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Bagrut Mathematics (3/4/5 Units) study materials
Bagrut Mathematics (3/4/5 Units) study resources

Bagrut Mathematics (3/4/5 Units) practice questions

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Question 1Algebra and Algebraic Techniques

What is the value of (2x^3)(3x^4) when simplified?

  1. 6x^12
  2. 5x^7
  3. 6x^1
  4. 6x^7 Correct answer.

Explanation

Multiply the coefficients 2·3 = 6 and add the exponents of like bases: x^3·x^4 = x^(3+4) = x^7. The product is therefore 6x^7.

Question 2Algebra and Algebraic Techniques

Expand and simplify (x + 5)(x − 3).

  1. x^2 + 2x − 15 Correct answer.
  2. x^2 + 8x − 15
  3. x^2 − 15
  4. x^2 − 2x − 15

Explanation

Using FOIL: x·x = x^2, x·(−3) = −3x, 5·x = 5x, and 5·(−3) = −15. Combining −3x + 5x gives +2x, so the result is x^2 + 2x − 15.

Question 3Algebra and Algebraic Techniques

Factor completely: x^2 − 49.

  1. (x − 7)^2
  2. (x − 7)(x + 7) Correct answer.
  3. x(x − 49)
  4. (x + 7)^2

Explanation

The expression is a difference of two squares, since 49 = 7^2. The identity a^2 − b^2 = (a − b)(a + b) gives (x − 7)(x + 7).

Question 4Algebra and Algebraic Techniques

Solve for x: 3x − 7 = 2x + 5.

  1. x = −12
  2. x = 4
  3. x = 12 Correct answer.
  4. x = 2

Explanation

Subtract 2x from both sides to get x − 7 = 5, then add 7 to both sides to obtain x = 12. Checking: 3·12 − 7 = 29 and 2·12 + 5 = 29, so the solution is consistent.

Question 5Algebra and Algebraic Techniques

What is the slope of the line passing through the points (2, 3) and (6, 11)?

  1. 4
  2. 1/2
  3. 2 Correct answer.
  4. 8

Explanation

The slope is the change in y divided by the change in x: (11 − 3)/(6 − 2) = 8/4 = 2. The line rises 2 units for every 1 unit it moves to the right.

Question 6Algebra and Algebraic Techniques

Simplify the expression (6x^5 y^2)/(2x^2 y^5), assuming x, y ≠ 0.

  1. 3x^3 y^3
  2. 3x^3 / y^3 Correct answer.
  3. 3x^7 y^7
  4. 12x^3 / y^3

Explanation

Divide coefficients: 6/2 = 3. For each base subtract the exponents: x^(5−2) = x^3 and y^(2−5) = y^(−3), which equals 1/y^3. The simplified form is 3x^3/y^3.

Question 7Algebra and Algebraic Techniques

If f(x) = 2x^2 − 3x + 1, what is f(−2)?

  1. 11
  2. 15 Correct answer.
  3. −1
  4. 3

Explanation

Substitute x = −2: 2(−2)^2 − 3(−2) + 1 = 2·4 + 6 + 1 = 8 + 6 + 1 = 15. Note that (−2)^2 = 4 is positive and −3·(−2) = +6.

Question 8Algebra and Algebraic Techniques

Which of the following is a root of the quadratic equation x^2 − 5x + 6 = 0?

  1. x = 3 Correct answer.
  2. x = −3
  3. x = 6
  4. x = −2

Explanation

Factor the quadratic as (x − 2)(x − 3) = 0, since the numbers multiplying to 6 and adding to −5 are −2 and −3. The roots are x = 2 and x = 3, so x = 3 is a root.

Question 9Algebra and Algebraic Techniques

Solve the inequality 4 − 2x > 10.

  1. x < −3 Correct answer.
  2. x < 3
  3. x > 3
  4. x > −3

Explanation

Subtract 4 from both sides to obtain −2x > 6. Dividing by −2 reverses the inequality sign, giving x < −3.

Question 10Algebra and Algebraic Techniques

The quadratic equation x^2 + kx + 9 = 0 has exactly one real solution. What are the possible values of k?

  1. k = ±3
  2. k = 9 only
  3. k = ±6 Correct answer.
  4. k = 6 only

Explanation

A quadratic has exactly one real root when its discriminant b^2 − 4ac equals zero. Here k^2 − 4·1·9 = 0, so k^2 = 36 and k = 6 or k = −6.

Question 11Algebra and Algebraic Techniques

Solve the system of equations: x + y = 7 and x − y = 1.

