos x$.

Example: $\int x e^{x}\,dx = x e^{x} - \int e^{x}\,dx = x e^{x} - e^{x} + C$.

Common mistakes

Dividing by $n$ instead of $n + 1$: $\int x^3\,dx = \frac{x^4}{4} + C$.
Forgetting to divide by $k$ after a linear substitution.
Not checking the answer, although the check by differentiation takes a minute.

Example

Definite integrals

The definite integral $\int_a^b f(x)\,dx$ is the signed area under the graph. It is computed without any areas at all: find an antiderivative and subtract its values at the ends of the segment.

Step by step

Find an antiderivative $F(x)$, as in the antiderivative tasks. The constant $C$ is not needed: it cancels in the subtraction.
Substitute the upper limit, then the lower one, and subtract: $F(b) - F(a)$.
Subtract $F(a)$ as a whole, in brackets: it may have minuses of its own.

The lower limit. The antiderivative's value there is subtracted.

The upper limit.

Example: $\int_0^2 3x^2\,dx = x^3 \Big|_0^2 = 8 - 0 = 8$.

Common mistakes

Subtracting the other way round: $F(a) - F(b)$ has the opposite sign.
Losing a minus when $F(a)$ is negative: $F(b) - (-3) = F(b) + 3$.
Making a mistake in the antiderivative itself. Check it by differentiating before substituting the limits.