Maths-
General
Easy

Question

If F(x) = ax + b, where a and b are integers, f(2) = 6 and F(3) = 7 then find the values of a and b.

Hint:

Put the given values of x and functions and find the required value.

The correct answer is: a = 1


    We have given the values of f(x)

    F(x) = ax + b

    F(2) = 6

    F(3) = 7

    When x = 2

    a(2) + b =6

    2a + b = 6      ------(i)

    when x = 3

    a(3) + b = 7

    3a + b = 7       -------(ii)
    Subtract equation (ii) from (i)

    3a – 2a + b – b = 7 – 6

    a = 1
    Put this value in equation (i)

    2(1) + b = 6

    b = 6 – 2

    b = 4
    Therefore the value of a = 1

    Related Questions to study

    General
    Maths-

    Find the GCF (GCD) of the given pair of monomials.
    4 x cubed space straight & space 9 y to the power of 5

    Solution:
    Hint:
    • The highest number that divides exactly into two or more numbers is known as GCF.
    • An algebraic expression consisting only one term is called monomial.
    Explanation:
    • We have been given a monomial in the question
    • We have to find the GCF of the given pair of monomials
    Step 1 of 1:
    We have given two monomials 4 x cubed comma space 9 y to the power of 5.
    4 x cubed equals 4 cross times x cross times x cross times x cross times 1
    9 y to the power of 5 equals 9 cross times y cross times y cross times y cross times y cross times y cross times 1
    As we can see there is only one common factor in this two monomials
    So, The highest factor in these two monomial is 1.
    So, The GCF 4x3 , 9y5 is 1.

    Find the GCF (GCD) of the given pair of monomials.
    4 x cubed space straight & space 9 y to the power of 5

    Maths-General
    Solution:
    Hint:
    • The highest number that divides exactly into two or more numbers is known as GCF.
    • An algebraic expression consisting only one term is called monomial.
    Explanation:
    • We have been given a monomial in the question
    • We have to find the GCF of the given pair of monomials
    Step 1 of 1:
    We have given two monomials 4 x cubed comma space 9 y to the power of 5.
    4 x cubed equals 4 cross times x cross times x cross times x cross times 1
    9 y to the power of 5 equals 9 cross times y cross times y cross times y cross times y cross times y cross times 1
    As we can see there is only one common factor in this two monomials
    So, The highest factor in these two monomial is 1.
    So, The GCF 4x3 , 9y5 is 1.
    General
    Maths-

    If the relation R: A → B, where A = {2, 3, 4} and B = {3, 5} is defined by R = {(x, y):x < y, x ϵ A, y ϵ B}, then find R-1

    We have given the Relation R: A → B

    A = {2, 3, 4}

    B = {3, 5}
    We have given the relation in Set- builder form ,

    R = {(x, y):x < y, x ϵ A, y ϵ B}
    We will first find the Cartesian product of set A and B

    A X B = {(2,3),(2,5),(3,3),(3,5)(4,3),(4,5)}

    R = {(2,3),(2,5),(3,5),(4,5)}

    Therefore, R-1 = (A X B) – R

    R-1 = {(3,3),(4,3)}

    R-1 = {(x, y):x >= y, x ϵ A, y ϵ B}

    If the relation R: A → B, where A = {2, 3, 4} and B = {3, 5} is defined by R = {(x, y):x < y, x ϵ A, y ϵ B}, then find R-1

    Maths-General
    We have given the Relation R: A → B

    A = {2, 3, 4}

    B = {3, 5}
    We have given the relation in Set- builder form ,

    R = {(x, y):x < y, x ϵ A, y ϵ B}
    We will first find the Cartesian product of set A and B

    A X B = {(2,3),(2,5),(3,3),(3,5)(4,3),(4,5)}

    R = {(2,3),(2,5),(3,5),(4,5)}

    Therefore, R-1 = (A X B) – R

    R-1 = {(3,3),(4,3)}

    R-1 = {(x, y):x >= y, x ϵ A, y ϵ B}

    General
    Maths-

    The shape of the window shown is a regular polygon. The window has a boundary made of silver wire. How many meters of silver wire are needed for this border if two sides are given to be left parenthesis x plus 7 right parenthesis text  and  end text left parenthesis 2 x minus 3 right parenthesis

