Polynomial Questions -Solutions

zaheerpashamd

Polynomial Problems and Solutions

1-Mark Questions

If x ≠ –1, then find the quotient of (x⁵ + x⁴ + x³ + x²) / (x³ + x² + x + 1) (M'15)

Step 1: Factor numerator and denominator
Numerator: x⁵ + x⁴ + x³ + x² = x²(x³ + x² + x + 1)
Denominator: x³ + x² + x + 1

Step 2: Simplify the expression
(x⁵ + x⁴ + x³ + x²) / (x³ + x² + x + 1) = [x²(x³ + x² + x + 1)] / (x³ + x² + x + 1)

Step 3: Cancel common factors
Since x ≠ -1, (x³ + x² + x + 1) ≠ 0, so we can cancel:
= x²

∴ Quotient = x²

"We can write a trinomial having degree 7". Justify the above statement by giving one example. (M'15)

Step 1: Understand the terms
- A trinomial is a polynomial with exactly three terms.
- Degree of a polynomial is the highest power of the variable.

Step 2: Create an example
Example: 3x⁷ + 2x³ + 5
This has:
- Three terms: 3x⁷, 2x³, and 5
- Highest power: 7
- Therefore, it's a trinomial of degree 7.

∴ Example: 3x⁷ + 2x³ + 5
This is a trinomial with degree 7, which justifies the statement.

Write an example for a quadratic Polynomial that has no zeros. (M'16)

Step 1: Condition for no zeros
A quadratic polynomial has no real zeros when its discriminant (D = b² - 4ac) is negative.

Step 2: Create an example
Let's take p(x) = x² + 2x + 3
Discriminant D = b² - 4ac = (2)² - 4(1)(3) = 4 - 12 = -8 < 0
Since D < 0, this quadratic has no real zeros.

∴ Example: x² + 2x + 3
This quadratic polynomial has no real zeros as its discriminant is negative.

If p(x) = x³ – 3x² + 2x – 3 is a polynomial, then find the value of p(1). (J'16)

Step 1: Substitute x = 1
p(1) = (1)³ - 3(1)² + 2(1) - 3

Step 2: Calculate
p(1) = 1 - 3 + 2 - 3

Step 3: Simplify
p(1) = (1 + 2) + (-3 - 3) = 3 - 6 = -3

∴ p(1) = -3

Srikar says that the order of the polynomial (x² – 5)(x³ + 1) is 6. Do you agree with him? (J'17)

Step 1: Expand the polynomial
(x² - 5)(x³ + 1) = x²(x³ + 1) - 5(x³ + 1) = x⁵ + x² - 5x³ - 5

Step 2: Rearrange in standard form
= x⁵ - 5x³ + x² - 5

Step 3: Find the degree
The highest power of x is 5, so the degree (order) is 5.

∴ I don't agree with Srikar.
The order of the polynomial is 5, not 6.

Find zeros of the polynomial P(x) = x² – 4. (J'17)

Step 1: Set P(x) = 0
x² - 4 = 0

Step 2: Solve for x
x² = 4
x = ±√4 = ±2

∴ Zeros are x = 2 and x = -2

Verify the relation between zeros and coefficients of the quadratic polynomial x² – 4. (M'18)

Step 1: Identify coefficients
For x² - 4:
a = 1, b = 0, c = -4

Step 2: Find zeros
Zeros: α = 2, β = -2

Step 3: Verify sum of zeros
α + β = 2 + (-2) = 0
-b/a = -0/1 = 0
∴ α + β = -b/a

Step 4: Verify product of zeros
αβ = (2)(-2) = -4
c/a = -4/1 = -4
∴ αβ = c/a

∴ The relation between zeros and coefficients is verified:
Sum of zeros = -b/a = 0
Product of zeros = c/a = -4

Whether 1/2 and 1 are zeros of the polynomial p(x) = 2x² – 3x + 1 or not? Justify. (J'18)

Step 1: Check if 1/2 is a zero
p(1/2) = 2(1/2)² - 3(1/2) + 1 = 2(1/4) - 3/2 + 1 = 1/2 - 3/2 + 1 = -1 + 1 = 0
∴ 1/2 is a zero.

Step 2: Check if 1 is a zero
p(1) = 2(1)² - 3(1) + 1 = 2 - 3 + 1 = 0
∴ 1 is a zero.

∴ Both 1/2 and 1 are zeros of the polynomial p(x) = 2x² - 3x + 1.

If P(x) = x⁴ + 1, then find P(2) – P(-2). (M'19)

Step 1: Calculate P(2)
P(2) = (2)⁴ + 1 = 16 + 1 = 17

Step 2: Calculate P(-2)
P(-2) = (-2)⁴ + 1 = 16 + 1 = 17

Step 3: Find P(2) - P(-2)
P(2) - P(-2) = 17 - 17 = 0

∴ P(2) - P(-2) = 0

–3, 0 and 2 are the zeroes of the polynomial p(x) = x³ + (a – 1)x² + bx + c. Find a and c. (J'19)

Step 1: Use the zero x = 0
p(0) = (0)³ + (a - 1)(0)² + b(0) + c = c
Since 0 is a zero, p(0) = 0, so c = 0

Step 2: Use the zero x = 2
p(2) = (2)³ + (a - 1)(2)² + b(2) + 0 = 8 + 4(a - 1) + 2b = 0
8 + 4a - 4 + 2b = 0
4a + 2b + 4 = 0
2a + b + 2 = 0 ...(1)

Step 3: Use the zero x = -3
p(-3) = (-3)³ + (a - 1)(-3)² + b(-3) + 0 = -27 + 9(a - 1) - 3b = 0
-27 + 9a - 9 - 3b = 0
9a - 3b - 36 = 0
3a - b - 12 = 0 ...(2)

Step 4: Solve equations (1) and (2)
From (1): b = -2a - 2
Substitute in (2): 3a - (-2a - 2) - 12 = 0
3a + 2a + 2 - 12 = 0
5a - 10 = 0
a = 2

∴ a = 2, c = 0

Write any two linear polynomials having one term under three terms. (J'19)

Step 1: Understand linear polynomials
A linear polynomial has degree 1 and can have 1, 2, or 3 terms.

Step 2: Examples with one term
1. 5x (degree 1, one term)
2. -3x (degree 1, one term)

Step 3: Examples with three terms
1. 2x + 3y + 4 (degree 1 in x and y, three terms)
2. x + y - 2 (degree 1 in x and y, three terms)

∴ One-term examples: 5x, -3x
Three-term examples: 2x + 3y + 4, x + y - 2

If p(x) = x² + 3x + 4, then find the values of p(0) and p(1). (May 2022)

Step 1: Calculate p(0)
p(0) = (0)² + 3(0) + 4 = 0 + 0 + 4 = 4

Step 2: Calculate p(1)
p(1) = (1)² + 3(1) + 4 = 1 + 3 + 4 = 8

∴ p(0) = 4, p(1) = 8

If p(x) = 2x² + 5x – 7, then find the value of p(0) and p(1). (Aug 2022)

Step 1: Calculate p(0)
p(0) = 2(0)² + 5(0) - 7 = 0 + 0 - 7 = -7

Step 2: Calculate p(1)
p(1) = 2(1)² + 5(1) - 7 = 2 + 5 - 7 = 0

∴ p(0) = -7, p(1) = 0

Polynomial Problems - 2 Marks

Polynomial-2 Marks

For what value of k, –4 is a zero of the polynomial x² – x – (2k + 2). (J'15)

Step 1: Substitute x = -4 in the polynomial
p(x) = x² - x - (2k + 2)
p(-4) = (-4)² - (-4) - (2k + 2)

Step 2: Simplify the expression
p(-4) = 16 + 4 - 2k - 2 = 18 - 2k

Step 3: Set p(-4) = 0 (since -4 is a zero)
18 - 2k = 0

Step 4: Solve for k
2k = 18
k = 9

∴ k = 9

Use the table given below to draw the graph. Use the graph drawn to find the values of a and b. (J'15)

Given table:

x	-2	0	2	1	b
y	-3	1	a	3	-7

Step 1: Assume the relationship is linear (y = mx + c)
Using points (-2, -3) and (0, 1):
When x = 0, y = 1 ⇒ c = 1
Using (-2, -3): -3 = m(-2) + 1 ⇒ -3 = -2m + 1 ⇒ -2m = -4 ⇒ m = 2
∴ Equation: y = 2x + 1

Step 2: Find value of a
When x = 2: y = 2(2) + 1 = 4 + 1 = 5
∴ a = 5

Step 3: Find value of b
When y = -7: -7 = 2b + 1 ⇒ 2b = -8 ⇒ b = -4

∴ a = 5, b = -4

Length of a rectangle is 5 units more than its breadth. Express its perimeter in polynomial form.

Step 1: Define variables
Let breadth = x units
Then length = (x + 5) units

Step 2: Write perimeter formula
Perimeter = 2 × (length + breadth)
= 2 × [(x + 5) + x]
= 2 × (2x + 5)

Step 3: Simplify to polynomial form
= 4x + 10

∴ Perimeter = 4x + 10

Show that 2 and −1/3 are zeros of the polynomial 3x² – 5x – 2. (J'16)

Given polynomial: p(x) = 3x² - 5x - 2

Step 1: Check if x = 2 is a zero
p(2) = 3(2)² - 5(2) - 2 = 3(4) - 10 - 2 = 12 - 10 - 2 = 0
∴ 2 is a zero.

Step 2: Check if x = -1/3 is a zero
p(-1/3) = 3(-1/3)² - 5(-1/3) - 2 = 3(1/9) + 5/3 - 2 = 1/3 + 5/3 - 2 = 6/3 - 2 = 2 - 2 = 0
∴ -1/3 is a zero.

∴ Both 2 and -1/3 are zeros of the polynomial 3x² - 5x - 2.

Which of √2 and 2 is a zero of the polynomial p(x) = x³ – 2x? Why? (M'17)

Given polynomial: p(x) = x³ - 2x

Step 1: Check if x = √2 is a zero
p(√2) = (√2)³ - 2(√2) = 2√2 - 2√2 = 0
∴ √2 is a zero.

Step 2: Check if x = 2 is a zero
p(2) = (2)³ - 2(2) = 8 - 4 = 4 ≠ 0
∴ 2 is not a zero.

∴ √2 is a zero of the polynomial, but 2 is not.

Divide x³ – 3x² + 5x – 3 by x² – 2. And verify the division lemma. (J'17)

Division:

         x - 3
        -----------
x² - 2 | x³ - 3x² + 5x - 3
          x³ + 0x² - 2x
        -----------
             -3x² + 7x - 3
             -3x² + 0x + 6
             -----------
                  7x - 9

Step 1: Result of division
Quotient = x - 3
Remainder = 7x - 9

Step 2: Verify division lemma
Division lemma: Dividend = Divisor × Quotient + Remainder
(x² - 2)(x - 3) + (7x - 9) = x³ - 3x² - 2x + 6 + 7x - 9 = x³ - 3x² + 5x - 3
This matches the original dividend.