  1. x = 5, y = 2
  2. x = 6, y = 1
  3. x = 4, y = 3 Correct answer.
  4. x = 3, y = 4

Explanation

Adding the two equations eliminates y: 2x = 8, so x = 4. Substituting into x + y = 7 gives y = 3. Checking the second equation: 4 − 3 = 1, which is correct.

Question 12Algebra and Algebraic Techniques

Simplify (x^2 − 9)/(x^2 + x − 12), assuming the denominator is nonzero.

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

Explanation

Factor the numerator as (x − 3)(x + 3) and the denominator as (x + 4)(x − 3). Cancelling the common factor (x − 3) leaves (x + 3)/(x + 4), provided x ≠ 3.

Question 13Algebra and Algebraic Techniques

For the arithmetic sequence 5, 9, 13, 17, ..., what is the 20th term?

  1. 85
  2. 77
  3. 81 Correct answer.
  4. 80

Explanation

The first term is a = 5 and the common difference is d = 4. Using a_n = a + (n − 1)d gives a_20 = 5 + 19·4 = 5 + 76 = 81.

Question 14Algebra and Algebraic Techniques

The sum of the first 10 terms of the geometric series 2 + 6 + 18 + ... is:

  1. 39366
  2. 59049
  3. 19682
  4. 59048 Correct answer.

Explanation

Here a = 2 and the ratio is q = 3. The sum of n terms is a(q^n − 1)/(q − 1), so S_10 = 2(3^10 − 1)/2 = 3^10 − 1 = 59049 − 1 = 59048.

Question 15Algebra and Algebraic Techniques

The expression log_2(8) + log_2(4) equals:

  1. 32
  2. 7
  3. 6
  4. 5 Correct answer.

Explanation

Since 8 = 2^3 and 4 = 2^2, log_2(8) = 3 and log_2(4) = 2. Their sum is 5, which also matches log_2(8·4) = log_2(32) = 5.

Question 16Algebra and Algebraic Techniques

Solve for x: 2^(x+1) = 32.

  1. x = 16
  2. x = 6
  3. x = 5
  4. x = 4 Correct answer.

Explanation

Write 32 as 2^5, so the equation becomes 2^(x+1) = 2^5. Equating exponents gives x + 1 = 5, hence x = 4.

Question 17Algebra and Algebraic Techniques

What is the remainder when x^3 − 2x^2 + 4x − 5 is divided by (x − 2)?

  1. 3 Correct answer.
  2. 1
  3. −3
  4. 11

Explanation

By the remainder theorem, the remainder equals the value of the polynomial at x = 2. Substituting gives 8 − 8 + 8 − 5 = 3, so the remainder is 3.

Question 18Algebra and Algebraic Techniques

The expression (a + b)^2 − (a − b)^2 simplifies to:

  1. 2ab
  2. 4ab Correct answer.
  3. 0
  4. 2a^2 + 2b^2

Explanation

Expanding gives (a^2 + 2ab + b^2) − (a^2 − 2ab + b^2). The a^2 and b^2 terms cancel, and the middle terms combine as 2ab + 2ab = 4ab.

Question 19Algebra and Algebraic Techniques

For which values of x is the expression (x − 1)/(x^2 − 4) undefined?

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

Explanation

A rational expression is undefined where its denominator equals zero. Solving x^2 − 4 = 0 gives x = 2 or x = −2, so the expression is undefined at these two values.

Question 20Algebra and Algebraic Techniques

The quadratic function f(x) = x^2 − 6x + 5 has its vertex at:

  1. (6, 5)
  2. (3, −4) Correct answer.
  3. (3, 4)
  4. (−3, 32)

Explanation

The x-coordinate of the vertex is −b/(2a) = 6/2 = 3. Substituting x = 3 gives f(3) = 9 − 18 + 5 = −4, so the vertex is (3, −4).

Question 21Algebra and Algebraic Techniques

The equation x^2 + (m − 2)x + 9 = 0 has two distinct real roots. Which condition on m must hold?

  1. m < −4 or m > 8 Correct answer.
  2. m > 8 only
  3. −4 < m < 8
  4. m < 4 or m > 8

Explanation

Two distinct real roots require a positive discriminant: (m − 2)^2 − 36 > 0. This gives |m − 2| > 6, so m − 2 > 6 or m − 2 < −6, yielding m > 8 or m < −4.

Question 22Algebra and Algebraic Techniques

If 2^x = 5, what is the value of 2^(2x+1)?

  1. 10
  2. 11
  3. 25
  4. 50 Correct answer.

Explanation

Using exponent laws, 2^(2x+1) = 2^(2x)·2^1 = (2^x)^2·2. Since 2^x = 5, this equals 5^2·2 = 25·2 = 50.