    Solution:
    Hint:
    • A regular polygon is a polygon with congruent sides and equal angles and are symmetrically placed about a common center.
    Explanation:
    • We have been given a figure of window in the question which is in the shape of a regular polygon, the boundary of the window is made up of silver wire.
    • We have also been given the two sides of it that is - (𝑥 + 7) 𝑎𝑛𝑑 (2𝑥 − 3)
    • We have to find out how many meters of sliver wire are needed for the border.
    Step 1 of 1:
    We have given a window is of shape of regular polygon
    Two of its side is represented by x + 7; 2x - 3
    Since, It is regular polygon, all the sides length will be equal.
    So,
    X + 7 = 2x - 3
    x = 7 + 3
    x = 10
    So,
    The side length is
    = x + 7
    = 7 + 10
    = 17
    Now the perimeter of octagon will be
    = 8 × side
    = 8 × 17
    = 136

    The shape of the window shown is a regular polygon. The window has a boundary made of silver wire. How many meters of silver wire are needed for this border if two sides are given to be left parenthesis x plus 7 right parenthesis text  and  end text left parenthesis 2 x minus 3 right parenthesis

    Maths-General
    Solution:
    Hint:
    • A regular polygon is a polygon with congruent sides and equal angles and are symmetrically placed about a common center.
    Explanation:
    • We have been given a figure of window in the question which is in the shape of a regular polygon, the boundary of the window is made up of silver wire.
    • We have also been given the two sides of it that is - (𝑥 + 7) 𝑎𝑛𝑑 (2𝑥 − 3)
    • We have to find out how many meters of sliver wire are needed for the border.
    Step 1 of 1:
    We have given a window is of shape of regular polygon
    Two of its side is represented by x + 7; 2x - 3
    Since, It is regular polygon, all the sides length will be equal.
    So,
    X + 7 = 2x - 3
    x = 7 + 3
    x = 10
    So,
    The side length is
    = x + 7
    = 7 + 10
    = 17
    Now the perimeter of octagon will be
    = 8 × side
    = 8 × 17
    = 136
    parallel
    General
    Maths-

    If f(x) = (25 – x2)1/2 then f( square root of 5) =…..

    We have given the function of x as

    f(x) = (25 – x2)1/2
    We have to find out the value of f(square root of 5)
    f(square root of 5) = (25 – (square root of 5)2)1/2
    = (25 – 5) ½
    = 20 1/2
    = 2square root of 5

    If f(x) = (25 – x2)1/2 then f( square root of 5) =…..

    Maths-General
    We have given the function of x as

    f(x) = (25 – x2)1/2
    We have to find out the value of f(square root of 5)
    f(square root of 5) = (25 – (square root of 5)2)1/2
    = (25 – 5) ½
    = 20 1/2
    = 2square root of 5

    General
    Maths-

    Given x = {(2, 7), (3, 9), (5, 13), (0, 3)} be a function from Z to Z defined by f(x) = ax + b for some integral a and b. What are the values of a and b?

    We have given a function from Z to Z
    Given x = {(2, 7), (3, 9), (5, 13), (0, 3)}
    And also we have given that

    f(x) = ax + b
    We have to find the value of a and b .
    First of all if the f is a function then its points will satisfy f(x) = ax + b

    f(2) = 7

    f(3) = 9

    f(5) = 13

    f(0) = 3

    i) (2,7)

    f(2) = a (2) + b

    7 = 2a + b

    ii) (3,9)

    f(3) = a(3) + b

    9 = 3a + b
    Subtract equation (i) from (ii)

    3a – 2a + b – b = 9 – 7

    a = 2
    Putting this value in equation (i)

    7 = 2(2) + b

    b = 7 – 4

    b = 3
    Therefore, value of a = 2 and b = 3.