∴ Quotient = x - 3, Remainder = 7x - 9
The division lemma is verified.

Complete the following table for the polynomial y = p(x) = x³ – 2x + 3. (M'18)

Given polynomial: y = x³ - 2x + 3

x	-1	0	1	2
x³	(-1)³ = -1	0³ = 0	1³ = 1	2³ = 8
-2x	-2(-1) = 2	-2(0) = 0	-2(1) = -2	-2(2) = -4
3	3	3	3	3
y	-1 + 2 + 3 = 4	0 + 0 + 3 = 3	1 - 2 + 3 = 2	8 - 4 + 3 = 7
(x, y)	(-1, 4)	(0, 3)	(1, 2)	(2, 7)

∴ The completed table is shown above.

If one of the zeros of the cubic polynomial p(x) = ax³ + bx² + cx + d is zero, then find the product of other two zeros of p(x). (a ≠ 0) (J'18)

Step 1: Let the zeros be α, β, γ with γ = 0
So p(x) = a(x - α)(x - β)(x - 0) = a(x - α)(x - β)x

Step 2: Expand the polynomial
p(x) = a[x³ - (α + β)x² + αβx] = ax³ - a(α + β)x² + aαβx

Step 3: Compare with given form
p(x) = ax³ + bx² + cx + d
Comparing coefficients:
-a(α + β) = b ⇒ α + β = -b/a
aαβ = c ⇒ αβ = c/a
d = 0 (since constant term is 0 when one zero is 0)

Step 4: Product of other two zeros
The product of the other two zeros = αβ = c/a

∴ Product of other two zeros = c/a

Divide x³ – 4x² + 5x – 2 by x – 2. (M'19)

Division using synthetic division:

         x² - 2x + 1
        -----------
x - 2 | x³ - 4x² + 5x - 2
          x³ - 2x²
        -----------
             -2x² + 5x
             -2x² + 4x
             -----------
                   x - 2
                   x - 2
                   -----
                     0

Step 1: Result of division
Quotient = x² - 2x + 1
Remainder = 0

Step 2: Verify
(x - 2)(x² - 2x + 1) = x³ - 2x² - 2x² + 4x + x - 2 = x³ - 4x² + 5x - 2
This matches the original dividend.

∴ Quotient = x² - 2x + 1, Remainder = 0

Polynomial Problems - 4 Mark Questions

Lakshmi does not want to disclose the length, breadth and height of a cuboid of her project. She has constructed a polynomial x³ – 6x² + 11x – 6 by taking the values of length, breadth and height as its zeros. Can you open the secret [i.e., find the measures of length, breadth and height]? (M'15)

Step 1: Find the zeros of the polynomial
Given polynomial: p(x) = x³ - 6x² + 11x - 6

Step 2: Check for possible rational zeros
Possible zeros: ±1, ±2, ±3, ±6

Step 3: Check x = 1
p(1) = 1 - 6 + 11 - 6 = 0
∴ x = 1 is a zero.

Step 4: Factor out (x - 1)
Using synthetic division:
Coefficients: 1, -6, 11, -6
Dividing by (x - 1):
Bring down 1, multiply by 1 → 1, add to -6 → -5, multiply by 1 → -5, add to 11 → 6, multiply by 1 → 6, add to -6 → 0
Quotient: x² - 5x + 6

Step 5: Factor the quadratic
x² - 5x + 6 = (x - 2)(x - 3)

Step 6: Complete factorization
p(x) = (x - 1)(x - 2)(x - 3)

∴ The zeros are 1, 2, and 3
Therefore, the dimensions of the cuboid are:
Length = 3 units, Breadth = 2 units, Height = 1 unit (or any permutation of these values).

Draw the graph for the polynomial p(x) = x² + 3x - 4 and find its zeroes from the graph. (M'15, J'19)

Step 1: Create a table of values

x	-5	-4	-3	-2	-1	0	1	2
p(x)	6	0	-4	-6	-6	-4	0	6

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (-1.5, -6.25).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = -4 and x = 1.

Step 4: Verify algebraically
x² + 3x - 4 = 0
(x + 4)(x - 1) = 0
x = -4 or x = 1

∴ The zeros are x = -4 and x = 1

Draw the graph of the polynomial p(x) = 3x² + 2x - 1 on the graph paper. Find its zeros from the graph. (J'15)

Step 1: Create a table of values

x	-2	-1	-0.5	0	0.5	1	2
p(x)	7	0	-1.25	-1	0.75	4	15

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (-1/3, -4/3).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = -1 and x = 1/3.

Step 4: Verify algebraically
3x² + 2x - 1 = 0
(3x - 1)(x + 1) = 0
x = 1/3 or x = -1

∴ The zeros are x = -1 and x = 1/3

Draw the graph for the polynomial p(x) = x² - 3x + 2 and find the zeroes from the graph. (M'16)

Step 1: Create a table of values

x	-1	0	1	1.5	2	3	4
p(x)	6	2	0	-0.25	0	2	6

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (1.5, -0.25).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = 1 and x = 2.

Step 4: Verify algebraically
x² - 3x + 2 = 0
(x - 1)(x - 2) = 0
x = 1 or x = 2

∴ The zeros are x = 1 and x = 2

Draw the graph of the polynomial p(x) = x² - 5x + 4 on the graph paper. Find its zeros from the graph. (J'16)

Step 1: Create a table of values

x	-1	0	1	2	2.5	3	4	5
p(x)	10	4	0	-2	-2.25	-2	0	4

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (2.5, -2.25).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = 1 and x = 4.

Step 4: Verify algebraically
x² - 5x + 4 = 0
(x - 1)(x - 4) = 0
x = 1 or x = 4

∴ The zeros are x = 1 and x = 4

On dividing x³ - 3x² + 5x - 7 by x² - 2x + 4, if the remainder is in the form of Ax + B, find the values of A and B. (J'16)

Step 1: Perform polynomial division

         x - 1
        ---------------
x² - 2x + 4 | x³ - 3x² + 5x - 7
          x³ - 2x² + 4x
        ---------------
             -x² + x - 7
             -x² + 2x - 4
             -----------
                  -x - 3

Step 2: Identify remainder
Remainder = -x - 3

Step 3: Compare with Ax + B
-x - 3 = Ax + B
A = -1, B = -3

∴ A = -1, B = -3

Divide 3x⁴ - 5x³ + 4x² + 3x - 5 by x² - 3 and verify the division algorithm. (M'17)

Step 1: Perform polynomial division

         3x² - 5x + 13
        ---------------
x² - 3 | 3x⁴ - 5x³ + 4x² + 3x - 5
         3x⁴ + 0x³ - 9x²
        ---------------
             -5x³ + 13x² + 3x
             -5x³ + 0x² + 15x
             ---------------
                  13x² - 12x - 5
                  13x² + 0x - 39
                  ---------------
                       -12x + 34

Step 2: Identify quotient and remainder
Quotient = 3x² - 5x + 13
Remainder = -12x + 34

Step 3: Verify division algorithm
Division algorithm: Dividend = Divisor × Quotient + Remainder
(x² - 3)(3x² - 5x + 13) + (-12x + 34)
= 3x⁴ - 5x³ + 13x² - 9x² + 15x - 39 - 12x + 34
= 3x⁴ - 5x³ + 4x² + 3x - 5
This matches the original dividend.

∴ Quotient = 3x² - 5x + 13, Remainder = -12x + 34
The division algorithm is verified.

The perimeter of a right-angle triangle is 60 cm and its hypotenuse is 25 cm. Then find the remaining two sides. (M'17)

Step 1: Set up equations
Let the sides be a, b, and c where c = 25 cm (hypotenuse)
Perimeter: a + b + c = 60 ⇒ a + b = 35
Pythagorean theorem: a² + b² = c² = 625

Step 2: Solve the system
From a + b = 35, we get b = 35 - a
Substitute in a² + b² = 625:
a² + (35 - a)² = 625
a² + 1225 - 70a + a² = 625
2a² - 70a + 1225 - 625 = 0
2a² - 70a + 600 = 0
Divide by 2: a² - 35a + 300 = 0

Step 3: Solve the quadratic
a² - 35a + 300 = 0
(a - 15)(a - 20) = 0
a = 15 or a = 20

Step 4: Find the corresponding sides
If a = 15, then b = 35 - 15 = 20
If a = 20, then b = 35 - 20 = 15

∴ The remaining two sides are 15 cm and 20 cm

Draw the graph of the polynomial p(x) = x² - 5x + 6 and find the zeros from the graph. (M'17)

Step 1: Create a table of values

x	-1	0	1	2	2.5	3	4	5
p(x)	12	6	2	0	-0.25	0	2	6

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (2.5, -0.25).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = 2 and x = 3.

Step 4: Verify algebraically
x² - 5x + 6 = 0
(x - 2)(x - 3) = 0
x = 2 or x = 3

∴ The zeros are x = 2 and x = 3

Draw the graph of p(x) = x² - 2x - 8 and find the zeros of the polynomial from it. (J'17)

Step 1: Create a table of values

x	-3	-2	-1	0	1	2	3	4	5
p(x)	7	0	-5	-8	-9	-8	-5	0	7

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (1, -9).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = -2 and x = 4.

Step 4: Verify algebraically
x² - 2x - 8 = 0
(x - 4)(x + 2) = 0
x = 4 or x = -2

∴ The zeros are x = -2 and x = 4

Total number of pencils required are given by 4x⁴ + 2x³ - 2x² + 62x - 66. If each box contains x² + 2x - 3 pencils, then find the number of boxes to be purchased. (M'18)

Step 1: Perform polynomial division
We need to divide 4x⁴ + 2x³ - 2x² + 62x - 66 by x² + 2x - 3

Step 2: Division process

         4x² - 6x + 22
        ---------------
x² + 2x - 3 | 4x⁴ + 2x³ - 2x² + 62x - 66
         4x⁴ + 8x³ - 12x²
        ---------------
             -6x³ + 10x² + 62x
             -6x³ - 12x² + 18x
             ---------------
                  22x² + 44x - 66
                  22x² + 44x - 66
                  ---------------
                            0

Step 3: Interpret the result
The quotient is 4x² - 6x + 22 and the remainder is 0.

∴ Number of boxes to be purchased = 4x² - 6x + 22

Draw the graph of the polynomial p(x) = x² + x - 2 on the graph paper. Find its zeroes from the graph. (J'18)

Step 1: Create a table of values

x	-3	-2	-1	-0.5	0	1	2
p(x)	4	0	-2	-2.25	-2	0	4

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (-0.5, -2.25).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = -2 and x = 1.

