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Czech Maturita Math study materials
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Czech Maturita Math practice questions

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Question 1Algebra Functions

A winter coat is initially discounted by 20%. During an end-of-season clearance, the already reduced price is further reduced by an additional 15%. If the final sale price of the coat is 1,224 CZK, what was its original full price before any discounts?

  1. 1,883 CZK
  2. 1,652 CZK
  3. 1,800 CZK Correct answer.
  4. 1,748 CZK

Explanation

Let P be the original price in CZK. After a 20% discount, the price becomes P * (1 - 0.20) = 0.80 * P. After a further 15% discount on this reduced price, the final price is (0.80 * P) * (1 - 0.15) = 0.80 * 0.85 * P = 0.68 * P. We are given 0.68 * P = 1,224 CZK. Solving for P: P = 1,224 / 0.68 = 1,800 CZK.

Question 2Algebra Functions

What is the simplified exact value of the numerical expression sqrt(72) - sqrt(18) + sqrt(50)?

  1. 8*sqrt(2) Correct answer.
  2. 10*sqrt(2)
  3. 4*sqrt(2)
  4. sqrt(104)

Explanation

Simplify each radical by factoring out perfect squares: sqrt(72) = sqrt(36 * 2) = 6*sqrt(2); sqrt(18) = sqrt(9 * 2) = 3*sqrt(2); sqrt(50) = sqrt(25 * 2) = 5*sqrt(2). Combining like terms gives: 6*sqrt(2) - 3*sqrt(2) + 5*sqrt(2) = (6 - 3 + 5)*sqrt(2) = 8*sqrt(2).

Question 3Algebra Functions

For all real numbers x where the expression is defined (x not in {-3, 0, 3}), simplify the algebraic expression: ((x^2 - 9) / (2x^2 + 6x)) * ((4x) / (x - 3)).

  1. 2x / (x + 3)
  2. (2x - 6) / (x + 3)
  3. 4
  4. 2 Correct answer.

Explanation

Factor all numerators and denominators: x^2 - 9 = (x - 3)(x + 3); 2x^2 + 6x = 2x(x + 3). Substituting these factorizations into the product: [((x - 3)(x + 3)) / (2x(x + 3))] * [(4x) / (x - 3)]. Cancel the common factors (x + 3) and (x - 3): [1 / (2x)] * (4x) = 4x / (2x) = 2.

Question 4Algebra Functions

Evaluate the following numerical expression and write the result in standard scientific notation: ((3.6 * 10^7) / (1.2 * 10^(-3))) * (2.0 * 10^(-5)).

  1. 6.0 * 10^9
  2. 6.0 * 10^5 Correct answer.
  3. 6.0 * 10^(-1)
  4. 6.0 * 10^4

Explanation

Separate the decimal coefficients from the powers of 10: Coefficient: (3.6 / 1.2) * 2.0 = 3.0 * 2.0 = 6.0. Power of 10: 10^(7 - (-3) + (-5)) = 10^(7 + 3 - 5) = 10^5. Combining gives 6.0 * 10^5.

Question 5Algebra Functions

For all real numbers a != 2 and a != -2, simplify the expression: ((a^3 - 8) / (a^2 - 4)) - ((a^2 + 2a + 4) / (a + 2)).

  1. 0 Correct answer.
  2. 1
  3. 4 / (a + 2)
  4. -4 / (a + 2)

Explanation

Factor the numerator and denominator of the first term using the difference of cubes and difference of squares formulas: a^3 - 8 = (a - 2)(a^2 + 2a + 4); a^2 - 4 = (a - 2)(a + 2). Thus, the first fraction becomes: [(a - 2)(a^2 + 2a + 4)] / [(a - 2)(a + 2)] = (a^2 + 2a + 4) / (a + 2). Now subtract the second fraction: (a^2 + 2a + 4)/(a + 2) - (a^2 + 2a + 4)/(a + 2) = 0.

Question 6Algebra Functions

Rationalize the denominator of the fraction 6 / (sqrt(7) - 1) and simplify the expression completely.

  1. sqrt(7) - 1
  2. 6*sqrt(7) + 6
  3. sqrt(7) + 1 Correct answer.
  4. (sqrt(7) + 1) / 2

Explanation

Multiply both numerator and denominator by the conjugate of the denominator, which is (sqrt(7) + 1): [6 * (sqrt(7) + 1)] / [(sqrt(7) - 1)(sqrt(7) + 1)]. The denominator becomes (sqrt(7))^2 - 1^2 = 7 - 1 = 6. Dividing numerator and denominator by 6: [6 * (sqrt(7) + 1)] / 6 = sqrt(7) + 1.

Question 7Algebra Functions

For any positive real number x > 0, simplify the radical expression: (x^(3/2) * (x^2)^(1/3)) / x^(1/6).