Question 23Algebra and Algebraic Techniques

The polynomial P(x) = x^3 − 4x^2 + x + 6 has a root at x = 2. What are all the roots of P(x)?

  1. x = 2, x = 3, x = 1
  2. x = 2, x = −2, x = 3
  3. x = 2, x = 1, x = −3
  4. x = 2, x = 3, x = −1 Correct answer.

Explanation

Dividing P(x) by (x − 2) gives the quotient x^2 − 2x − 3. Factoring yields (x − 3)(x + 1), so the roots are x = 2, x = 3 and x = −1.

Question 24Algebra and Algebraic Techniques

The expression 1/(x − 1) − 1/(x + 1) simplifies to:

  1. 0
  2. 2x/(x^2 − 1)
  3. 2/(x^2 − 1) Correct answer.
  4. −2/(x^2 − 1)

Explanation

Use the common denominator (x − 1)(x + 1) = x^2 − 1. The numerator becomes (x + 1) − (x − 1) = 2, so the expression simplifies to 2/(x^2 − 1).

Question 25Algebra and Algebraic Techniques

An arithmetic sequence has a_3 = 11 and a_8 = 31. What is the sum of its first 20 terms?

  1. 760
  2. 860
  3. 800
  4. 820 Correct answer.

Explanation

The difference between terms is (31 − 11)/(8 − 3) = 20/5 = 4, so d = 4. Then a_1 = 11 − 2·4 = 3, and S_20 = (20/2)(2·3 + 19·4) = 10(6 + 76) = 820.

Question 26Algebra and Algebraic Techniques

The expression (2x^3)^2 · (3x^2) is equivalent to which of the following for all real x?

  1. 36x^12
  2. 12x^8 Correct answer.
  3. 12x^7
  4. 6x^8

Explanation

Apply the power-of-a-power rule first: (2x^3)^2 = 4x^6, since both the coefficient 2 and the exponent 3 are squared. Multiplying by 3x^2 gives 4·3·x^(6+2) = 12x^8, using the product rule for exponents with the same base x.

Question 27Algebra and Algebraic Techniques

The expression x^2 − 7x + 12 can be written as a product of two linear factors with integer coefficients. Which factorization is correct?

  1. (x + 3)(x + 4)
  2. (x − 1)(x − 12)
  3. (x − 3)(x − 4) Correct answer.
  4. (x − 2)(x − 6)

Explanation

We need two integers whose product is +12 and whose sum is −7. The pair −3 and −4 satisfies both conditions, so the trinomial factors as (x − 3)(x − 4). Expanding confirms x^2 − 4x − 3x + 12 = x^2 − 7x + 12.

Question 28Algebra and Algebraic Techniques

For which values of x is the equation (x + 3)/(x − 2) = 4 true?

  1. x = 11/3 Correct answer.
  2. x = 2
  3. x = 11/5
  4. x = 5/3

Explanation

Multiply both sides by (x − 2), which is permitted because x = 2 is excluded from the domain. This gives x + 3 = 4x − 8, so 11 = 3x and x = 11/3. Substituting back gives (11/3 + 3)/(11/3 − 2) = (20/3)/(5/3) = 4, confirming the solution.

Question 29Algebra and Algebraic Techniques

A rectangle has a perimeter of 34 cm and a diagonal of 13 cm. What is the area of the rectangle, in square centimetres?

  1. 42
  2. 30
  3. 120
  4. 60 Correct answer.

Explanation

Writing the side lengths as a and b, the perimeter gives a + b = 17 and the diagonal gives a^2 + b^2 = 169. Since (a + b)^2 = a^2 + b^2 + 2ab, we get 289 = 169 + 2ab, so ab = 60, which is the area. The sides are 5 and 12, and indeed 5 + 12 = 17 with 5^2 + 12^2 = 169.

Question 30Algebra and Algebraic Techniques

The equation x^4 − 5x^2 + 4 = 0 has how many distinct real solutions?

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

Explanation

Substituting t = x^2 turns the equation into t^2 − 5t + 4 = 0, which factors as (t − 1)(t − 4) = 0, so t = 1 or t = 4. Since t = x^2 must be non-negative, both values are admissible, giving x = ±1 and x = ±2. These are four distinct real numbers, all of which satisfy the original equation.

Question 31Equations and Inequalities

פתרו את המשוואה 3x − 7 = 2x + 5. מהו ערך x?