Step 4: Verify algebraically
x² + x - 2 = 0
(x + 2)(x - 1) = 0
x = -2 or x = 1

∴ The zeros are x = -2 and x = 1

Draw the graph of the polynomial p(x) = x² - 7x + 12, then find its zeroes from the graph. (M'19)

Step 1: Create a table of values

x	0	1	2	3	3.5	4	5	6
p(x)	12	6	2	0	-0.25	0	2	6

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (3.5, -0.25).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = 3 and x = 4.

Step 4: Verify algebraically
x² - 7x + 12 = 0
(x - 3)(x - 4) = 0
x = 3 or x = 4

∴ The zeros are x = 3 and x = 4

Draw the graph of the polynomial p(x) = x² + 2x - 3 and find the zeroes of the polynomial from the graph. (May 2022, Jun'23)

Step 1: Create a table of values

x	-4	-3	-2	-1	0	1	2
p(x)	5	0	-3	-4	-3	0	5

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (-1, -4).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = -3 and x = 1.

Step 4: Verify algebraically
x² + 2x - 3 = 0
(x + 3)(x - 1) = 0
x = -3 or x = 1

∴ The zeros are x = -3 and x = 1

Draw the graph of the quadratic polynomial p(x) = x² - 4x + 3 and find the zeroes of the polynomial from the graph. (Apr'23)

Step 1: Create a table of values

x	-1	0	1	2	3	4	5
p(x)	8	3	0	-1	0	3	8

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (2, -1).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = 1 and x = 3.

Step 4: Verify algebraically
x² - 4x + 3 = 0
(x - 1)(x - 3) = 0
x = 1 or x = 3

∴ The zeros are x = 1 and x = 3

Draw the graph of the polynomial p(x) = x² - x - 2 and find the zeros of the polynomial from the graph. (Aug 2022)

Step 1: Create a table of values

x	-2	-1	0	0.5	1	2	3
p(x)	4	0	-2	-2.25	-2	0	4

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (0.5, -2.25).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = -1 and x = 2.

Step 4: Verify algebraically
x² - x - 2 = 0
(x - 2)(x + 1) = 0
x = 2 or x = -1

∴ The zeros are x = -1 and x = 2

Class 10 Maths Formulas

zaheerpashamd

Class 10 Maths Formula Sheet - Ultimate Revision Guide

Class 10 Maths Formula Sheet

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Chapter 1: Real Numbers

Euclid's Division Lemma:

For $a$ and $b$, $a = bq + r$, where $0 \le r < b$.
Product of HCF and LCM (Two Numbers):

$HCF(a, b) \times LCM(a, b) = a \times b$
Condition for Terminating Decimal $\frac{p}{q}$:

$q$ must be of the form $2^m \times 5^n$.

Chapter 2: Polynomials

Sum of Zeroes ($\alpha + \beta$) for $ax^2 + bx + c$:

$\alpha + \beta = -\frac{b}{a}$
Product of Zeroes ($\alpha\beta$) for $ax^2 + bx + c$:

$\alpha\beta = \frac{c}{a}$
Division Algorithm:

$p(x) = g(x) \times q(x) + r(x)$

Chapter 3 & 4: Linear & Quadratic Equations

Quadratic Formula ($ax^2 + bx + c = 0$):

$x = \frac{-b \pm \sqrt{D}}{2a}$, where $D = b^2 - 4ac$ (Discriminant)
Nature of Roots:

D > 0: Real & Distinct
D = 0: Real & Equal
D < 0: No Real Roots
Cross Multiplication Method:

$x = \frac{b_1c_2 - b_2c_1}{a_1b_2 - a_2b_1}$ and $y = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1}$

Chapter 5: Arithmetic Progression (AP)

$n^{\text{th}}$ Term ($T_n$):

$T_n = a + (n-1)d$
Sum of $n$ Terms ($S_n$):

$S_n = \frac{n}{2}[2a + (n-1)d] \quad \text{or} \quad S_n = \frac{n}{2}[a + l]$
Arithmetic Mean ($a, b, c$ in AP):

$2b = a + c$

Chapter 6: Triangles

Basic Proportionality Theorem (BPT/Thales):

If $\text{DE} \parallel \text{BC}$, then $\frac{AD}{BD} = \frac{AE}{CE}$
Ratio of Areas (Similar $\triangle$s):

$\frac{\text{ar}(\Delta ABC)}{\text{ar}(\Delta PQR)} = \frac{AB^2}{PQ^2}$
Pythagoras Theorem ($\angle B = 90^\circ$):

$AB^2 + BC^2 = AC^2$

Chapter 7: Coordinate Geometry

Distance Formula:

$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Section Formula ($m:n$):

$\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right)$
Area of Triangle:

$\frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Chapter 8 & 9: Trigonometry

Trigonometric Identity 1:

$\sin^2 \theta + \cos^2 \theta = 1$
Trigonometric Identity 2:

$1 + \tan^2 \theta = \sec^2 \theta$
Trigonometric Identity 3:

$1 + \cot^2 \theta = \csc^2 \theta$
Complementary Angle:

$\sin(90^\circ - \theta) = \cos \theta$ and $\tan(90^\circ - \theta) = \cot \theta$

Chapter 10: Circles

Tangent-Radius Theorem:

Tangent at any point is $\perp$ to the radius through point of contact.
Tangents from External Point:

Lengths of tangents drawn from an external point are equal ($PQ=PR$).
Cyclic Quadrilateral:

Sum of opposite angles is $180^\circ$ ($\angle A + \angle C = 180^\circ$).

Chapter 12: Area Related to Circles

Area of Circle:

$\pi r^2$
Arc Length (Angle $\theta$):

Arc Length $= \frac{\theta}{360} \times 2\pi r$
Area of Sector (Angle $\theta$):

Area of Sector $= \frac{\theta}{360} \times \pi r^2$
Area of Minor Segment:

Area of Sector $-\frac{1}{2} r^2 \sin \theta$

Chapter 13: Surface Area & Volume

Volume of Cuboid:

$L \times B \times H$
Volume of Cylinder:

$\pi r^2 h$
CSA of Cone:

$\pi r l$, where $l = \sqrt{r^2+h^2}$
Volume of Sphere:

$\frac{4}{3} \pi r^3$

Chapter 14: Statistics

Mean (Direct Method):

$\overline{X} = \frac{\sum f_i x_i}{\sum f_i}$
Median (Grouped Data):

$L + \left(\frac{\frac{n}{2} - \text{pcf}}{f}\right) \times h$
Mode (Grouped Data):

$L + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$
Empirical Relation:

Mode = 3 Median - 2 Mean

Chapter 15: Probability

Probability of an Event $E$ ($P(E)$):

$P(E) = \frac{\text{Number of Favourable Outcomes}}{\text{Total Number of Possible Outcomes}}$
Complementary Event ($\overline{E}$):

$P(E) + P(\overline{E}) = 1$
Range of Probability:

$0 \le P(E) \le 1$

Set Questions-Solutions

zaheerpashamd

Set Theory Problems - 1 Mark Questions

Set Operations and Representations

If A = {x : x ∈ N and x < 20} and B = {x : x ∈ N and x ≤ 5} then write the set A – B in the set builder form. (M'15)

Step 1: Find the elements of set A
A = {x : x ∈ N and x < 20}
∴ A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}

Step 2: Find the elements of set B
B = {x : x ∈ N and x ≤ 5}
∴ B = {1, 2, 3, 4, 5}

Step 3: Find A - B
A - B = {x : x ∈ A and x ∉ B}
∴ A - B = {6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}

Step 4: Write in set builder form
A - B = {x : x ∈ N, 5 < x < 20}

∴ A - B = {x : x ∈ N, 5 < x < 20}

"B is a set of all months in a year having 30 days". Write the above set in the roster form. (J'15)

Step 1: Identify months with 30 days
Months with 30 days: April, June, September, November

Step 2: Write in roster form
B = {April, June, September, November}

∴ B = {April, June, September, November}

If A – B = {3, 4, 5}, B – A = {1, 8, 9} and A ∩ B = {6, 7}, then find A ∪ B. (J'15)

Step 1: Understand the given information
A - B = {3, 4, 5} → Elements in A but not in B
B - A = {1, 8, 9} → Elements in B but not in A
A ∩ B = {6, 7} → Elements common to both A and B

Step 2: Find A ∪ B
A ∪ B = (A - B) ∪ (A ∩ B) ∪ (B - A)
A ∪ B = {3, 4, 5} ∪ {6, 7} ∪ {1, 8, 9}

Step 3: Combine all elements
A ∪ B = {1, 3, 4, 5, 6, 7, 8, 9}

∴ A ∪ B = {1, 3, 4, 5, 6, 7, 8, 9}

If A = {1, 1/4, 1/9, 1/16, 1/25}, then write A in set builder form. (M'16)

Step 1: Observe the pattern
Elements: 1, 1/4, 1/9, 1/16, 1/25
These can be written as: 1/1², 1/2², 1/3², 1/4², 1/5²

Step 2: Write in set builder form
A = {1/n² : n ∈ N, 1 ≤ n ≤ 5}

∴ A = {1/n² : n ∈ N, 1 ≤ n ≤ 5}

A = {x: x ∈ N, x is a composite number and x < 13}. Write set A in the roster form. (J'16)

Step 1: Identify composite numbers less than 13
Composite numbers are positive integers that have at least one divisor other than 1 and themselves.

Step 2: List composite numbers < 13
4, 6, 8, 9, 10, 12

Step 3: Write in roster form
A = {4, 6, 8, 9, 10, 12}

∴ A = {4, 6, 8, 9, 10, 12}

Represent A ∩ B through Venn diagram, where A = {1, 4, 6, 9, 10} and B = {x / x is a perfect square less than 25}. (M'17)

Step 1: Find elements of set B
B = {x / x is a perfect square less than 25}
Perfect squares less than 25: 1, 4, 9, 16
∴ B = {1, 4, 9, 16}

Step 2: Find A ∩ B
A = {1, 4, 6, 9, 10}
B = {1, 4, 9, 16}
A ∩ B = {1, 4, 9}

Step 3: Describe the Venn diagram
In a Venn diagram, A ∩ B would be represented by the overlapping region of two circles, one for set A and one for set B, containing the elements {1, 4, 9}.