  1. x^(7/6)
  2. x^2 Correct answer.
  3. x^(5/3)
  4. x^(8/3)

Explanation

Convert all components into fractional powers of x: x^(3/2) has exponent 3/2; (x^2)^(1/3) = x^(2/3); x^(1/6) has exponent 1/6. Apply the rules of exponents: x^(3/2 + 2/3 - 1/6). Find a common denominator for the exponents (common denominator is 6): 3/2 = 9/6; 2/3 = 4/6; 1/6 = 1/6. The exponent is (9 + 4 - 1)/6 = 12/6 = 2. Thus, the expression equals x^2.

Question 8Algebra Functions

Given two intervals of real numbers A = (-4, 3] and B = [1, 6), determine the set difference A \ B (elements that belong to A but not to B).

  1. (-4, 1]
  2. [-4, 1)
  3. (-4, 6)
  4. (-4, 1) Correct answer.

Explanation

The set difference A \ B consists of all numbers x such that x in A and x not in B. A = {x in R: -4 < x <= 3} and B = {x in R: 1 <= x < 6}. Since B contains 1, 1 is excluded from the set difference. The elements remaining in A are all x satisfying -4 < x < 1. In interval notation, this is (-4, 1).

Question 9Algebra Functions

For any real number x in the open interval (-3, -1), simplify the absolute value expression: |x + 3| - |2x - 4| + |x|.

  1. 2x - 1 Correct answer.
  2. 4x + 7
  3. -1
  4. 2x + 7

Explanation

Determine the sign of each argument when x in (-3, -1): 1) Since x > -3, x + 3 > 0, so |x + 3| = x + 3. 2) Since x < -1, 2x - 4 < 2(-1) - 4 = -6 < 0, so |2x - 4| = -(2x - 4) = -2x + 4. 3) Since x < -1 < 0, x is negative, so |x| = -x. Substituting these into the expression: (x + 3) - (-2x + 4) + (-x) = x + 3 + 2x - 4 - x = 2x - 1.

Question 10Algebra Functions

Find the greatest common divisor (GCD / NSD) and the least common multiple (LCM / NSN) of the positive integers 84 and 126.

  1. GCD = 14, LCM = 504
  2. GCD = 21, LCM = 252
  3. GCD = 42, LCM = 252 Correct answer.
  4. GCD = 42, LCM = 504

Explanation

Write the prime factorizations: 84 = 4 * 21 = 2^2 * 3^1 * 7^1; 126 = 2 * 63 = 2^1 * 3^2 * 7^1. GCD takes the lowest power of each common prime factor: GCD(84, 126) = 2^1 * 3^1 * 7^1 = 42. LCM takes the highest power of each prime factor: LCM(84, 126) = 2^2 * 3^2 * 7^1 = 4 * 9 * 7 = 252. Verify: GCD * LCM = 42 * 252 = 10,584 = 84 * 126.

Question 11Algebra Functions

Determine the remainder when the polynomial P(x) = 2x^3 - 5x^2 + 4x - 7 is divided by the binomial (x - 2).

  1. 5
  2. -3 Correct answer.
  3. 1
  4. -7

Explanation

According to the Polynomial Remainder Theorem, the remainder when a polynomial P(x) is divided by (x - c) is equal to P(c). Here c = 2. Compute P(2): P(2) = 2*(2^3) - 5*(2^2) + 4*(2) - 7 = 2*(8) - 5*(4) + 8 - 7 = 16 - 20 + 8 - 7 = -3.

Question 12Algebra Functions

A laboratory flask contains 6.0 liters of a 40% acid solution. How many liters of pure distilled water (0% acid) must be added to dilute the solution to a 24% acid solution?

  1. 2.5 liters
  2. 3.0 liters
  3. 3.6 liters
  4. 4.0 liters Correct answer.

Explanation

Calculate the volume of pure acid in the initial solution: V_acid = 6.0 * 0.40 = 2.4 liters. Let x be the volume of water added (in liters). The total volume becomes 6.0 + x liters, while the amount of pure acid remains 2.4 liters. We require: 2.4 / (6.0 + x) = 0.24. Multiplying both sides by (6.0 + x): 2.4 = 0.24 * (6.0 + x) => 2.4 = 1.44 + 0.24x => 0.24x = 0.96 => x = 0.96 / 0.24 = 4.0 liters.

Question 13Algebra Functions

Simplify the nested radical expression sqrt(9 + 4*sqrt(5)) into an equivalent form without nested square roots.

  1. 2 + sqrt(5) Correct answer.
  2. 3 + 2*sqrt(5)
  3. 1 + 2*sqrt(5)
  4. sqrt(5) + 4

Explanation

Assume sqrt(9 + 4*sqrt(5)) = a + b*sqrt(5) with positive integers a and b. Squaring both sides: (a + b*sqrt(5))^2 = a^2 + 5b^2 + 2ab*sqrt(5). Equating rational and irrational parts: 1) 2ab = 4 => ab = 2; 2) a^2 + 5b^2 = 9. If a = 2 and b = 1, then a^2 + 5b^2 = 2^2 + 5*(1^2) = 4 + 5 = 9. This matches perfectly. Therefore, sqrt(9 + 4*sqrt(5)) = 2 + sqrt(5).