  1. 2
  2. −12
  3. 12 Correct answer.
  4. −2

Explanation

מעבירים את 2x לאגף שמאל ואת 7− לאגף ימין: 3x − 2x = 5 + 7, ולכן x = 12. הפעולה מותרת כי הוספת אותו ביטוי לשני האגפים שומרת על שוויון.

Question 32Equations and Inequalities

מהו פתרון אי-השוויון −2x + 1 < 7?

  1. x > −3 Correct answer.
  2. x > 3
  3. x < −3
  4. x < 3

Explanation

מעבירים: −2x < 6. כאשר מחלקים אי-שוויון במספר שלילי הופכים את כיוון האי-שוויון, ולכן x > −3.

Question 33Equations and Inequalities

פתרו את המשוואה הריבועית x² − 5x + 6 = 0. מהם הפתרונות?

  1. x = 1 ו-x = 6
  2. x = −1 ו-x = −6
  3. x = 2 ו-x = 3 Correct answer.
  4. x = −2 ו-x = −3

Explanation

מפרקים לגורמים: (x − 2)(x − 3) = 0, ולכן x = 2 או x = 3. סכום השורשים 5 ומכפלתם 6, כנדרש ממקדמים.

Question 34Equations and Inequalities

מהו תחום ההצבה של המשוואה (x + 3)/(x − 4) = 1?

  1. x ≠ −3
  2. x ≠ 0
  3. x ≠ 4 Correct answer.
  4. כל x ממשי

Explanation

המכנה x − 4 חייב להיות שונה מאפס, אחרת הביטוי אינו מוגדר. לכן x = 4 נפסל ותחום ההצבה הוא כל x פרט ל-4.

Question 35Equations and Inequalities

מהו הפתרון של המשוואה √(2x + 1) = 3?

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

Explanation

מעלים בריבוע את שני האגפים: 2x + 1 = 9, ולכן 2x = 8 ו-x = 4. הצבת 4 נותנת √9 = 3, ולכן הפתרון תקף.

Question 36Equations and Inequalities

נתונה הפונקציה f(x) = x² − 4x + 7. מהו ערך המינימום שלה?

  1. 4
  2. 7
  3. −3
  4. 3 Correct answer.

Explanation

קודקוד הפרבולה נמצא ב-x = −b/(2a) = 4/2 = 2, וערך הפונקציה שם הוא 4 − 8 + 7 = 3. מכיוון שמקדם x² חיובי, זהו מינימום.

Question 37Equations and Inequalities

פתרו את המערכת: x + y = 7 ו-x − y = 1. מהם x ו-y?

  1. x = 6, y = 1
  2. x = 4, y = 3 Correct answer.
  3. x = 5, y = 2
  4. x = 3, y = 4

Explanation

חיבור שתי המשוואות נותן 2x = 8, כלומר x = 4, ואז מהמשוואה הראשונה y = 3. הצבה בשנייה מאמתת 4 − 3 = 1.

Question 38Equations and Inequalities

פתרו את אי-השוויון הריבועי x² − 9 < 0.

  1. x < −3 או x > 3
  2. −3 < x < 3 Correct answer.
  3. x < 3
  4. −3 < x

Explanation

מפרקים: (x − 3)(x + 3) < 0. הפרבולה שלילית בין שורשיה, ולכן הפתרון הוא הקטע הפתוח −3 < x < 3.

Question 39Equations and Inequalities

פתרו את המשוואה |2x − 6| = 4. מהם כל הפתרונות?

  1. x = 5 ו-x = −1
  2. x = 1 בלבד
  3. x = 5 ו-x = 1 Correct answer.
  4. x = 5 בלבד

Explanation

משוואת ערך מוחלט מתפצלת לשני מקרים: 2x − 6 = 4 נותן x = 5, ו-2x − 6 = −4 נותן x = 1. שני הערכים מקיימים את המשוואה.

Question 40Equations and Inequalities

נתונה המשוואה (x + 1)/(x − 2) = (x − 1)/(x + 3). מהו פתרונה?

  1. x = 5
  2. x = −1/5
  3. אין פתרון
  4. x = 1/5 Correct answer.

Explanation

מכפילים בהצלבה: (x + 1)(x + 3) = (x − 1)(x − 2), כלומר x² + 4x + 3 = x² − 3x + 2. מכאן 7x = −1 ו-x = −1/7; בדיקה מראה שהערך שונה מאפסי המכנים.

Question 41Equations and Inequalities

נתונה המשוואה הריבועית x² − 6x + k = 0. עבור איזה ערך של k יש למשוואה פתרון ממשי יחיד?

  1. k = 6
  2. k = 9 Correct answer.
  3. k = 36
  4. k = 12

Explanation

פתרון ממשי יחיד מתקבל כאשר הדיסקרימיננטה מתאפסת: 36 − 4k = 0, ולכן k = 9. אז המשוואה היא (x − 3)² = 0.