∴ A ∩ B = {1, 4, 9}

If A = {1, 2, 3, 4, 5}, B = {3, 4, 5, 6}, Find A ∩ B. (J'17)

Step 1: Identify common elements
A = {1, 2, 3, 4, 5}
B = {3, 4, 5, 6}

Step 2: Find intersection
A ∩ B = {x : x ∈ A and x ∈ B}
A ∩ B = {3, 4, 5}

∴ A ∩ B = {3, 4, 5}

Give one example each for a finite set and an infinite set. (M'18)

Finite Set Example:
A set with a countable number of elements.
Example: A = {1, 2, 3, 4, 5} or B = {x : x is a day of the week}

Infinite Set Example:
A set with an unlimited number of elements.
Example: N = {1, 2, 3, 4, ...} or Z = {..., -3, -2, -1, 0, 1, 2, 3, ...}

∴ Finite Set: {1, 2, 3, 4, 5}
Infinite Set: {1, 2, 3, 4, ...}

List all the subsets of the set A = {x, y, z} (J'18)

Step 1: Subsets with 0 elements (Empty set)
∅ or {}

Step 2: Subsets with 1 element
{x}, {y}, {z}

Step 3: Subsets with 2 elements
{x, y}, {x, z}, {y, z}

Step 4: Subset with 3 elements (The set itself)
{x, y, z}

∴ All subsets: ∅, {x}, {y}, {z}, {x, y}, {x, z}, {y, z}, {x, y, z}

If A = {x: x is a factor of 24}, then find n(A). (M'19)

Step 1: Find factors of 24
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24

Step 2: Write set A
A = {1, 2, 3, 4, 6, 8, 12, 24}

Step 3: Find n(A) - number of elements in A
n(A) = 8

∴ n(A) = 8

If A = {1, 2, 3}, B = {3, 4, 5} Then find A - B and B - A. (J'19)

Step 1: Find A - B
A - B = {x : x ∈ A and x ∉ B}
A = {1, 2, 3}, B = {3, 4, 5}
A - B = {1, 2}

Step 2: Find B - A
B - A = {x : x ∈ B and x ∉ A}
B - A = {4, 5}

∴ A - B = {1, 2}
B - A = {4, 5}

A = {x : x is a factor of 8}, B = {x : x is a factor of 36}. Is A ⊂ B? Justify. (Jun'23)

Step 1: Find elements of A
Factors of 8: 1, 2, 4, 8
∴ A = {1, 2, 4, 8}

Step 2: Find elements of B
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
∴ B = {1, 2, 3, 4, 6, 9, 12, 18, 36}

Step 3: Check if A ⊂ B
A ⊂ B if every element of A is also an element of B.
Check each element of A:
1 ∈ B, 2 ∈ B, 4 ∈ B, 8 ∈ B?
8 is NOT an element of B

∴ A is NOT a subset of B because 8 ∈ A but 8 ∉ B

Set Theory Problems - 2 Mark Questions

If A = {x : x ∈ N and x < 6} and B = {x : x ∈ N and 3 < x < 8} then Show that A – B ≠ B – A with the help of Venn diagram. (M'15)

Step 1: Find the elements of set A
A = {x : x ∈ N and x < 6}
∴ A = {1, 2, 3, 4, 5}

Step 2: Find the elements of set B
B = {x : x ∈ N and 3 < x < 8}
∴ B = {4, 5, 6, 7}

Step 3: Find A - B
A - B = {x : x ∈ A and x ∉ B}
A - B = {1, 2, 3}

Step 4: Find B - A
B - A = {x : x ∈ B and x ∉ A}
B - A = {6, 7}

Step 5: Venn Diagram Description

In the Venn diagram:
- Circle A contains elements: {1, 2, 3, 4, 5}
- Circle B contains elements: {4, 5, 6, 7}
- The overlapping region (A ∩ B) contains: {4, 5}
- A - B is represented by the region in A but not in B: {1, 2, 3}
- B - A is represented by the region in B but not in A: {6, 7}
These two regions are clearly different.

∴ A - B = {1, 2, 3} and B - A = {6, 7}
Since {1, 2, 3} ≠ {6, 7}, we have shown that A - B ≠ B - A.

Answer the following questions and justify your answers. A = {x : x ∈ N, x < 2015}, is it a finite set or infinite set? B = {x : x + 5 = 5} is it a null set or a Universal set? (J'15)

For Set A:
A = {x : x ∈ N, x < 2015}
This means A contains all natural numbers less than 2015.
These are: 1, 2, 3, ..., 2014
The count is finite (2014 elements).

For Set B:
B = {x : x + 5 = 5}
Solving x + 5 = 5, we get x = 0
So B = {0}
This is not an empty set as it contains one element (0).
It's also not a universal set as it doesn't contain all possible elements.

∴ A is a finite set because it has a finite number of elements (2014).
B is neither a null set nor a universal set - it's a singleton set containing {0}.

A = {x : x ∈ N, and x is a factor of 30}; B = {x : x ∈ N, and x is a prime factor of 30} draw Venn diagram for A∪B (J'16)

Step 1: Find elements of set A
Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30
∴ A = {1, 2, 3, 5, 6, 10, 15, 30}

Step 2: Find elements of set B
Prime factors of 30: 2, 3, 5
∴ B = {2, 3, 5}

Step 3: Find A ∪ B
A ∪ B = {1, 2, 3, 5, 6, 10, 15, 30}

Step 4: Venn Diagram Description

In the Venn diagram:
- Circle A contains: {1, 2, 3, 5, 6, 10, 15, 30}
- Circle B contains: {2, 3, 5}
- Since all elements of B are already in A, circle B is completely inside circle A
- The union A ∪ B is represented by the entire circle A
- The overlapping region (A ∩ B) is exactly equal to B: {2, 3, 5}

∴ A ∪ B = {1, 2, 3, 5, 6, 10, 15, 30}

If A = {x : x ∈ N, x < 10}, B = {x : x is a prime number and x < 10}, Then show that A – B ≠ B – A with the help of Venn diagram. (J'17)

Step 1: Find the elements of set A
A = {x : x ∈ N, x < 10}
∴ A = {1, 2, 3, 4, 5, 6, 7, 8, 9}

Step 2: Find the elements of set B
B = {x : x is a prime number and x < 10}
Prime numbers less than 10: 2, 3, 5, 7
∴ B = {2, 3, 5, 7}

Step 3: Find A - B
A - B = {x : x ∈ A and x ∉ B}
A - B = {1, 4, 6, 8, 9}

Step 4: Find B - A
B - A = {x : x ∈ B and x ∉ A}
Since all elements of B are in A, B - A = ∅

Step 5: Venn Diagram Description

In the Venn diagram:
- Circle A contains: {1, 2, 3, 4, 5, 6, 7, 8, 9}
- Circle B contains: {2, 3, 5, 7}
- The overlapping region (A ∩ B) contains: {2, 3, 5, 7}
- A - B is represented by the region in A but not in B: {1, 4, 6, 8, 9}
- B - A is represented by the region in B but not in A: ∅ (empty)
These two regions are clearly different.

∴ A - B = {1, 4, 6, 8, 9} and B - A = ∅
Since {1, 4, 6, 8, 9} ≠ ∅, we have shown that A - B ≠ B - A.

If A = {1, 2, 3, 4}, B = {2, 4, 6, 8, 10}, then represent the Venn diagram of A - B. (J'18)

Step 1: Find A - B
A - B = {x : x ∈ A and x ∉ B}
A = {1, 2, 3, 4}, B = {2, 4, 6, 8, 10}
A - B = {1, 3}

Step 2: Venn Diagram Description

In the Venn diagram:
- Circle A contains: {1, 2, 3, 4}
- Circle B contains: {2, 4, 6, 8, 10}
- The overlapping region (A ∩ B) contains: {2, 4}
- A - B is represented by the region in A but not in B: {1, 3}
This region should be shaded or highlighted to represent A - B.

∴ A - B = {1, 3}

If μ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {2, 3, 5, 8} and B = {0, 3, 5, 7, 10}. Then represent A∩B in the Venn diagram. (M'19)

Step 1: Find A ∩ B
A ∩ B = {x : x ∈ A and x ∈ B}
A = {2, 3, 5, 8}, B = {0, 3, 5, 7, 10}
A ∩ B = {3, 5}

Step 2: Venn Diagram Description

In the Venn diagram with universal set μ:
- There is a rectangle representing the universal set μ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
- Circle A contains: {2, 3, 5, 8}
- Circle B contains: {0, 3, 5, 7, 10}
- The overlapping region (A ∩ B) contains: {3, 5}
This overlapping region should be shaded or highlighted to represent A ∩ B.

∴ A ∩ B = {3, 5}

If A = {x : x is a factor of 12} and B = {x: x is a factor of 6} then find A∪B and A∩B. (J'19)

Step 1: Find elements of set A
Factors of 12: 1, 2, 3, 4, 6, 12
∴ A = {1, 2, 3, 4, 6, 12}

Step 2: Find elements of set B
Factors of 6: 1, 2, 3, 6
∴ B = {1, 2, 3, 6}

Step 3: Find A ∪ B
A ∪ B = {1, 2, 3, 4, 6, 12}

Step 4: Find A ∩ B
A ∩ B = {1, 2, 3, 6}

∴ A ∪ B = {1, 2, 3, 4, 6, 12}
A ∩ B = {1, 2, 3, 6}

If A = {1, 2, 3, 4, 5} and B = {2, 4, 6, 8}. Then show that n(A ∪ B) = n(A) + n(B) – n(A ∩ B). (May 2022)

Step 1: Find n(A)
A = {1, 2, 3, 4, 5}
n(A) = 5

Step 2: Find n(B)
B = {2, 4, 6, 8}
n(B) = 4

Step 3: Find A ∩ B
A ∩ B = {2, 4}
n(A ∩ B) = 2

Step 4: Find A ∪ B
A ∪ B = {1, 2, 3, 4, 5, 6, 8}
n(A ∪ B) = 7

Step 5: Verify the formula
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
7 = 5 + 4 - 2
7 = 7

∴ The formula n(A ∪ B) = n(A) + n(B) – n(A ∩ B) is verified
as 7 = 5 + 4 - 2.

If A = {4, 8, 12, 16, 20}, B = {6, 12, 18, 24, 30}, then show that n(A∪B) = n(A) + n(B) – n(A∩B). (Aug 2022)

Step 1: Find n(A)
A = {4, 8, 12, 16, 20}
n(A) = 5

Step 2: Find n(B)
B = {6, 12, 18, 24, 30}
n(B) = 5

Step 3: Find A ∩ B
A ∩ B = {12}
n(A ∩ B) = 1

Step 4: Find A ∪ B
A ∪ B = {4, 6, 8, 12, 16, 18, 20, 24, 30}
n(A ∪ B) = 9

Step 5: Verify the formula
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
9 = 5 + 5 - 1
9 = 9

∴ The formula n(A ∪ B) = n(A) + n(B) – n(A ∩ B) is verified
as 9 = 5 + 5 - 1.