Question 14Algebra Functions

A country's annual inflation rate increased from 4.0% to 5.0%. What was the increase in percentage points, and what was the relative percentage increase in the inflation rate?

  1. 1.0 percentage point and 20.0% relative increase
  2. 0.25 percentage points and 25.0% relative increase
  3. 1.0 percentage point and 25.0% relative increase Correct answer.
  4. 1.25 percentage points and 25.0% relative increase

Explanation

The difference in percentage points is simply the arithmetic subtraction: 5.0% - 4.0% = 1.0 percentage point. The relative percentage increase measures the change relative to the initial value: ((5.0 - 4.0) / 4.0) * 100% = (1.0 / 4.0) * 100% = 0.25 * 100% = 25.0%.

Question 15Algebra Functions

Evaluate the exact numerical value of the logarithmic expression: log_3(54) + log_3(1/2) - log_3(root(3, 27)).

  1. 0
  2. 1
  3. 3
  4. 2 Correct answer.

Explanation

Use the logarithmic rules log_b(u) + log_b(v) = log_b(u * v) and log_b(u) - log_b(w) = log_b(u / w): log_3(54) + log_3(1/2) = log_3(54 * 1/2) = log_3(27). For the third term, root(3, 27) = 3, so log_3(root(3, 27)) = log_3(3) = 1. Therefore: log_3(27) - 1 = 3 - 1 = 2.

Question 16Algebra Functions

Eight construction workers can assemble a prefabricated steel frame in 15 days working at a constant rate. Assuming all workers work at the exact same pace, how many days would it take 12 workers to assemble the same frame?

  1. 22.5 days
  2. 10 days Correct answer.
  3. 12 days
  4. 7.5 days

Explanation

This is an inverse proportion problem: more workers require fewer days. The total amount of labor required is: Total labor = 8 workers * 15 days = 120 worker-days. With 12 workers working at the same pace: Days = 120 worker-days / 12 workers = 10 days.

Question 17Algebra Functions

A principal of 50,000 CZK is deposited into a savings account with an annual interest rate of 4.0% compounded annually. In accordance with Czech tax law, a 15% withholding tax is deducted from the earned interest at the end of each year. What is the total balance on the account at the end of 2 years?

  1. 53,457.80 CZK Correct answer.
  2. 54,080.00 CZK
  3. 53,400.00 CZK
  4. 52,890.50 CZK

Explanation

The effective net annual interest rate after deducting the 15% withholding tax is: r_net = 4.0% * (1 - 0.15) = 4.0% * 0.85 = 3.4% = 0.034. Balance at end of Year 1: 50,000 * (1 + 0.034) = 50,000 * 1.034 = 51,700 CZK. Balance at end of Year 2: 51,700 * 1.034 = 50,000 * (1.034)^2 = 50,000 * 1.069156 = 53,457.80 CZK.

Question 18Algebra Functions

For all real x where the expression is defined (x not in {-1, 0, 0.5, 1}), simplify the complex fraction: ((1 / x) - (1 / (x + 1))) / ((1 / x) + (1 / (x - 1))).

  1. 1 / (2x - 1)
  2. (x + 1) / (2x - 1)
  3. (x - 1) / ((x + 1)(2x - 1)) Correct answer.
  4. (x - 1) / (x + 1)

Explanation

Simplify numerator and denominator separately: Numerator: 1/x - 1/(x + 1) = [(x + 1) - x] / [x(x + 1)] = 1 / [x(x + 1)]. Denominator: 1/x + 1/(x - 1) = [(x - 1) + x] / [x(x - 1)] = (2x - 1) / [x(x - 1)]. Now divide the numerator by the denominator: [1 / (x(x + 1))] * [[x(x - 1)] / (2x - 1)] = (x - 1) / [(x + 1)(2x - 1)].

Question 19Algebra Functions

Calculate the exact value of the numerical expression: (16 / 81)^(-3/4) * (27 / 8)^(-2/3).

  1. 9 / 4
  2. 2 / 3
  3. 1
  4. 3 / 2 Correct answer.

Explanation

Apply negative exponents by taking reciprocals: (16/81)^(-3/4) = (81/16)^(3/4) = [(3/2)^4]^(3/4) = (3/2)^3 = 27/8. Similarly: (27/8)^(-2/3) = (8/27)^(2/3) = [(2/3)^3]^(2/3) = (2/3)^2 = 4/9. Now multiply the two results: (27/8) * (4/9) = (27/9) * (4/8) = 3 * (1/2) = 3/2.

Question 20Algebra Functions

Which interval represents the set of all real numbers x whose distance from the number 3 on the real number line is strictly less than 5?

  1. (-2, 8) Correct answer.
  2. [-2, 8]
  3. (-8, 2)
  4. (-5, 5)

Explanation

The distance between x and 3 is given by the absolute value |x - 3|. The condition that this distance is strictly less than 5 translates to the inequality: |x - 3| < 5. This double inequality is: -5 < x - 3 < 5. Adding 3 to all parts: -5 + 3 < x < 5 + 3 => -2 < x < 8. In interval notation, this is the open interval (-2, 8).