    Given x = {(2, 7), (3, 9), (5, 13), (0, 3)} be a function from Z to Z defined by f(x) = ax + b for some integral a and b. What are the values of a and b?

    Maths-General
    We have given a function from Z to Z
    Given x = {(2, 7), (3, 9), (5, 13), (0, 3)}
    And also we have given that

    f(x) = ax + b
    We have to find the value of a and b .
    First of all if the f is a function then its points will satisfy f(x) = ax + b

    f(2) = 7

    f(3) = 9

    f(5) = 13

    f(0) = 3

    i) (2,7)

    f(2) = a (2) + b

    7 = 2a + b

    ii) (3,9)

    f(3) = a(3) + b

    9 = 3a + b
    Subtract equation (i) from (ii)

    3a – 2a + b – b = 9 – 7

    a = 2
    Putting this value in equation (i)

    7 = 2(2) + b

    b = 7 – 4

    b = 3
    Therefore, value of a = 2 and b = 3.

    General
    Maths-

    Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y and z are distinct elements.

    We have given the two sets A and B

    And n( A)= 3

    n(B) = 2

    Therefore, n(AXB) = 3X2 = 6

    Since, {(x,1),(y, 2),(z,1)} are the elements of A×B.

    It follows that the elements of

    set A={x, y, z} 

    and B={1,2}

    Hence,

    A×B={(x,1),(x,2),(y,1),(y,2),(z,1),(z,2)}

    Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y and z are distinct elements.

    Maths-General
    We have given the two sets A and B

    And n( A)= 3

    n(B) = 2

    Therefore, n(AXB) = 3X2 = 6

    Since, {(x,1),(y, 2),(z,1)} are the elements of A×B.

    It follows that the elements of

    set A={x, y, z} 

    and B={1,2}

    Hence,

    A×B={(x,1),(x,2),(y,1),(y,2),(z,1),(z,2)}

    parallel
    General
    Maths-

     Fundamental Algebra:
    1. Add:
    (ii) –7y2 + 8y + 2 ; 3y2 – 7y + 1

    • Hint:
    ○        The concept used in this question is the addition of polynomials.
    ○        Arrange the term in the decreasing order of their power.
    ○        To add polynomials simply add any like terms together.
    ○        Like terms are terms whose variables and exponents are the same.
    • Step by step explanation:
    ○        Given:
               Expression:   –7y2 + 8y + 2 ;
     3y2 – 7y + 1 ;
    ○        Step 1:
    ○        Simplify the expression.
    (–7y2 + 8y + 2  ) + ( 3y2 – 7y + 1 )
    rightwards double arrow (–7y2 + 8y + 2  ) + ( 3y2 – 7y + 1 )
    ○        Step 2:
    ○        Group the like terms.
    rightwards double arrow ( –7y2. + 3y2 ) + ( 8y - 7y ) + ( 2 + 1 )

    ○        Step 3:
    ○        Add like terms,
    rightwards double arrow – 4y2 + y + 3
    • Final Answer:
                  – 4y2 + y + 3

     Fundamental Algebra:
    1. Add:
    (ii) –7y2 + 8y + 2 ; 3y2 – 7y + 1

    Maths-General
    • Hint:
    ○        The concept used in this question is the addition of polynomials.
    ○        Arrange the term in the decreasing order of their power.
    ○        To add polynomials simply add any like terms together.
    ○        Like terms are terms whose variables and exponents are the same.
    • Step by step explanation:
    ○        Given:
               Expression:   –7y2 + 8y + 2 ;
     3y2 – 7y + 1 ;
    ○        Step 1:
    ○        Simplify the expression.
    (–7y2 + 8y + 2  ) + ( 3y2 – 7y + 1 )
    rightwards double arrow (–7y2 + 8y + 2  ) + ( 3y2 – 7y + 1 )
    ○        Step 2:
    ○        Group the like terms.
    rightwards double arrow ( –7y2. + 3y2 ) + ( 8y - 7y ) + ( 2 + 1 )

    ○        Step 3:
    ○        Add like terms,
    rightwards double arrow – 4y2 + y + 3
    • Final Answer:
                  – 4y2 + y + 3
    General
    Maths-

    A regular polygon has 9 diagonals. Find the number of sides and classify it based on the number of sides.