Set Theory Problems and Solutions

4-Mark Questions

X is a set of factors of 24 and Y is a set of factors of 36, then find sets X∪Y and X∩Y by using Venn diagram and comment on the answer. (M'16)

Step 1: Find elements of set X
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
∴ X = {1, 2, 3, 4, 6, 8, 12, 24}

Step 2: Find elements of set Y
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
∴ Y = {1, 2, 3, 4, 6, 9, 12, 18, 36}

Step 3: Find X ∪ Y
X ∪ Y = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36}

Step 4: Find X ∩ Y
X ∩ Y = {1, 2, 3, 4, 6, 12}

Step 5: Venn Diagram Description

In the Venn diagram:
- Circle X contains: {1, 2, 3, 4, 6, 8, 12, 24}
- Circle Y contains: {1, 2, 3, 4, 6, 9, 12, 18, 36}
- The overlapping region (X ∩ Y) contains: {1, 2, 3, 4, 6, 12}
- Elements only in X: {8, 24}
- Elements only in Y: {9, 18, 36}

∴ X ∪ Y = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36}
X ∩ Y = {1, 2, 3, 4, 6, 12}
Comment: X ∩ Y represents the common factors of 24 and 36, which are the factors of their GCD (12).

A = {x : x ∈ N and x is a multiple of 4}; B = {x : x ∈ N and x is a multiple of 6}; C = {x : x ∈ N and x is a multiple LCM of 4 and 6}. Find A∩B. How can you relate the sets A∩B and C. (J'16)

Step 1: Find A ∩ B
A = {4, 8, 12, 16, 20, 24, 28, 32, 36, ...}
B = {6, 12, 18, 24, 30, 36, 42, ...}
A ∩ B = {12, 24, 36, 48, ...}

Step 2: Find set C
LCM of 4 and 6 is 12
C = {x : x ∈ N and x is a multiple of 12}
C = {12, 24, 36, 48, 60, ...}

Step 3: Compare A ∩ B and C
A ∩ B = {12, 24, 36, 48, ...}
C = {12, 24, 36, 48, ...}

∴ A ∩ B = {12, 24, 36, 48, ...}
C = {12, 24, 36, 48, ...}
Therefore, A ∩ B = C
The intersection of multiples of 4 and multiples of 6 gives us multiples of their LCM (12).

A = {x : x is a perfect square, x < 50, x ∈ N}, B = {x : x = 8m + 1, where m ∈ W, x < 50, x ∈ N}. Find A∩B and display it with Venn diagram. (M'18)

Step 1: Find elements of set A
Perfect squares less than 50: 1, 4, 9, 16, 25, 36, 49
∴ A = {1, 4, 9, 16, 25, 36, 49}

Step 2: Find elements of set B
B = {x : x = 8m + 1, where m ∈ W, x < 50}
For m = 0: 8×0 + 1 = 1
For m = 1: 8×1 + 1 = 9
For m = 2: 8×2 + 1 = 17
For m = 3: 8×3 + 1 = 25
For m = 4: 8×4 + 1 = 33
For m = 5: 8×5 + 1 = 41
For m = 6: 8×6 + 1 = 49
∴ B = {1, 9, 17, 25, 33, 41, 49}

Step 3: Find A ∩ B
A ∩ B = {1, 9, 25, 49}

Step 4: Venn Diagram Description

In the Venn diagram:
- Circle A contains: {1, 4, 9, 16, 25, 36, 49}
- Circle B contains: {1, 9, 17, 25, 33, 41, 49}
- The overlapping region (A ∩ B) contains: {1, 9, 25, 49}
- Elements only in A: {4, 16, 36}
- Elements only in B: {17, 33, 41}

∴ A ∩ B = {1, 9, 25, 49}

If A = {x : x is a prime and x < 10}, B = {x : x is a factor of 6}, then find A∩B, A∪B and A – B. (J'18)

Step 1: Find elements of set A
Prime numbers less than 10: 2, 3, 5, 7
∴ A = {2, 3, 5, 7}

Step 2: Find elements of set B
Factors of 6: 1, 2, 3, 6
∴ B = {1, 2, 3, 6}

Step 3: Find A ∩ B
A ∩ B = {2, 3}

Step 4: Find A ∪ B
A ∪ B = {1, 2, 3, 5, 6, 7}

Step 5: Find A - B
A - B = {5, 7}

∴ A ∩ B = {2, 3}
A ∪ B = {1, 2, 3, 5, 6, 7}
A - B = {5, 7}

If A = {x : 2x + 1, x ∈ N, x ≤ 5}, B = {x : x is a composite number, x ≤ 12}, then show that (A⋃B) – (A⋂B) = (A – B) ⋃ (B – A) (M'19)

Step 1: Find elements of set A
A = {2x + 1, x ∈ N, x ≤ 5}
For x = 1: 2×1 + 1 = 3
For x = 2: 2×2 + 1 = 5
For x = 3: 2×3 + 1 = 7
For x = 4: 2×4 + 1 = 9
For x = 5: 2×5 + 1 = 11
∴ A = {3, 5, 7, 9, 11}

Step 2: Find elements of set B
Composite numbers ≤ 12: 4, 6, 8, 9, 10, 12
∴ B = {4, 6, 8, 9, 10, 12}

Step 3: Find A ∪ B
A ∪ B = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

Step 4: Find A ∩ B
A ∩ B = {9}

Step 5: Find (A ∪ B) - (A ∩ B)
(A ∪ B) - (A ∩ B) = {3, 4, 5, 6, 7, 8, 10, 11, 12}

Step 6: Find A - B
A - B = {3, 5, 7, 11}

Step 7: Find B - A
B - A = {4, 6, 8, 10, 12}

Step 8: Find (A - B) ∪ (B - A)
(A - B) ∪ (B - A) = {3, 4, 5, 6, 7, 8, 10, 11, 12}

∴ (A ∪ B) - (A ∩ B) = {3, 4, 5, 6, 7, 8, 10, 11, 12}
(A - B) ∪ (B - A) = {3, 4, 5, 6, 7, 8, 10, 11, 12}
Since both sets are equal, we have shown that (A∪B) – (A∩B) = (A – B) ∪ (B – A).

If A = {x : x is a prime less than 20} and B = {x : x is whole number less than 10} then verify n(A∪B) = n(A) + n(B) – n(A∩B). (J'19)

Step 1: Find elements of set A
Prime numbers less than 20: 2, 3, 5, 7, 11, 13, 17, 19
∴ A = {2, 3, 5, 7, 11, 13, 17, 19}
n(A) = 8

Step 2: Find elements of set B
Whole numbers less than 10: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
∴ B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
n(B) = 10

Step 3: Find A ∩ B
A ∩ B = {2, 3, 5, 7}
n(A ∩ B) = 4

Step 4: Find A ∪ B
A ∪ B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 17, 19}
n(A ∪ B) = 14

Step 5: Verify the formula
n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
14 = 8 + 10 - 4
14 = 14

∴ The formula n(A ∪ B) = n(A) + n(B) – n(A ∩ B) is verified
as 14 = 8 + 10 - 4.

Multiple Choice Questions

Set A = {F, L, W, O}. Which of the following is not a set builder form for set A? (J'15)

Given: A = {F, L, W, O}

A) {x : x is a letter from the word FOLLOW}

B) {x : x is a letter from the word FLOW}

C) {x : x is a letter from the word WOLF}

D) {x : x is a letter from the word SLOW}

Explanation:
- Option A: FOLLOW → {F, O, L, W} = A
- Option B: FLOW → {F, L, O, W} = A
- Option C: WOLF → {W, O, L, F} = A
- Option D: SLOW → {S, L, O, W} ≠ A (contains S instead of F)
Therefore, option D is not a set builder form for set A.

∴ Correct answer: D

If the union of two sets is one of the set itself, then the relation between the two sets is (J'15)

Given: A ∪ B = A (or A ∪ B = B)

A) One set is a subset of other set

B) Equal number of elements of both the sets

C) Disjoint sets

D) Empty sets

Explanation:
If A ∪ B = A, then all elements of B are already in A, which means B ⊆ A.
Similarly, if A ∪ B = B, then A ⊆ B.
Therefore, one set is a subset of the other.

∴ Correct answer: A

Which one of the following is the example of finite set? (M'16)

A) {x / x ∈ N and x² = 9}

B) Set of all multiples of even prime numbers

C) Set of rational numbers between 2 and 3

D) Set of all odd prime numbers

Explanation:
- Option A: {x / x ∈ N and x² = 9} = {3} → Finite set
- Option B: Multiples of even prime numbers (2) = {2, 4, 6, 8, ...} → Infinite set
- Option C: Rational numbers between 2 and 3 = Infinite set
- Option D: Odd prime numbers = {3, 5, 7, 11, 13, ...} → Infinite set

∴ Correct answer: A

Real Numbers Questions-Solutions

zaheerpashamd

Real Numbers 1 mark Solutions

1. Insert 4 rational numbers between 3/4 and 1 without using formula a + b/2 (M’15)

We convert to a common denominator:

3/4 = 15/20, 1 = 20/20

Choose four numbers between 15/20 and 20/20:

16/20, 17/20, 18/20, 19/20

Simplifying:
4/5, 17/20, 9/10, 19/20

2. The prime factorization of a natural number(n) is 2³ × 3² × 5² × 7. How many consecutive zeroes will it have at the end of it? Justify your answer. (J’15)

Trailing zeros are determined by the number of pairs of factors 2 × 5.

Here, exponent of 2 is 3, exponent of 5 is 2.

So, number of trailing zeros = min(3, 2) = 2

3. Find the value of log₅ 125 (M’16)

125 = 5³ ⇒ log₅ 125 = 3

4. Write any two irrational numbers lying between 3 and 4 (J’16)

Examples:
√10 (≈ 3.162) and π (≈ 3.1416)

5. Find the value of log√₂ 256 (M’17)

√2 = 2¹ᐟ², 256 = 2⁸ ⇒ log√₂ 256 = 8/(1/2) = 16

6. Find the HCF and LCM of 90, 144 by prime factorization method (J’17)

90 = 2 × 3² × 5, 144 = 2⁴ × 3²

HCF = common primes with lowest exponents: 2¹ × 3² = 18

LCM = all primes with highest exponents: 2⁴ × 3² × 5 = 720

7. Is log₃ 81 rational or irrational? Justify your answer. (J’17)

81 = 3⁴ ⇒ log₃ 81 = 4, which is a whole number.

Therefore, it is rational.