Question 21Algebra Functions

Find the real solution to the linear equation: (2x - 1) / 3 - (x + 2) / 4 = (x - 1) / 2.

  1. x = 4
  2. x = -4 Correct answer.
  3. x = -10
  4. x = 2

Explanation

Multiply the entire equation by the least common denominator 12: 12 * [(2x - 1)/3] - 12 * [(x + 2)/4] = 12 * [(x - 1)/2] => 4(2x - 1) - 3(x + 2) = 6(x - 1). Expand brackets carefully: 8x - 4 - 3x - 6 = 6x - 6. Combine like terms on the left: 5x - 10 = 6x - 6. Subtract 5x from both sides: -10 = x - 6. Add 6 to both sides: x = -4.

Question 22Algebra Functions

Let x_1 and x_2 be the two real roots of the quadratic equation 2x^2 - 7x + 3 = 0. Without solving for the individual roots, find the exact value of (x_1)^2 + (x_2)^2.

  1. 49 / 4
  2. 25 / 4
  3. 37 / 4 Correct answer.
  4. 43 / 4

Explanation

According to Vieta's formulas for ax^2 + bx + c = 0: x_1 + x_2 = -b / a = -(-7) / 2 = 7/2, and x_1 * x_2 = c / a = 3/2. Using the algebraic identity (x_1 + x_2)^2 = (x_1)^2 + 2*x_1*x_2 + (x_2)^2, we get: (x_1)^2 + (x_2)^2 = (x_1 + x_2)^2 - 2*(x_1 * x_2). Substitute the values: (7/2)^2 - 2*(3/2) = 49/4 - 3 = 49/4 - 12/4 = 37/4.

Question 23Algebra Functions

Solve the rational equation in the domain of real numbers: (x + 3) / (x - 2) - (x - 1) / (x + 2) = 16 / (x^2 - 4).

  1. x = 3/2 Correct answer.
  2. x = 2
  3. x = -2 and x = 3/2
  4. No real solution

Explanation

The domain requires x != 2 and x != -2, since x^2 - 4 = (x - 2)(x + 2). Multiply through by the common denominator (x - 2)(x + 2): (x + 3)(x + 2) - (x - 1)(x - 2) = 16. Expand both products: (x^2 + 5x + 6) - (x^2 - 3x + 2) = 16 => x^2 + 5x + 6 - x^2 + 3x - 2 = 16 => 8x + 4 = 16 => 8x = 12 => x = 12/8 = 3/2. Since 3/2 is in the domain, it is the unique solution.

Question 24Algebra Functions

Find all real solutions to the radical equation: sqrt(2x + 7) - x = 2.

  1. x = -3 and x = 1
  2. x = -3 only
  3. x = 3
  4. x = 1 only Correct answer.

Explanation

Isolate the square root: sqrt(2x + 7) = x + 2. Note the non-negativity condition for the right side: x + 2 >= 0 => x >= -2. Square both sides: 2x + 7 = (x + 2)^2 => 2x + 7 = x^2 + 4x + 4. Rearrange into standard quadratic form: x^2 + 2x - 3 = 0 => (x + 3)(x - 1) = 0. Potential roots: x = -3 or x = 1. Testing x = -3: sqrt(2(-3) + 7) - (-3) = sqrt(1) + 3 = 4 != 2 (extraneous). Testing x = 1: sqrt(2(1) + 7) - 1 = sqrt(9) - 1 = 3 - 1 = 2 (valid). Thus x = 1 is the only real solution.

Question 25Algebra Functions

Solve the absolute value equation in R: |2x - 6| + |x + 1| = 8.

  1. x = 1 and x = 13/3
  2. x = -1 and x = 13/3 Correct answer.
  3. x = -1 and x = 5
  4. x = 13/3 only

Explanation

The critical points where arguments change sign are x = -1 and x = 3, creating three intervals: 1) For x < -1: -(2x - 6) - (x + 1) = 8 => -3x + 5 = 8 => -3x = 3 => x = -1 (not in x < -1). 2) For -1 <= x <= 3: -(2x - 6) + (x + 1) = 8 => -2x + 6 + x + 1 = 8 => -x + 7 = 8 => -x = 1 => x = -1 (valid, since -1 is in [-1, 3]). 3) For x > 3: (2x - 6) + (x + 1) = 8 => 3x - 5 = 8 => 3x = 13 => x = 13/3 (valid, since 13/3 ≈ 4.33 > 3). Thus, the solution set is {-1, 13/3}.

Question 26Algebra Functions

Solve the quadratic inequality in R: x^2 - 4x - 5 <= 0.

  1. (-inf, -1] U [5, inf)
  2. (-1, 5)
  3. [-1, 5] Correct answer.
  4. [-5, 1]

Explanation

Factor the corresponding quadratic equation: x^2 - 4x - 5 = (x - 5)(x + 1) = 0, which gives the roots x = -1 and x = 5. The graph of y = x^2 - 4x - 5 is a parabola opening upward (leading coefficient a = 1 > 0). Therefore, the quadratic expression is non-positive (<= 0) between and including its roots: [-1, 5].