    Solution:
    Hint:
    • A regular polygon is a polygon with congruent sides and equal angles and are symmetrically placed about a common center
    Explanation:
    • We have been given in the question information about a regular polygon that has 9 diagonals which means it is a hexagon.
    •  We have to find the number of sides and classify it based on the number of sides.
    Step 1 of 1:
    We know that the number of diagonal in n - side polygon  equals fraction numerator n left parenthesis n minus 3 right parenthesis over denominator 2 end fraction
    Here, the number of diagonals is 9
    So,
    fraction numerator n left parenthesis n minus 3 right parenthesis over denominator 2 end fraction equals 9
    n squared minus 3 n equals 18
    n squared minus 3 n minus 18 equals 0
    (n - 6) (n + 3) = 0
    n = 6
    The number of sides of the given polygon are 6
    And it is Hexagon.

    A regular polygon has 9 diagonals. Find the number of sides and classify it based on the number of sides.

    Maths-General
    Solution:
    Hint:
    • A regular polygon is a polygon with congruent sides and equal angles and are symmetrically placed about a common center
    Explanation:
    • We have been given in the question information about a regular polygon that has 9 diagonals which means it is a hexagon.
    •  We have to find the number of sides and classify it based on the number of sides.
    Step 1 of 1:
    We know that the number of diagonal in n - side polygon  equals fraction numerator n left parenthesis n minus 3 right parenthesis over denominator 2 end fraction
    Here, the number of diagonals is 9
    So,
    fraction numerator n left parenthesis n minus 3 right parenthesis over denominator 2 end fraction equals 9
    n squared minus 3 n equals 18
    n squared minus 3 n minus 18 equals 0
    (n - 6) (n + 3) = 0
    n = 6
    The number of sides of the given polygon are 6
    And it is Hexagon.
    General
    Maths-

    Find the GCF (GCD) of the given pair of monomials.
    8 a squared space straight & space 28 a to the power of 5

    Solution:
    Hint:
    • The highest number that divides exactly into two or more numbers is known as GCF.
    • An algebraic expression consisting only one term is called monomial.
    Explanation:
    • We have been given a monomial in the question
    • We have to find the GCF of the given pair of monomials
    Step 1 of 1:
    We have given two monomials 8 a squared comma 28 a to the power of 5.
    8 a squared equals 8 cross times a cross times a
    28 a to the power of 5 equals 5 cross times 5 cross times a cross times a cross times a cross times a cross times a
    The highest factor in these two monomial is a × a
    Ie, a2.
    So, The GCF 8a2, 28a5 is a2.

    Find the GCF (GCD) of the given pair of monomials.
    8 a squared space straight & space 28 a to the power of 5

    Maths-General
    Solution:
    Hint:
    • The highest number that divides exactly into two or more numbers is known as GCF.
    • An algebraic expression consisting only one term is called monomial.
    Explanation:
    • We have been given a monomial in the question
    • We have to find the GCF of the given pair of monomials
    Step 1 of 1:
    We have given two monomials 8 a squared comma 28 a to the power of 5.
    8 a squared equals 8 cross times a cross times a
    28 a to the power of 5 equals 5 cross times 5 cross times a cross times a cross times a cross times a cross times a
    The highest factor in these two monomial is a × a
    Ie, a2.
    So, The GCF 8a2, 28a5 is a2.
    parallel
    General
    Maths-

    A regular polygon has 20 diagonals. Find the number of sides and classify it based on the number of sides.