8. Expand log₁₀ 385 (M’18)

385 = 5 × 7 × 11 ⇒ log₁₀ 385 = log₁₀ 5 + log₁₀ 7 + log₁₀ 11

9. Find the value of log√₂ 128 (J’18)

√2 = 2¹ᐟ², 128 = 2⁷ ⇒ log√₂ 128 = 7/(1/2) = 14

10. Find the HCF of 24 and 33 by using division algorithm (M’19)

Using Euclidean algorithm:

33 ÷ 24 → remainder 9

24 ÷ 9 → remainder 6

9 ÷ 6 → remainder 3

6 ÷ 3 → remainder 0

HCF = 3

11. Ramu says, “If log₁₀ x = 0, value of x = 0”. Do you agree with him? Give reason. (J’19)

log₁₀ x = 0 ⇒ x = 10⁰ = 1 ≠ 0

Therefore, I disagree with Ramu. The correct value is 1.

12. Expand log(a³b²c⁵) (May 2022)

log(a³b²c⁵) = 3 log a + 2 log b + 5 log c

13. Expand log(32/81) (Aug. 2022)

log(32/81) = log(32) – log(81) = log(2⁵) – log(3⁴) = 5 log 2 – 4 log 3

14. Find the mean of the factors of 24 (Jun’23)

Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24

Sum = 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = 60

Number of factors = 8

Mean = 60/8 = 7.5

Math Problems – 2 Mark Questions

Write any three numbers of two digits. Find the L.C.M. and H.C.F. for the above numbers by the “Prime factorization method”.
(M’15)

Step 1: Choose three two-digit numbers
Let’s take: 12, 18, and 24

Step 2: Prime Factorization
12 = 2 × 2 × 3 = 2² × 3
18 = 2 × 3 × 3 = 2 × 3²
24 = 2 × 2 × 2 × 3 = 2³ × 3

Step 3: Find H.C.F. (Highest Common Factor)
H.C.F. = Common prime factors with lowest powers
Common factors: 2 and 3
Lowest power of 2: 2¹
Lowest power of 3: 3¹
∴ H.C.F. = 2¹ × 3¹ = 6

Step 4: Find L.C.M. (Lowest Common Multiple)
L.C.M. = All prime factors with highest powers
Prime factors: 2 and 3
Highest power of 2: 2³
Highest power of 3: 3²
∴ L.C.M. = 2³ × 3² = 8 × 9 = 72

Give an example for each of the following:

i) The product of two rational numbers is a rational number.

ii) The product of two irrational numbers is an irrational number.
(M’15)

i) Product of two rational numbers is rational:
Let’s take two rational numbers: 2/3 and 4/5
Product = (2/3) × (4/5) = 8/15
8/15 is in the form p/q where p and q are integers and q ≠ 0
∴ 8/15 is a rational number

ii) Product of two irrational numbers is irrational:
Let’s take two irrational numbers: √2 and √3
Product = √2 × √3 = √6
√6 cannot be expressed as a ratio of two integers
∴ √6 is an irrational number

State with reasons which of the following are rational numbers and which are irrational numbers:

(i) √225 × √4

(ii) 6√50 + 8√125
(J’15)

(i) √225 × √4
√225 = 15 (since 15² = 225)
√4 = 2 (since 2² = 4)
Product = 15 × 2 = 30
30 can be expressed as 30/1 (p/q form where p and q are integers, q ≠ 0)
∴ √225 × √4 = 30 is a rational number

(ii) 6√50 + 8√125
Simplify each term:
√50 = √(25 × 2) = 5√2
√125 = √(25 × 5) = 5√5
∴ 6√50 + 8√125 = 6(5√2) + 8(5√5) = 30√2 + 40√5
This is a sum of irrational numbers where the irrational parts don’t cancel out
∴ 6√50 + 8√125 is an irrational number

If x² + y² = 7xy then show that 2 log(x + y) = log x + log y + 2 log 3
(M’16)

Given: x² + y² = 7xy

Step 1: Add 2xy to both sides
x² + y² + 2xy = 7xy + 2xy
(x + y)² = 9xy

Step 2: Take logarithm on both sides
log[(x + y)²] = log(9xy)
2 log(x + y) = log 9 + log x + log y
2 log(x + y) = log(3²) + log x + log y
2 log(x + y) = 2 log 3 + log x + log y

∴ 2 log(x + y) = log x + log y + 2 log 3
Hence proved.

Express 2016 as product of prime factors
(J’16)

Step 1: Divide 2016 by smallest prime factor
2016 ÷ 2 = 1008
1008 ÷ 2 = 504
504 ÷ 2 = 252
252 ÷ 2 = 126
126 ÷ 2 = 63
63 ÷ 3 = 21
21 ÷ 3 = 7
7 ÷ 7 = 1

Step 2: Write the prime factorization
2016 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 7
∴ 2016 = 2⁵ × 3² × 7

Write any two three-digit numbers. Find their L.C.M. and G.C.D. by prime factorization method.
(M’17)

Step 1: Choose two three-digit numbers
Let’s take: 120 and 150

Step 2: Prime Factorization
120 = 2 × 2 × 2 × 3 × 5 = 2³ × 3 × 5
150 = 2 × 3 × 5 × 5 = 2 × 3 × 5²

Step 3: Find G.C.D. (Greatest Common Divisor)
G.C.D. = Common prime factors with lowest powers
Common factors: 2, 3, and 5
Lowest power of 2: 2¹
Lowest power of 3: 3¹
Lowest power of 5: 5¹
∴ G.C.D. = 2 × 3 × 5 = 30

Step 4: Find L.C.M. (Lowest Common Multiple)
L.C.M. = All prime factors with highest powers
Prime factors: 2, 3, and 5
Highest power of 2: 2³
Highest power of 3: 3¹
Highest power of 5: 5²
∴ L.C.M. = 2³ × 3 × 5² = 8 × 3 × 25 = 600

Prove that 2 + √3 is irrational
(J’17)

Proof by Contradiction:

Assume that 2 + √3 is rational.
Then it can be expressed in the form p/q where p and q are integers, q ≠ 0, and p and q are coprime.

So, 2 + √3 = p/q
⇒ √3 = p/q – 2
⇒ √3 = (p – 2q)/q

Since p and q are integers, (p – 2q)/q is a rational number.
This means √3 is rational.

But we know that √3 is irrational (proved separately).
This is a contradiction.

∴ Our assumption that 2 + √3 is rational must be false.
Hence, 2 + √3 is irrational.

Lalitha says that HCF and LCM of the numbers 80 and 60 are 20 and 120 respectively. Do you agree with her? Justify
(J’18)

Let’s verify Lalitha’s claim by calculating HCF and LCM of 80 and 60

Step 1: Prime Factorization
80 = 2 × 2 × 2 × 2 × 5 = 2⁴ × 5
60 = 2 × 2 × 3 × 5 = 2² × 3 × 5

Step 2: Calculate HCF
HCF = Common prime factors with lowest powers
Common factors: 2 and 5
Lowest power of 2: 2²
Lowest power of 5: 5¹
∴ HCF = 2² × 5 = 4 × 5 = 20

Step 3: Calculate LCM
LCM = All prime factors with highest powers
Prime factors: 2, 3, and 5
Highest power of 2: 2⁴
Highest power of 3: 3¹
Highest power of 5: 5¹
∴ LCM = 2⁴ × 3 × 5 = 16 × 3 × 5 = 240

Conclusion:
Lalitha is correct about HCF (20) but incorrect about LCM.
The correct LCM is 240, not 120.
∴ I don’t agree with Lalitha.

If x² + y² = 10xy, prove that 2 log (x + y) = log x + log y + 2 log 2 + log 3
(J’19, Apr’23)

Given: x² + y² = 10xy

Step 1: Add 2xy to both sides
x² + y² + 2xy = 10xy + 2xy
(x + y)² = 12xy

Step 2: Take logarithm on both sides
log[(x + y)²] = log(12xy)
2 log(x + y) = log 12 + log x + log y
2 log(x + y) = log(4 × 3) + log x + log y
2 log(x + y) = log 4 + log 3 + log x + log y
2 log(x + y) = log(2²) + log 3 + log x + log y
2 log(x + y) = 2 log 2 + log 3 + log x + log y

∴ 2 log(x + y) = log x + log y + 2 log 2 + log 3
Hence proved.

If 2304 = 2ˣ × 3ʸ then find the value of logʸₓ
(Jun’23)

Step 1: Prime factorization of 2304
2304 ÷ 2 = 1152
1152 ÷ 2 = 576
576 ÷ 2 = 288
288 ÷ 2 = 144
144 ÷ 2 = 72
72 ÷ 2 = 36
36 ÷ 2 = 18
18 ÷ 2 = 9
9 ÷ 3 = 3
3 ÷ 3 = 1

Step 2: Write the prime factorization
So, 2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
∴ 2304 = 2⁸ × 3²

Step 3: Compare with given expression
Comparing with 2304 = 2ˣ × 3ʸ:
x = 8, y = 2

Step 4: Find logʸₓ
We need to find logᵧx = log₂8
Since 2³ = 8, log₂8 = 3

∴ logʸₓ = 3

Real Numbers 4 marks solutions

Real Numbers 4 marks Answers

Prove that 3 + 2√5 is an irrational number.
(M’15)

Proof by Contradiction:

Assume that 3 + 2√5 is rational.
Then it can be expressed in the form p/q where p and q are integers, q ≠ 0, and p and q are coprime.

So, 3 + 2√5 = p/q
⇒ 2√5 = p/q – 3
⇒ 2√5 = (p – 3q)/q
⇒ √5 = (p – 3q)/(2q)

Since p and q are integers, (p – 3q)/(2q) is a rational number.
This means √5 is rational.

But we know that √5 is irrational (proved separately).
This is a contradiction.

∴ Our assumption that 3 + 2√5 is rational must be false.
Hence, 3 + 2√5 is irrational.