Question 27Algebra Functions

Determine the complete set of all real solutions to the rational inequality: (2x - 4) / (x + 3) >= 1.

  1. (-inf, -3) U [7, inf) Correct answer.
  2. (-3, 7]
  3. [-3, 7]
  4. (-inf, -3] U [7, inf)

Explanation

Subtract 1 from both sides to compare with 0: (2x - 4)/(x + 3) - 1 >= 0 => [(2x - 4) - (x + 3)] / (x + 3) >= 0 => (x - 7) / (x + 3) >= 0. The critical numbers are x = -3 (denominator zero, excluded) and x = 7 (numerator zero, included). Test signs in the intervals: 1) For x < -3: numerator is (-), denominator is (-), quotient is (+) >= 0 (satisfies). 2) For -3 < x < 7: numerator is (-), denominator is (+), quotient is (-) < 0 (does not satisfy). 3) For x >= 7: numerator is (+), denominator is (+), quotient is (+) >= 0 (satisfies). Hence, x in (-inf, -3) U [7, inf).

Question 28Algebra Functions

Solve the system of linear equations in R x R: 3x + 2y = 12 5x - 4y = -2

  1. (x, y) = (4, 0)
  2. (x, y) = (0, 6)
  3. (x, y) = (1, 4.5)
  4. (x, y) = (2, 3) Correct answer.

Explanation

Multiply the first equation by 2 to eliminate y: 2 * (3x + 2y) = 2 * 12 => 6x + 4y = 24. Now add this to the second equation: (6x + 4y) + (5x - 4y) = 24 + (-2) => 11x = 22 => x = 2. Substitute x = 2 into the first equation: 3(2) + 2y = 12 => 6 + 2y = 12 => 2y = 6 => y = 3. Verify in second equation: 5(2) - 4(3) = 10 - 12 = -2. The unique solution is (x, y) = (2, 3).

Question 29Algebra Functions

Solve the exponential equation in R: 4^(2x - 1) = (1 / 8)^(x - 2).

  1. x = 4/5
  2. x = 8/7 Correct answer.
  3. x = 7/8
  4. x = 2

Explanation

Express both sides as powers of the common base 2: 4 = 2^2, so 4^(2x - 1) = (2^2)^(2x - 1) = 2^(4x - 2). 1/8 = 2^(-3), so (1/8)^(x - 2) = (2^(-3))^(x - 2) = 2^(-3x + 6). Since the bases are equal and strictly positive (and != 1), equate the exponents: 4x - 2 = -3x + 6 => 4x + 3x = 6 + 2 => 7x = 8 => x = 8/7.

Question 30Algebra Functions

Find the set of all real solutions to the exponential equation: 3^(2x) - 4 * 3^x + 3 = 0.

  1. x in {0, 1} Correct answer.
  2. x in {1, 3}
  3. x in {-1, 0}
  4. x in {0, 3}

Explanation

Let u = 3^x (with u > 0). Then 3^(2x) = (3^x)^2 = u^2. The equation becomes quadratic: u^2 - 4u + 3 = 0. Factoring: (u - 1)(u - 3) = 0 => u_1 = 1, u_2 = 3. Returning to x: 1) 3^x = 1 = 3^0 => x = 0; 2) 3^x = 3 = 3^1 => x = 1. Both roots are valid real numbers, so x in {0, 1}.

Question 31Algebra Functions

Solve the logarithmic equation in R: log_2(x + 3) + log_2(x - 1) = 5.

  1. x in {-7, 5}
  2. x = 4
  3. x = 5 Correct answer.
  4. x = 7

Explanation

Domain restrictions: x + 3 > 0 => x > -3, and x - 1 > 0 => x > 1. Thus, x > 1. Apply log properties: log_2((x + 3)(x - 1)) = 5 => (x + 3)(x - 1) = 2^5 = 32. Expand: x^2 + 2x - 3 = 32 => x^2 + 2x - 35 = 0. Factor: (x + 7)(x - 5) = 0 => x = -7 or x = 5. Since x must be strictly greater than 1, x = -7 is extraneous. Therefore, x = 5 is the sole solution.

Question 32Algebra Functions

For which real values of the parameter m does the quadratic equation x^2 - 2(m - 1)x + (m + 5) = 0 have exactly one real solution (a double root)?

  1. m in {-4, 1}
  2. m in {-1, 4} Correct answer.
  3. m in {-5, 2}
  4. m = 4 only

Explanation

A quadratic equation ax^2 + bx + c = 0 has exactly one real root if and only if its discriminant D equals zero: D = b^2 - 4ac = [-2(m - 1)]^2 - 4(1)(m + 5) = 0. Expand: 4(m^2 - 2m + 1) - 4(m + 5) = 0. Divide by 4: (m^2 - 2m + 1) - (m + 5) = 0 => m^2 - 3m - 4 = 0. Factor this quadratic equation in m: (m - 4)(m + 1) = 0 => m = 4 or m = -1.