    Solution:
    Hint:
    • A regular polygon is a polygon with congruent sides and equal angles and are symmetrically placed about a common center.
    Explanation:
    • We have been given in the question information about a regular polygon that has 20 diagonals which means it is an octagon.
    • We have to find the number of sides and classify it based on the number of sides.
    Step 1 of 1:
    We know that the number of diagonal in n - side polygon  equals fraction numerator n left parenthesis n minus 3 right parenthesis over denominator 2 end fraction
    Here, the number of diagonals is 20
    So,
    fraction numerator n left parenthesis n minus 3 right parenthesis over denominator 2 end fraction equals 20
    n squared minus 3 n equals 40
    n squared minus 3 n minus 40 equals 0
    (n - 8) (n + 5) = 0
    n = 8
    So, The number of sides are 8.
    And It is Octagon.

    A regular polygon has 20 diagonals. Find the number of sides and classify it based on the number of sides.

    Maths-General
    Solution:
    Hint:
    • A regular polygon is a polygon with congruent sides and equal angles and are symmetrically placed about a common center.
    Explanation:
    • We have been given in the question information about a regular polygon that has 20 diagonals which means it is an octagon.
    • We have to find the number of sides and classify it based on the number of sides.
    Step 1 of 1:
    We know that the number of diagonal in n - side polygon  equals fraction numerator n left parenthesis n minus 3 right parenthesis over denominator 2 end fraction
    Here, the number of diagonals is 20
    So,
    fraction numerator n left parenthesis n minus 3 right parenthesis over denominator 2 end fraction equals 20
    n squared minus 3 n equals 40
    n squared minus 3 n minus 40 equals 0
    (n - 8) (n + 5) = 0
    n = 8
    So, The number of sides are 8.
    And It is Octagon.
    General
    Maths-

    A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form.

    We have given the two sets
    A = {1, 2, 3, 5} and B = {4, 6, 9}
    And
    R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}
    First of all we will find the cartesian product of set A and set B
    A X B = {(1,4),(1,6),(1,9),(2,4),(2,6),(2,9),(3,4),(3,6),(3,9),(5,4),(5,6),(5,9)}
    We have relation R , with condition difference between x and y is odd, in this cartesian product we will find out the points satisfying the condition.
    So, R in roaster form is
    R = {(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)}

    A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form.

    Maths-General
    We have given the two sets
    A = {1, 2, 3, 5} and B = {4, 6, 9}
    And
    R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}
    First of all we will find the cartesian product of set A and set B
    A X B = {(1,4),(1,6),(1,9),(2,4),(2,6),(2,9),(3,4),(3,6),(3,9),(5,4),(5,6),(5,9)}
    We have relation R , with condition difference between x and y is odd, in this cartesian product we will find out the points satisfying the condition.
    So, R in roaster form is
    R = {(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)}
    General
    Maths-

    Which of the following relations are functions? Give reasons.
    (i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}
    (ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}

    (i) We have given the data
    As {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}
    Domain:-  { 2 , 5, 8, 11, 14, 17}
    Co- Domain:- {1}
    From the given data we can analyse that each element in the domain input set has exactly one output.
    Therefore, the given data is a function.

    Which of the following relations are functions? Give reasons.
    (i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}
    (ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}

    Maths-General
    (i) We have given the data
    As {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}
    Domain:-  { 2 , 5, 8, 11, 14, 17}
    Co- Domain:- {1}
    From the given data we can analyse that each element in the domain input set has exactly one output.
    Therefore, the given data is a function.
    parallel
    General
    Maths-

    Find the GCF (GCD) of the given pair of monomials.
    x cubed y squared straight & x to the power of 5 y

    Solution:
    Hint:
    • The highest number that divides exactly into two or more numbers is known as GCF.
    • An algebraic expression consisting only one term is called monomial.
    Explanation:
    • We have been given a monomial in the question
    • We have to find the GCF of the given pair of monomials.
    Step 1 of 1:
    We have given two monomials x cubed y squared space comma space x to the power of 5 y.
    x cubed y squared equals x cross times x cross times x cross times y cross times y
    x to the power of 5 y equals x cross times x cross times x cross times x cross times x cross times y
    The highest factor in these two monomial is x × x × x × y
    Ie, x3y.
    So, The GCF x3y2 , x5y  is x3y.