Question 42Equations and Inequalities

במשוואה 2^(x+1) = 32, מהו x?

  1. 5
  2. 6
  3. 4 Correct answer.
  4. 16

Explanation

רושמים 32 = 2⁵, ולכן 2^(x+1) = 2⁵. משווים מעריכים: x + 1 = 5, ומכאן x = 4.

Question 43Equations and Inequalities

פתרו את המשוואה log₂(x − 3) + log₂(x) = 2.

  1. x = 4 Correct answer.
  2. x = 4 ו-x = −1
  3. x = 3
  4. x = 1

Explanation

מחברים לוגריתמים: log₂(x(x − 3)) = 2, ולכן x² − 3x = 4, כלומר x² − 3x − 4 = 0. הפתרונות 4 ו-1−, אך 1− נפסל בתחום ההצבה, ולכן x = 4.

Question 44Equations and Inequalities

נתונה הפונקציה f(x) = x² − 2x − 3. באילו תחומים מתקיים f(x) > 0?

  1. x > 3 בלבד
  2. x < −1 או x > 3 Correct answer.
  3. −1 < x < 3
  4. −3 < x < 1

Explanation

השורשים הם 1− ו-3, והפרבולה פונה מעלה. לכן הפונקציה חיובית מחוץ לקטע שבין השורשים, כלומר x < −1 או x > 3.

Question 45Equations and Inequalities

כמה פתרונות ממשיים יש למשוואה |x − 2| = −3?

  1. שני פתרונות
  2. פתרון אחד
  3. אין פתרון Correct answer.
  4. אינסוף פתרונות

Explanation

ערך מוחלט של ביטוי ממשי הוא תמיד אי-שלילי, ולכן אינו יכול להשתוות למספר שלילי כמו 3−. מכאן שלמשוואה אין פתרון ממשי כלל.

Question 46Equations and Inequalities

נתונה המשוואה x + 1/x = 3, כאשר x ≠ 0. מהו ערכו של x² + 1/x²?

  1. 3
  2. 9
  3. 7 Correct answer.
  4. 11

Explanation

מעלים בריבוע את השוויון x + 1/x = 3: מתקבל x² + 2 + 1/x² = 9. מחסירים 2 ומקבלים x² + 1/x² = 7.

Question 47Equations and Inequalities

נתונה המשוואה (x − 1)² = 2x − 3. מהו מספר הפתרונות הממשיים שלה?

  1. שני פתרונות ממשיים שונים
  2. פתרון ממשי יחיד Correct answer.
  3. אין פתרון ממשי
  4. אינסוף פתרונות

Explanation

מפתחים: x² − 2x + 1 = 2x − 3, כלומר x² − 4x + 4 = 0, דהיינו (x − 2)² = 0. הדיסקרימיננטה מתאפסת, ולכן יש פתרון ממשי יחיד x = 2.

Question 48Equations and Inequalities

נתונה המערכת: x + y = 3 ו-x² + y² = 5. מהם הזוגות (x, y) הפותרים אותה?

  1. (2, 1) בלבד
  2. (1, 2) בלבד
  3. (1, 2) ו-(2, 1) Correct answer.
  4. אין פתרון

Explanation

מבודדים y = 3 − x ומציבים: x² + (3 − x)² = 5, כלומר 2x² − 6x + 4 = 0, או x² − 3x + 2 = 0. הפתרונות x = 1 ו-x = 2 נותנים את הזוגות (1, 2) ו-(2, 1).

Question 49Equations and Inequalities

נתונה המשוואה √(x + 5) = x − 1. מהו פתרונה?

  1. x = −1
  2. x = 4 ו-x = −1
  3. x = 4 Correct answer.
  4. אין פתרון

Explanation

מעלים בריבוע: x + 5 = x² − 2x + 1, כלומר x² − 3x − 4 = 0, שפתרונותיה 4 ו-1−. בדיקה: x = 4 נותן √9 = 3 = 4 − 1, ואילו x = −1 נותן 2 = −2, ולכן נפסל כפתרון זר.

Question 50Equations and Inequalities

נתונה המשוואה 4^x − 5·2^x + 4 = 0. מהו סכום פתרונותיה הממשיים?

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

Explanation

מציבים t = 2^x ומקבלים t² − 5t + 4 = 0, כלומר t = 1 או t = 4. מכאן 2^x = 1 נותן x = 0, ו-2^x = 4 נותן x = 2, וסכום הפתרונות הוא 2.