Question 33Algebra Functions

Two cyclists set out at the same moment towards each other from two towns located 72 km apart. The first cyclist rides at a constant speed of 16 km/h, and the second cyclist rides at a constant speed of 20 km/h. How many hours after departing will they meet?

  1. 3.0 hours
  2. 2.5 hours
  3. 1.8 hours
  4. 2.0 hours Correct answer.

Explanation

Since they are moving directly toward each other, their relative speed of approach is the sum of their individual speeds: v_rel = 16 + 20 = 36 km/h. The time t required to cover the total distance of 72 km is: t = d / v_rel = 72 / 36 = 2.0 hours.

Question 34Algebra Functions

Find all real solutions (x, y) to the system of equations: x + y = 5 x^2 + y^2 = 13

  1. {(2, 3), (3, 2)} Correct answer.
  2. {(1, 4), (4, 1)}
  3. {(-2, 7), (7, -2)}
  4. {(2.5, 2.5)}

Explanation

From the first equation, express y in terms of x: y = 5 - x. Substitute this into the second equation: x^2 + (5 - x)^2 = 13 => x^2 + 25 - 10x + x^2 = 13 => 2x^2 - 10x + 12 = 0. Divide by 2: x^2 - 5x + 6 = 0. Factor: (x - 2)(x - 3) = 0 => x_1 = 2, x_2 = 3. If x = 2, y = 5 - 2 = 3. If x = 3, y = 5 - 3 = 2. Both solutions satisfy both equations: 2^2 + 3^2 = 4 + 9 = 13.

Question 35Algebra Functions

Find the set of all real numbers x satisfying the compound inequality: 1 < |2x - 3| <= 5.

  1. (-1, 1) U (2, 4)
  2. [-1, 4]
  3. [-1, 1) U (2, 4] Correct answer.
  4. [1, 2]

Explanation

Break the compound inequality into two parts: 1) |2x - 3| <= 5: -5 <= 2x - 3 <= 5 => -2 <= 2x <= 8 => -1 <= x <= 4, which gives the interval [-1, 4]. 2) |2x - 3| > 1: Either 2x - 3 > 1 => 2x > 4 => x > 2, OR 2x - 3 < -1 => 2x < 2 => x < 1. This excludes the closed interval [1, 2]. Intersecting both conditions: [-1, 4] intersected with ((-inf, 1) U (2, inf)) yields [-1, 1) U (2, 4].

Question 36Algebra Functions

Solve the logarithmic equation in R: (log_5(x))^2 - 3 * log_5(x) - 4 = 0.

  1. x in {4, -1}
  2. x in {20, 1/5}
  3. x in {125, 1/25}
  4. x in {0.2, 625} Correct answer.

Explanation

Substitute y = log_5(x), with x > 0: y^2 - 3y - 4 = 0. Factor the quadratic: (y - 4)(y + 1) = 0 => y_1 = 4 or y_2 = -1. Return to x: 1) log_5(x) = 4 => x = 5^4 = 625; 2) log_5(x) = -1 => x = 5^(-1) = 1/5 = 0.2. Both solutions are positive and valid, so x in {0.2, 625}.

Question 37Algebra Functions

Solve the linear inequality in R: 3 - 2x >= 5x + 17.

  1. [-2, inf)
  2. (-inf, -2] Correct answer.
  3. (-inf, 2]
  4. [2, inf)

Explanation

Rearrange terms by moving x terms to the left and constants to the right: -2x - 5x >= 17 - 3 => -7x >= 14. Divide both sides by -7. Remember to reverse the inequality direction when dividing by a negative number: x <= 14 / (-7) => x <= -2. In interval notation, this is (-inf, -2].

Question 38Algebra Functions

Find all real solutions to the rational equation: (2x) / (x - 3) - 6 / (x - 3) = 4.

  1. No real solution (K = emptyset) Correct answer.
  2. x = 3
  3. x = 0
  4. All real numbers except x = 3

Explanation

State the domain condition: the denominator x - 3 != 0, so x != 3. Combine fractions on the left-hand side over their common denominator: (2x - 6) / (x - 3) = 4. Factor the numerator: 2(x - 3) / (x - 3) = 4. For all x != 3, the factor (x - 3) cancels: 2 = 4. This statement is a contradiction (never true). Therefore, the equation has no real solutions (K = emptyset).

Question 39Algebra Functions

Solve the equation with square roots in R: sqrt(3x + 1) = sqrt(x + 4) + 1.