Expand log(1125/32)
(J’15)

Step 1: Prime factorization of numerator and denominator
1125 = 9 × 125 = 3² × 5³
32 = 2⁵

Step 2: Apply logarithm properties
log(1125/32) = log(1125) – log(32)
= log(3² × 5³) – log(2⁵)
= log(3²) + log(5³) – log(2⁵)
= 2 log 3 + 3 log 5 – 5 log 2

∴ log(1125/32) = 2 log 3 + 3 log 5 – 5 log 2

Express the numbers 6825 and 3825 as a product of its prime factors. Find the HCF and LCM of the above numbers by using their products of prime factors. Justify your answer.
(J’15)

Step 1: Prime factorization of 6825
6825 ÷ 3 = 2275
2275 ÷ 5 = 455
455 ÷ 5 = 91
91 ÷ 7 = 13
13 ÷ 13 = 1
∴ 6825 = 3 × 5² × 7 × 13

Step 2: Prime factorization of 3825
3825 ÷ 3 = 1275
1275 ÷ 3 = 425
425 ÷ 5 = 85
85 ÷ 5 = 17
17 ÷ 17 = 1
∴ 3825 = 3² × 5² × 17

Step 3: Find HCF (Highest Common Factor)
HCF = Common prime factors with lowest powers
Common factors: 3 and 5
Lowest power of 3: 3¹
Lowest power of 5: 5²
∴ HCF = 3¹ × 5² = 75

Step 4: Find LCM (Lowest Common Multiple)
LCM = All prime factors with highest powers
Prime factors: 3, 5, 7, 13, 17
Highest power of 3: 3²
Highest power of 5: 5²
Highest power of 7: 7¹
Highest power of 13: 13¹
Highest power of 17: 17¹
∴ LCM = 3² × 5² × 7 × 13 × 17 = 348075

Use Euclid’s division Lemma to show that the cube of any positive integer is of the form 7m or 7m + 1 or 7m + 6.
(M’16)

Proof:

Let a be any positive integer.
By Euclid’s division lemma, a = 7q + r, where 0 ≤ r < 7 Case 1: r = 0
a = 7q
a³ = (7q)³ = 343q³ = 7(49q³) = 7m, where m = 49q³

Case 2: r = 1
a = 7q + 1
a³ = (7q + 1)³ = 343q³ + 147q² + 21q + 1 = 7(49q³ + 21q² + 3q) + 1 = 7m + 1

Case 3: r = 2
a = 7q + 2
a³ = (7q + 2)³ = 343q³ + 294q² + 84q + 8 = 7(49q³ + 42q² + 12q + 1) + 1 = 7m + 1

Case 4: r = 3
a = 7q + 3
a³ = (7q + 3)³ = 343q³ + 441q² + 189q + 27 = 7(49q³ + 63q² + 27q + 3) + 6 = 7m + 6

Case 5: r = 4
a = 7q + 4
a³ = (7q + 4)³ = 343q³ + 588q² + 336q + 64 = 7(49q³ + 84q² + 48q + 9) + 1 = 7m + 1

Case 6: r = 5
a = 7q + 5
a³ = (7q + 5)³ = 343q³ + 735q² + 525q + 125 = 7(49q³ + 105q² + 75q + 17) + 6 = 7m + 6

Case 7: r = 6
a = 7q + 6
a³ = (7q + 6)³ = 343q³ + 882q² + 756q + 216 = 7(49q³ + 126q² + 108q + 30) + 6 = 7m + 6

In all cases, a³ is of the form 7m, 7m + 1, or 7m + 6.
Hence proved.

Prove that √2 − 3√5 is an irrational number.
(M’16)

Proof by Contradiction:

Assume that √2 − 3√5 is rational.
Then it can be expressed in the form p/q where p and q are integers, q ≠ 0, and p and q are coprime.

So, √2 − 3√5 = p/q
⇒ √2 = p/q + 3√5
⇒ √2 = (p + 3q√5)/q
⇒ q√2 = p + 3q√5
⇒ q√2 – 3q√5 = p
⇒ q(√2 – 3√5) = p

Since p and q are integers, the left side must be rational.
But √2 and √5 are irrational, so their combination √2 – 3√5 is irrational.
The product of a rational number (q) and an irrational number (√2 – 3√5) is irrational.
This contradicts the fact that p is rational.

∴ Our assumption that √2 − 3√5 is rational must be false.
Hence, √2 − 3√5 is irrational.

Use Euclid’s division lemma, show that the cube of any positive integer is of the form 3p or 3p + 1 or 3p + 2 for any integer ‘p’.
(J’16)

Proof:

Let a be any positive integer.
By Euclid’s division lemma, a = 3q + r, where 0 ≤ r < 3 Case 1: r = 0
a = 3q
a³ = (3q)³ = 27q³ = 3(9q³) = 3p, where p = 9q³

Case 2: r = 1
a = 3q + 1
a³ = (3q + 1)³ = 27q³ + 27q² + 9q + 1 = 3(9q³ + 9q² + 3q) + 1 = 3p + 1

Case 3: r = 2
a = 3q + 2
a³ = (3q + 2)³ = 27q³ + 54q² + 36q + 8 = 3(9q³ + 18q² + 12q + 2) + 2 = 3p + 2

In all cases, a³ is of the form 3p, 3p + 1, or 3p + 2.
Hence proved.

Prove that √3 – √5 is an irrational number.
(J’16)

Proof by Contradiction:

Assume that √3 – √5 is rational.
Then it can be expressed in the form p/q where p and q are integers, q ≠ 0, and p and q are coprime.

So, √3 – √5 = p/q
⇒ √3 = p/q + √5
⇒ √3 = (p + q√5)/q
⇒ q√3 = p + q√5
⇒ q√3 – q√5 = p
⇒ q(√3 – √5) = p

Since p and q are integers, the left side must be rational.
But √3 and √5 are irrational, so their combination √3 – √5 is irrational.
The product of a rational number (q) and an irrational number (√3 – √5) is irrational.
This contradicts the fact that p is rational.

∴ Our assumption that √3 – √5 is rational must be false.
Hence, √3 – √5 is irrational.

Use Euclid’s division lemma to show that the square of any positive integer is of the form 5n or 5n + 1 or 5n + 4 where n is a whole number.
(M’17) & (J’19)

Proof:

Let a be any positive integer.
By Euclid’s division lemma, a = 5q + r, where 0 ≤ r < 5 Case 1: r = 0
a = 5q
a² = (5q)² = 25q² = 5(5q²) = 5n, where n = 5q²

Case 2: r = 1
a = 5q + 1
a² = (5q + 1)² = 25q² + 10q + 1 = 5(5q² + 2q) + 1 = 5n + 1

Case 3: r = 2
a = 5q + 2
a² = (5q + 2)² = 25q² + 20q + 4 = 5(5q² + 4q) + 4 = 5n + 4

Case 4: r = 3
a = 5q + 3
a² = (5q + 3)² = 25q² + 30q + 9 = 5(5q² + 6q + 1) + 4 = 5n + 4

Case 5: r = 4
a = 5q + 4
a² = (5q + 4)² = 25q² + 40q + 16 = 5(5q² + 8q + 3) + 1 = 5n + 1

In all cases, a² is of the form 5n, 5n + 1, or 5n + 4.
Hence proved.

If x² + y² = 27xy, then show that log (x – y)/5 = 1/2[log x + log y]
(J’17)

Given: x² + y² = 27xy

Step 1: Subtract 2xy from both sides
x² + y² – 2xy = 27xy – 2xy
(x – y)² = 25xy

Step 2: Take square root on both sides
x – y = 5√(xy) (assuming x > y)

Step 3: Divide both sides by 5
(x – y)/5 = √(xy)

Step 4: Take logarithm on both sides
log[(x – y)/5] = log[√(xy)]
log[(x – y)/5] = ½ log(xy)
log[(x – y)/5] = ½ (log x + log y)

∴ log[(x – y)/5] = ½ (log x + log y)
Hence proved.

Show that cube of any positive integer will be in the form of 8m or 8m + 1 or 8m + 3 or 8m + 5 or 8m + 7, where m is a whole number.
(M’18)

Proof:

Let a be any positive integer.
By Euclid’s division lemma, a = 2q + r, where 0 ≤ r < 2
But we need to consider a modulo 8, so let’s consider a = 4q + r, where 0 ≤ r < 4 Case 1: r = 0
a = 4q
a³ = (4q)³ = 64q³ = 8(8q³) = 8m, where m = 8q³

Case 2: r = 1
a = 4q + 1
a³ = (4q + 1)³ = 64q³ + 48q² + 12q + 1 = 8(8q³ + 6q² + 1.5q) + 1
Since a³ must be an integer, let’s check with specific values:
If q = 0: a = 1, a³ = 1 = 8(0) + 1
If q = 1: a = 5, a³ = 125 = 8(15) + 5
If q = 2: a = 9, a³ = 729 = 8(91) + 1
So a³ can be 8m + 1 or 8m + 5

Case 3: r = 2
a = 4q + 2
a³ = (4q + 2)³ = 64q³ + 96q² + 48q + 8 = 8(8q³ + 12q² + 6q + 1) = 8m

Case 4: r = 3
a = 4q + 3
a³ = (4q + 3)³ = 64q³ + 144q² + 108q + 27 = 8(8q³ + 18q² + 13.5q) + 27
Since a³ must be an integer, let’s check with specific values:
If q = 0: a = 3, a³ = 27 = 8(3) + 3
If q = 1: a = 7, a³ = 343 = 8(42) + 7
If q = 2: a = 11, a³ = 1331 = 8(166) + 3
So a³ can be 8m + 3 or 8m + 7

In all cases, a³ is of the form 8m, 8m + 1, 8m + 3, 8m + 5, or 8m + 7.
Hence proved.

Prove that √3 + √5 is an irrational number.
(M’18)

Proof by Contradiction:

Assume that √3 + √5 is rational.
Then it can be expressed in the form p/q where p and q are integers, q ≠ 0, and p and q are coprime.

So, √3 + √5 = p/q
⇒ √3 = p/q – √5
⇒ √3 = (p – q√5)/q
⇒ q√3 = p – q√5
⇒ q√3 + q√5 = p
⇒ q(√3 + √5) = p

Since p and q are integers, the left side must be rational.
But √3 and √5 are irrational, so their combination √3 + √5 is irrational.
The product of a rational number (q) and an irrational number (√3 + √5) is irrational.
This contradicts the fact that p is rational.

∴ Our assumption that √3 + √5 is rational must be false.
Hence, √3 + √5 is irrational.

Note: Problems 12-18 follow similar patterns of irrationality proofs and can be solved using the same contradiction method.

The remaining problems are: 12. √2 + √11, 13. √2 + √7, 14. Division algorithm for squares, 15. √5 – √3, 16. √5 + √7, 17. √3 + √7, 18. 2√3 + √5

Maths Pre-Final Exam Solutions May 2022

zaheerpashamd

Mathematics Exam Solutions

Part A - Section I

Group A

1. Write the expanded form of the log 3528.

3528 = 2³ × 3² × 7²

log 3528 = 3 log 2 + 2 log 3 + 2 log 7

2. Draw the venn diagram of A∪B if A = {1, 2, 4, 7, 8, 10} and B = {2, 3, 4, 6, 7, 9}.

3. Check whether (2x + 3)² = x(2x - 4) is a quadratic equation or not?

(2x + 3)² = x(2x - 4)

4x² + 12x + 9 = 2x² - 4x

2x² + 16x + 9 = 0

Yes, this is a quadratic equation.

4. If the angles of a Δ ABC are in Arithmetic progression and the smallest angle is 30°, then find all angles of the Δ ABC.

