Mathematics RU

Russian state exam, profile level

The tasks with equations, inequalities, values of expressions, the derivative and a parameter. Each task comes with a guide to the method and a trainer that never runs out of tasks.

Here are the tasks solved by algebra and calculus: No. 6, 7, 12, 13, 15 and 18. Geometry, probability and word problems are not here yet. The answer is always the full one: all roots, the whole set.

No. 6 Simple equations OpenClose

Solve an equation of one of six kinds. First work out which kind it is: the kind tells you the method.

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

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

Equations with a root

In an equation with a root the unknown is under a square root: $\sqrt{x + 2} = x$. Squaring removes the root, but it can also add extra, «extraneous» roots. So the last step is always a check.

Step by step

Isolate the root: leave it alone on one side of the equation.
Note the condition: a square root is never negative, so the other side must be $\ge 0$ too.
Square both sides and solve the new equation.
Drop the roots that make the other side negative. The safest way is to substitute each root into the original equation.

The expression under the root. After squaring it equals a square, so it is automatically non-negative.

The other side: it equals a root, so it cannot be negative. This condition removes the extraneous roots.

Example: $\sqrt{x + 2} = x$ gives $x + 2 = x^2$ with roots $2$ and $-1$. But the right side $x$ cannot be negative, so only $x = 2$ remains.

Common mistakes

Not checking the roots after squaring and keeping an extraneous one.
Squaring a difference wrongly: $(x - 1)^2 = x^2 - 2x + 1$, not $x^2 + 1$.
Squaring before the root is isolated: the root does not go away.

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

No. 7 Values of expressions OpenClose

Find the value of an expression with logarithms, powers or trigonometry. No calculator: everything cancels by the rules.

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

No. 12 The greatest and least value of a function OpenClose

Find the greatest or least value of a function on a segment. The derivative helps.

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

No. 13 An equation with roots on a segment OpenClose

A trigonometric equation from part 2: solve it and pick the roots that lie on a segment.

How to solve it

Trigonometric equations with roots on a segment

This is a part 2 task: solve a trigonometric equation such as $2\cos^2 x - \cos x - 1 = 0$, then pick the roots that lie on a given segment. The equation has infinitely many solutions, the segment holds only a few.

Step by step

Reduce the equation to one function. For example, replace $\sin^2 x$ with $1 - \cos^2 x$ so that only cosine remains.
Substitute $t = \cos x$ and solve an ordinary, often quadratic, equation. Remember that cosine and sine lie between $-1$ and $1$: drop any other values of $t$.
Go back to $x$ with the formulas for the basic equations.
Pick the roots on the segment: try integer $n$ one by one, or mark the points on the unit circle.

A number from $-1$ to $1$. Outside this range there are no roots.

Any integer: the roots repeat every full turn.

Example: $\cos x = \frac{1}{2}$ on $[0, 2\pi]$. The general solution is $x = \pm\frac{\pi}{3} + 2\pi n$; the segment contains $\frac{\pi}{3}$ and $\frac{5\pi}{3}$.

Common mistakes

Keeping a value of $t$ outside $[-1, 1]$ and hunting for roots that do not exist.
Forgetting the second series $-\arccos a + 2\pi n$ and losing half the roots.
Missing roots at the ends of the segment. The ends belong to it.

Example

No. 15 Inequalities OpenClose

An inequality from part 2: quadratic, with fractions, exponential or logarithmic. The answer is an interval or a union of intervals.

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

No. 18 A problem with a parameter OpenClose

The equation has a letter a in it. Find every a for which the equation has exactly two roots.

How to solve it

Equations with a parameter

An equation with a parameter has a letter besides $x$, usually $a$: $x^2 - 2ax + 9 = 0$. Its value is not given, and you need every value of $a$ for which the equation behaves as asked, for example has exactly two roots. The answer is not a number but a set of values of $a$.

Step by step

See where the parameter sits. If it is in the coefficient of $x^2$, deal separately with the case where that coefficient is zero: the equation becomes linear.
A quadratic equation has two roots exactly when its discriminant is positive. Write $D$ in terms of $a$.
Solve the inequality $D > 0$ for $a$.
Write the answer as intervals of $a$.

The coefficient of $x$. The parameter may sit here.

The constant term. The parameter may sit here as well.

Example: $x^2 - 2ax + 9 = 0$. Here $D = 4a^2 - 36 > 0$, so $a^2 > 9$: $a < -3$ or $a > 3$. Answer: $(-\infty, -3) \cup (3, +\infty)$.

Common mistakes

Solving $D \ge 0$ instead of $D > 0$. At $D = 0$ there is only one root.
Getting only $a > 3$ out of $a^2 > 9$ and losing the negative values.
Skipping the case where the parameter makes the coefficient of $x^2$ zero.