  1. x in {0, 5}
  2. x = 0 only
  3. x = 5 only Correct answer.
  4. x = 7

Explanation

Domain restrictions require 3x + 1 >= 0 => x >= -1/3 and x + 4 >= 0 => x >= -4, so x >= -1/3. Square both sides: 3x + 1 = (sqrt(x + 4) + 1)^2 => 3x + 1 = (x + 4) + 2*sqrt(x + 4) + 1. Simplify: 3x + 1 = x + 5 + 2*sqrt(x + 4) => 2x - 4 = 2*sqrt(x + 4) => x - 2 = sqrt(x + 4). For the left side to equal a principal root, we must have x - 2 >= 0 => x >= 2. Squaring again: (x - 2)^2 = x + 4 => x^2 - 4x + 4 = x + 4 => x^2 - 5x = 0 => x(x - 5) = 0. The potential roots are x = 0 and x = 5. Checking x = 0: 0 - 2 = -2 != sqrt(4) = 2 (extraneous). Checking x = 5: sqrt(16) = 4, and sqrt(9) + 1 = 3 + 1 = 4 (valid). Thus, x = 5 is the sole solution.

Question 40Algebra Functions

How many integers x satisfy both inequalities simultaneously: 2x - 3 > x / 2 + 3 and x^2 - 9x + 14 <= 0?

  1. 2
  2. 3 Correct answer.
  3. 4
  4. 6

Explanation

Solve the first inequality: 2x - x/2 > 3 + 3 => (3/2)x > 6 => x > 6 * (2/3) => x > 4. Solve the second quadratic inequality: x^2 - 9x + 14 = (x - 2)(x - 7) <= 0. The quadratic is non-positive between its roots: 2 <= x <= 7. Intersect the two solution sets: (4, inf) intersected with [2, 7] = (4, 7], which means 4 < x <= 7. The integer values in this interval are 5, 6, and 7. Thus, there are exactly 3 integer solutions.

Question 41Algebra Functions

Find the domain of definition D(f) of the real function: f(x) = sqrt(x - 4) / (x - 7).

  1. [4, 7) U (7, inf) Correct answer.
  2. (4, 7) U (7, inf)
  3. [4, inf)
  4. (-inf, 4] U (7, inf)

Explanation

Two conditions must be met simultaneously for f(x) to be defined: 1) The expression under the square root must be non-negative: x - 4 >= 0 => x >= 4. 2) The denominator cannot be zero: x - 7 != 0 => x != 7. Combining these two restrictions yields all real numbers x >= 4 except x = 7. In interval notation: D(f) = [4, 7) U (7, inf).

Question 42Algebra Functions

Find the coordinates of the vertex V and the range H(f) of the quadratic function: f(x) = -2x^2 + 8x - 3.

  1. V = [-2, 5], H(f) = (-inf, 5]
  2. V = [2, -3], H(f) = [-3, inf)
  3. V = [2, 5], H(f) = (-inf, 5] Correct answer.
  4. V = [4, 5], H(f) = [5, inf)

Explanation

For f(x) = ax^2 + bx + c with a = -2, b = 8, c = -3: The x-coordinate of the vertex is: x_v = -b / (2a) = -8 / (2 * (-2)) = -8 / (-4) = 2. The y-coordinate is: y_v = f(2) = -2(2^2) + 8(2) - 3 = -2(4) + 16 - 3 = -8 + 16 - 3 = 5. So the vertex is V = [2, 5]. Since a = -2 < 0, the parabola opens downward, meaning the vertex is a global maximum. Hence the range is H(f) = (-inf, 5].

Question 43Algebra Functions

Determine the formula of the linear function f(x) = k*x + q whose graph passes through the points A[1, 5] and B[4, -1].

  1. f(x) = 2x + 3
  2. f(x) = -2x + 9
  3. f(x) = -0.5x + 5.5
  4. f(x) = -2x + 7 Correct answer.

Explanation

The slope k is given by: k = (y_B - y_A) / (x_B - x_A) = (-1 - 5) / (4 - 1) = -6 / 3 = -2. Using point A[1, 5] in the equation y = k*x + q: 5 = -2(1) + q => 5 = -2 + q => q = 7. Verify with point B[4, -1]: f(4) = -2(4) + 7 = -8 + 7 = -1. Therefore, f(x) = -2x + 7.

Question 44Algebra Functions

Find the equations of the vertical and horizontal asymptotes of the rational function: f(x) = (3x + 5) / (x - 2).

  1. Vertical: x = -2; Horizontal: y = 3
  2. Vertical: x = 2; Horizontal: y = 3 Correct answer.
  3. Vertical: x = 2; Horizontal: y = 5
  4. Vertical: x = 3; Horizontal: y = 2

Explanation

1) Vertical asymptote occurs where the denominator is zero (and numerator non-zero): x - 2 = 0 => x = 2. 2) Horizontal asymptote describes behavior as x -> +-inf: lim (x -> inf) (3x + 5)/(x - 2) = 3/1 = 3. Alternatively, decompose the rational function: (3x + 5)/(x - 2) = (3(x - 2) + 11)/(x - 2) = 3 + 11/(x - 2). This clearly shows a hyperbola with center S[2, 3] and asymptotes x = 2 and y = 3.

Question 45Algebra Functions

Which of the following functions defined on R is an odd function (lichá funkce)?