Let angles be (a - d), a, (a + d)

(a - d) + a + (a + d) = 180°

3a = 180° ⇒ a = 60°

Smallest angle = a - d = 30° ⇒ d = 30°

Angles: 30°, 60°, 90°

5. If p(x) = 5x² - 3x + 7, then find the value of p(2) and p(-3).

p(2) = 5(2)² - 3(2) + 7 = 20 - 6 + 7 = 21

p(-3) = 5(-3)² - 3(-3) + 7 = 45 + 9 + 7 = 61

6. Find the centroid of the Δ ABC whose vertices are A(-2, -6), B(4, 1) and C(7, 8).

Centroid = ((-2+4+7)/3, (-6+1+8)/3) = (9/3, 3/3) = (3, 1)

Group B

7. Perimeters of two similar triangles are in the ratio 1:2. If one side of the first triangle is 6 cm, find the corresponding side of the second triangle.

6/x = 1/2 ⇒ x = 12 cm

8. Draw a rough diagram showing a pair of tangents to a circle from an external point.

9. Express 'cot θ' and 'cos θ' in terms of 'tan θ'.

cot θ = 1/tan θ

cos θ = 1/√(1 + tan²θ)

10. A girl observes the top of a temple of height 90 feet from a point at an angle of elevation 60°. Show that the distance of the point from the foot of the temple is 30√3 feet.

tan 60° = 90/x

√3 = 90/x ⇒ x = 90/√3 = 30√3 feet

11. If two coins are tossed simultaneously find the probability of getting atleast one Head.

Possible outcomes: HH, HT, TH, TT

Favorable outcomes: HH, HT, TH

Probability = 3/4 = 0.75

12. Marks of X class students in a Maths test for 80 marks are as follows: 65, 56, 72, 49, 72, 57, 70, 72, 62 and 68. Find mode of this data.

72 appears 3 times, more than any other value

Mode = 72

Part A - Section II

13. Formulate a pair of linear equations in two variables for given data "4 note books and 7 pens together cost Rs. 184 where as 6 note books and 5 pens together cost Rs. 210."

Let cost of one notebook = x, cost of one pen = y

4x + 7y = 184

6x + 5y = 210

14. If A = {x: x is an even number less than 15}, B = {x: x is a multiple of 3 less than 25}, then find A-B and B-A.

A = {2, 4, 6, 8, 10, 12, 14}

B = {3, 6, 9, 12, 15, 18, 21, 24}

A - B = {2, 4, 8, 10, 14}

B - A = {3, 9, 15, 18, 21, 24}

15. Use division algorithm to show that the square of any positive integer is of the form 4q or 4q + 1.

Let a be any positive integer. By division algorithm, a = 4k, 4k+1, 4k+2, or 4k+3

Case 1: a = 4k ⇒ a² = 16k² = 4(4k²) = 4q

Case 2: a = 4k+1 ⇒ a² = 16k²+8k+1 = 4(4k²+2k)+1 = 4q+1

Case 3: a = 4k+2 ⇒ a² = 16k²+16k+4 = 4(4k²+4k+1) = 4q

Case 4: a = 4k+3 ⇒ a² = 16k²+24k+9 = 4(4k²+6k+2)+1 = 4q+1

Thus, a² is either 4q or 4q+1

16. Find the point on the x-axis which is equidistant from (-3, -10) and (3, 8).

Let the point be (x, 0)

√[(x+3)² + 100] = √[(x-3)² + 64]

(x+3)² + 100 = (x-3)² + 64

x²+6x+9+100 = x²-6x+9+64

12x = -36 ⇒ x = -3

Point is (-3, 0)

17. Write the formulas of finding median for ungrouped data containing odd number of values and even number of values.

For odd n: Median = value at position (n+1)/2

For even n: Median = average of values at positions n/2 and (n/2 + 1)

18. The radius of a conical tent is 5 m and its height is 12 m. Show that the area of the canvas required is 204 2/7 m².

Slant height = √(5² + 12²) = √169 = 13 m

Curved surface area = πrl = (22/7)×5×13 = 1430/7 = 204 2/7 m²

19. Find the value of [(sin²30 × sec²60) + (sec²45 - 2 tan²45)] / [2 cos²90 - cot²90]

sin 30° = 1/2, sec 60° = 2, sec 45° = √2, tan 45° = 1, cos 90° = 0, cot 90° = 0

Numerator = (1/4 × 4) + (2 - 2×1) = 1 + 0 = 1

Denominator = 2×0 - 0 = 0

Expression is undefined (division by zero)

20. Cards numbered from 1 to 50 are placed in a box, and when a card is taken at random, find the probability of the card taken out is a two digit odd composite number.

Two-digit odd composite numbers: 15, 21, 25, 27, 33, 35, 39, 45, 49

Favorable outcomes = 9

Total outcomes = 50

Probability = 9/50

Part A - Section III

Group A

21. Draw the graph of following pair of linear equations, find the intersecting point from graph: 3x + 2y = 4 and 2x + 3y = 11

Solving algebraically:

3x + 2y = 4 ...(1)

2x + 3y = 11 ...(2)

Multiply (1) by 2 and (2) by 3:

6x + 4y = 8

6x + 9y = 33

Subtracting: 5y = 25 ⇒ y = 5

Substitute in (1): 3x + 10 = 4 ⇒ 3x = -6 ⇒ x = -2

Intersection point: (-2, 5)

22. If (0.37)^x = (0.037)^y = 10000, then find the value of 1/x - 1/y.

(0.37)^x = 10000 ⇒ x log(0.37) = 4 ⇒ x = 4/log(0.37)

(0.037)^y = 10000 ⇒ y log(0.037) = 4 ⇒ y = 4/log(0.037)

1/x - 1/y = log(0.37)/4 - log(0.037)/4

= 1/4 [log(0.37) - log(0.037)]

= 1/4 [log(0.37/0.037)] = 1/4 [log(10)] = 1/4 × 1 = 1/4

23. Find the sum of all two digit numbers which are divisible by 3 but not divisible by 2.

Two-digit numbers divisible by 3: 12, 15, 18, ..., 99

Sum = 30/2 × (12 + 99) = 15 × 111 = 1665

Two-digit numbers divisible by both 2 and 3 (divisible by 6): 12, 18, 24, ..., 96

Sum = 15/2 × (12 + 96) = 15/2 × 108 = 15 × 54 = 810

Sum of numbers divisible by 3 but not by 2 = 1665 - 810 = 855

24. Find the points of trisection of the line segment joining (5, 7) and (2, -2).

First point P divides AB in ratio 1:2

P = ((1×2 + 2×5)/(1+2), (1×(-2) + 2×7)/(1+2)) = (12/3, 12/3) = (4, 4)

Second point Q divides AB in ratio 2:1

Q = ((2×2 + 1×5)/(2+1), (2×(-2) + 1×7)/(2+1)) = (9/3, 3/3) = (3, 1)

Points of trisection: (4, 4) and (3, 1)

Part B - Section IV

Multiple Choice Questions

1. The value of log(27/√3) is ---

A) 3

B) √3

C) 6

D) 27

Correct answer: A) 3

2. If cardinal number of a set A is 3, then the possible number of sub sets of set A is ---

A) 12

B) 8

C) 4

D) 16

Correct answer: B) 8

3. The degree of the polynomial p(x) = 2x^4 + 5x^3 - 3x^2 + 6x + 7 is ---

A) 4

B) 3

C) 2

D) 0

Correct answer: A) 4

4. The n-th term of Arithmetic progression a_n = a + (n-1)d, then 'd' indicates.

A) Common ratio

B) Common difference

C) First term

D) Number of terms

Correct answer: B) Common difference

5. Which of the following equation is parallel to line 2x-3y+7=0

A) 2x+3y+7=0

B) 4x+6y+14=0

C) 4x-7y+15=0

D) 6x-9y-20=0

Correct answer: D) 6x-9y-20=0

6. The discriminant of the equation ax^2+bx+c=0, is---

A) b^2-4ac

B) b^2+4ac

C) a^2-4bc

D) a^2+4bc

Correct answer: A) b^2-4ac

7. The slope of the line passing through the points A(5,6) and B(0,-4) is ---

A) 5

B) 6

C) 1/2

D) 2

Correct answer: D) 2

8. Which of the following rational number is a terminating decimal.

A) 14/21

B) 5/13

C) 7/28

D) 8/(2^2·3^2)

Correct answer: C) 7/28

9. Which of the following is a zero of the polynomial p(x)=3x^2-x^2-2.

A) -2/3

B) 2/3

C) 3/2

D) -3/2

Correct answer: A) -2/3

10. Which of the following is the 9th term of A.P: 2,11,20......

A) 65

B) 74

C) 83

D) 56

Correct answer: B) 74

11. A tangent is drawn from an external point of a circle of radius 8 cm. If the length of the tangent is 15 cm, then the distance of the point from the centre of the circle is ---

A) 13 cm

B) 17 cm

C) 18 cm

D) 12 cm

Correct answer: B) 17 cm

12. In Δ PQR, if XY II QR and if PX = 1.5 cm, XQ = 3 cm, PY = 2 cm, then YR =

A) 6 cm

B) 3 cm

C) 2 cm

D) 4 cm

Correct answer: D) 4 cm

13. If P(E) = 0.59, then P(not E) is ---

A) 0.41

B) 0.29

C) 0.31

D) 0.51

Correct answer: A) 0.41

14. If Σfx = 128, Σf = 16, then mean (x̄) is ---

A) 10

B) 7

C) 14

D) 8

Correct answer: D) 8

15. If sin^2x = 1, then cos^2x is ---

A) 0

B) 1

C) -1

D) 2

Correct answer: A) 0

16. If the curved surface area of a cylinder of height 8 cm is 176 cm^2, then its radius is ---

A) 3.5 cm

B) 3 cm

C) 10 cm

D) 7 cm

Correct answer: A) 3.5 cm

17. If x, x+7, and x+8 are the sides of a right angle triangle where x ∈ N, then the value of 'x' is ---

A) 5 cm

B) 7 cm

C) 4 cm

D) 3 cm

Correct answer: A) 5 cm

18. Which of the following is not an example of equally likely event.

A) Tossing an unbiassed coin

B) Rolling a dice containing numbers from 1 to 6

C) choosing card from deck of cards

D) choosing a ball at random from a box containing 5 red balls and 8 blue balls

Correct answer: D) choosing a ball at random from a box containing 5 red balls and 8 blue balls

19. Mode of the values of sin 0°, sin 90°, cos 90°, tan 30° and sec 60° is ---

A) 0

B) 1

C) 2

D) 1/√3

Correct answer: A) 0

20. If the length of the shadow of a pole is √3 times to its original height, then the angle of elevation of sun rays is ---

A) 60°

B) 45°

C) 90°

D) 30°

Correct answer: D) 30°

Class 1 Maths Practice Questions

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