    Find the GCF (GCD) of the given pair of monomials.
    x cubed y squared straight & x to the power of 5 y

    Maths-General
    Solution:
    Hint:
    • The highest number that divides exactly into two or more numbers is known as GCF.
    • An algebraic expression consisting only one term is called monomial.
    Explanation:
    • We have been given a monomial in the question
    • We have to find the GCF of the given pair of monomials.
    Step 1 of 1:
    We have given two monomials x cubed y squared space comma space x to the power of 5 y.
    x cubed y squared equals x cross times x cross times x cross times y cross times y
    x to the power of 5 y equals x cross times x cross times x cross times x cross times x cross times y
    The highest factor in these two monomial is x × x × x × y
    Ie, x3y.
    So, The GCF x3y2 , x5y  is x3y.
    General
    Maths-

    A square has lines of symmetry

    Solution:
    Hint:
    • A square is a regular quadrilateral which has four equal sides and four equal angles or four right angles.
    Explanation:
    • We have been given a statement in the question for which we have to fill the blank from the given four options.
    Step 1 of 1:
    We know that in regular polygon, the number of line of symmetry is equal to number of sides of polygon.
    A square is regular polygon with 4 sides.
    So, The number of line of symmetry will be 4.

    A square has lines of symmetry

    Maths-General
    Solution:
    Hint:
    • A square is a regular quadrilateral which has four equal sides and four equal angles or four right angles.
    Explanation:
    • We have been given a statement in the question for which we have to fill the blank from the given four options.
    Step 1 of 1:
    We know that in regular polygon, the number of line of symmetry is equal to number of sides of polygon.
    A square is regular polygon with 4 sides.
    So, The number of line of symmetry will be 4.
    General
    Maths-

    Let R be a relation from N-N defined by R= { (a,b) : a, b belong to N and a= b2 }.Are the following true ?
    a) (a,a) ∈R , for all a ∈N
    b) (a,b) ∈R implies (b,a) ∈R
    c) (a,b) ∈R , (b,c) ∈R implies (a,c) ∈R
    Justify your answer in each case.

    R is relation from N to N defined by R = { (a, b) : a, b belong to N and a= b2 }.
    We will consider all the given cases
    i) It is not true because a = a2

    a2-a = 0

    a(a-1) = 0

    a = 0 ,1
    Hence the given condition is only true for a = 1
    Not for all natural numbers.
    ii) It is not true because if a = b2 , then b = a2 is not possible
    Eg. 32  = 9  but 92 > 3
    iii) It is not true because
    If left parenthesis a comma b right parenthesis element of R comma left parenthesis b comma c right parenthesis element of R comma left parenthesis a comma c right parenthesis element of R then
    It means a = b2 and b = c2 , a = c2
    This means a = (c2)2 = c4
    But a = c2
    So the given statement is wrong.

    Let R be a relation from N-N defined by R= { (a,b) : a, b belong to N and a= b2 }.Are the following true ?
    a) (a,a) ∈R , for all a ∈N
    b) (a,b) ∈R implies (b,a) ∈R
    c) (a,b) ∈R , (b,c) ∈R implies (a,c) ∈R
    Justify your answer in each case.

    Maths-General
    R is relation from N to N defined by R = { (a, b) : a, b belong to N and a= b2 }.
    We will consider all the given cases
    i) It is not true because a = a2

    a2-a = 0

    a(a-1) = 0

    a = 0 ,1
    Hence the given condition is only true for a = 1
    Not for all natural numbers.
    ii) It is not true because if a = b2 , then b = a2 is not possible
    Eg. 32  = 9  but 92 > 3
    iii) It is not true because
    If left parenthesis a comma b right parenthesis element of R comma left parenthesis b comma c right parenthesis element of R comma left parenthesis a comma c right parenthesis element of R then
    It means a = b2 and b = c2 , a = c2
    This means a = (c2)2 = c4
    But a = c2
    So the given statement is wrong.

    parallel

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