  1. f(x) = x^3 - 4x Correct answer.
  2. f(x) = x^4 + 2x^2
  3. f(x) = cos(x) + 1
  4. f(x) = x^2 + x

Explanation

A function f is odd if f(-x) = -f(x) for all x in its symmetric domain. Test f(x) = x^3 - 4x: f(-x) = (-x)^3 - 4(-x) = -x^3 + 4x = -(x^3 - 4x) = -f(x). Thus, f(x) = x^3 - 4x is an odd function. For comparison: x^4 + 2x^2 and cos(x) + 1 are even functions (f(-x) = f(x)), while x^2 + x is neither even nor odd.

Question 46Algebra Functions

Determine the fundamental period T and the range H(f) of the trigonometric function: f(x) = 3*sin(2x - pi/3) + 1.

  1. T = 2*pi, H(f) = [-2, 4]
  2. T = pi, H(f) = [-3, 3]
  3. T = pi/2, H(f) = [-2, 4]
  4. T = pi, H(f) = [-2, 4] Correct answer.

Explanation

The standard sine function sin(u) has a period of 2*pi and a range of [-1, 1]. 1) Period: For sin(B*x - C), the fundamental period is T = (2*pi) / |B| = (2*pi) / 2 = pi. 2) Range: Since -1 <= sin(2x - pi/3) <= 1, multiplying by 3 gives -3 <= 3*sin(...) <= 3. Adding 1 shifts the entire range upward: -3 + 1 <= f(x) <= 3 + 1 => -2 <= f(x) <= 4. Thus H(f) = [-2, 4].

Question 47Algebra Functions

Find the domain of definition of the real logarithmic function: f(x) = ln((x - 2) / (5 - x)).

  1. (-inf, 2) U (5, inf)
  2. (2, 5) Correct answer.
  3. [2, 5]
  4. (2, 5]

Explanation

The natural logarithm ln(u) requires its argument to be strictly positive: (x - 2) / (5 - x) > 0. The critical points are x = 2 and x = 5. Note that x != 5 to prevent division by zero, and x != 2 because ln(0) is undefined. Test signs in the intervals: 1) For x < 2: numerator (-), denominator (+), quotient (-) < 0. 2) For 2 < x < 5: numerator (+), denominator (+), quotient (+) > 0. 3) For x > 5: numerator (+), denominator (-), quotient (-) < 0. Thus the quotient is strictly positive only on the open interval (2, 5).

Question 48Algebra Functions

Find the inverse function f^(-1)(x) for the bijective function f: R \ {-3} -> R \ {2} defined by f(x) = (2x - 1) / (x + 3).

  1. f^(-1)(x) = (x + 3) / (2x - 1)
  2. f^(-1)(x) = (2x + 1) / (x - 3)
  3. f^(-1)(x) = (3x + 1) / (2 - x) Correct answer.
  4. f^(-1)(x) = (3x - 1) / (x - 2)

Explanation

Set y = (2x - 1) / (x + 3) and solve for x in terms of y: y*(x + 3) = 2x - 1 => yx + 3y = 2x - 1. Group terms involving x on one side: yx - 2x = -3y - 1 => x*(y - 2) = -(3y + 1). Divide by (y - 2): x = -(3y + 1) / (y - 2) = (3y + 1) / (2 - y). Swapping variables to write as a function of x gives: f^(-1)(x) = (3x + 1) / (2 - x).

Question 49Algebra Functions

Calculate the exact numerical value of the trigonometric expression: cos(7*pi / 6) + tan(4*pi / 3).

  1. sqrt(3) / 2 Correct answer.
  2. -sqrt(3) / 2
  3. (3*sqrt(3)) / 2
  4. 0

Explanation

Use unit circle values and symmetry: 1) 7*pi/6 = pi + pi/6 (in Quadrant III). cos(pi + alpha) = -cos(alpha), so cos(7*pi/6) = -cos(pi/6) = -sqrt(3)/2. 2) 4*pi/3 = pi + pi/3 (in Quadrant III). tan(pi + alpha) = tan(alpha), so tan(4*pi/3) = tan(pi/3) = sqrt(3). Summing the two values: -sqrt(3)/2 + sqrt(3) = (-1/2 + 1)*sqrt(3) = (1/2)*sqrt(3) = sqrt(3)/2.

Question 50Algebra Functions

The graph of a function y = f(x) is transformed in three consecutive steps: shifted 3 units to the left, stretched vertically by a factor of 2, and shifted 4 units downward. What is the equation of the resulting function g(x)?

  1. g(x) = 2*f(x - 3) - 4
  2. g(x) = 0.5*f(x + 3) - 4
  3. g(x) = 2*f(x + 3) - 4 Correct answer.
  4. g(x) = 2*f(x + 3) + 4

Explanation

Trace each transformation: 1) Horizontal shift 3 units to the left replaces x with (x + 3), yielding f(x + 3). 2) Vertical stretch by a factor of 2 multiplies the function value by 2, yielding 2*f(x + 3). 3) Vertical shift 4 units downward subtracts 4 from the entire function, yielding g(x) = 2*f(x + 3) - 4.