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What are the definitions of a function, the value of f(-2) for f(x) = 4x + 3, the maximum number of roots for a quadratic equation, and an identity matrix?

Mathematics

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Graph of f(x) = 4x + 3

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Key points of f(x) = 4x + 3
Pointxy
x-intercept−0.750
y-intercept03

Step-by-step solution

31 steps

Here are the solutions to all the questions:

1)

  1. Define a function.

    A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. It describes how one quantity depends on another.

    2)

  2. Substitute the value of xx into the function.

    Given f(x)=4x+3f(x) = 4x + 3.
    To find f(−2)f(-2), substitute x=−2x = -2:

    f(−2)=4(−2)+3f(-2) = 4(-2) + 3
  3. Calculate the value.

    f(−2)=−8+3f(-2) = -8 + 3 f(−2)=−5f(-2) = -5

    The value of f(−2)f(-2) is −5\boxed{-5}.

    3)

  4. Recall the nature of roots for a quadratic equation.

    A quadratic equation is of the form ax2+bx+c=0ax^2 + bx + c = 0, where a≠0a \neq 0. The number of roots depends on the discriminant Δ=b2−4ac\Delta = b^2 - 4ac.

    • If Δ>0\Delta > 0, there are two distinct real roots.
    • If Δ=0\Delta = 0, there is one real root (a repeated root).
    • If Δ<0\Delta < 0, there are two distinct complex roots.
  5. State the maximum number of roots.

    A quadratic equation can have at most two roots (real or complex).
    A quadratic equation can have 2 roots\boxed{2 \text{ roots}}.

    4)

  6. Define an identity matrix.

    An identity matrix is a square matrix where all the elements on the main diagonal are 11s and all other elements are 00s. It is denoted by II or InI_n, where nn is the dimension of the square matrix. When an identity matrix multiplies another matrix, the other matrix remains unchanged.

    5)

  7. Check if matrix multiplication is possible.

    Matrix A has order 2×32 \times 3.
    Matrix B has order 3×13 \times 1.
    For matrix multiplication AB to be possible, the number of columns in A must be equal to the number of rows in B.
    Number of columns in A = 3.
    Number of rows in B = 3.
    Since 3=33 = 3, multiplication AB is possible.

  8. Determine the order of the resulting matrix AB.

    If A is an m×nm \times n matrix and B is an n×pn \times p matrix, then the product AB is an m×pm \times p matrix.
    For A (2×32 \times 3) and B (3×13 \times 1), the order of AB will be 2×12 \times 1.
    Yes, AB is possible. The order of matrix AB will be 2×1\boxed{2 \times 1}.

    6)

  9. Recall the formula for the slope of a line given its inclination.

    The slope mm of a line with inclination θ\theta is given by m=tan⁡θm = \tan \theta.

  10. Substitute the given inclination and calculate the slope.

    Given inclination θ=135∘\theta = 135^\circ.

    m=tan⁡(135∘)m = \tan(135^\circ) m=tan⁡(180∘−45∘)m = \tan(180^\circ - 45^\circ) m=−tan⁡(45∘)m = -\tan(45^\circ) m=−1m = -1

    The slope of the line is −1\boxed{-1}.

    7)

  11. State the formula for external division.

    The coordinates of point P (x,y)(x, y) that divides the join of R (x1,y1)(x_1, y_1) and S (x2,y2)(x_2, y_2) in the ratio m1:m2m_1:m_2 externally are given by:

    P(x,y)=(m1x2−m2x1m1−m2,m1y2−m2y1m1−m2)P(x, y) = \left( \frac{m_1 x_2 - m_2 x_1}{m_1 - m_2}, \frac{m_1 y_2 - m_2 y_1}{m_1 - m_2} \right)

    The coordinates of point P are (m1x2−m2x1m1−m2,m1y2−m2y1m1−m2)\boxed{\left( \frac{m_1 x_2 - m_2 x_1}{m_1 - m_2}, \frac{m_1 y_2 - m_2 y_1}{m_1 - m_2} \right)}.

    8)

  12. Write the given vectors in column form.

    Given a⃗=(1,2)\vec{a} = (1, 2) and b⃗=(4,5)\vec{b} = (4, 5).
    In column vector form:

    a⃗=[12]\vec{a} = \begin{bmatrix} 1 \\ 2 \end{bmatrix} b⃗=[45]\vec{b} = \begin{bmatrix} 4 \\ 5 \end{bmatrix}
  13. Add the vectors.

    a⃗+b⃗=[12]+[45]\vec{a} + \vec{b} = \begin{bmatrix} 1 \\ 2 \end{bmatrix} + \begin{bmatrix} 4 \\ 5 \end{bmatrix} a⃗+b⃗=[1+42+5]\vec{a} + \vec{b} = \begin{bmatrix} 1+4 \\ 2+5 \end{bmatrix} a⃗+b⃗=[57]\vec{a} + \vec{b} = \begin{bmatrix} 5 \\ 7 \end{bmatrix}

    The sum a⃗+b⃗\vec{a} + \vec{b} in column vector form is [57]\boxed{\begin{bmatrix} 5 \\ 7 \end{bmatrix}}.

    9)

  14. Recall the formula for Quartile Deviation (Q.D.).

    The Quartile Deviation is given by the formula:

    Q.D.=Q3−Q12Q.D. = \frac{Q_3 - Q_1}{2}

    where Q3Q_3 is the third quartile and Q1Q_1 is the first quartile.

  15. Substitute the given values and solve for Q3Q_3.

    Given Q.D.=30Q.D. = 30 and Q1=10Q_1 = 10.

    30=Q3−10230 = \frac{Q_3 - 10}{2}
  16. Isolate Q3Q_3.

    Multiply both sides by 2:

    30×2=Q3−1030 \times 2 = Q_3 - 10 60=Q3−1060 = Q_3 - 10

    Add 10 to both sides:

    Q3=60+10Q_3 = 60 + 10 Q3=70Q_3 = 70

    The value of Q3Q_3 is 70\boxed{70}.

    10)

  17. Define a box and whisker plot.

    A box and whisker plot (or box plot) is a standardized way of displaying the distribution of data based on five key numbers: the minimum value, the first quartile (Q1Q_1), the median (Q2Q_2), the third quartile (Q3Q_3), and the maximum value. The "box" represents the interquartile range (Q1Q_1 to Q3Q_3), and the "whiskers" extend to the minimum and maximum values within a certain range.

    11)

  18. Find f(x)f(x).

    Given f(2x−3)=4x+5f(2x - 3) = 4x + 5.
    Let y=2x−3y = 2x - 3.
    Then y+3=2xy + 3 = 2x, so x=y+32x = \frac{y+3}{2}.
    Substitute this expression for xx into the function:

    f(y)=4(y+32)+5f(y) = 4\left(\frac{y+3}{2}\right) + 5 f(y)=2(y+3)+5f(y) = 2(y+3) + 5 f(y)=2y+6+5f(y) = 2y + 6 + 5 f(y)=2y+11f(y) = 2y + 11

    Replacing yy with xx, we get:

    f(x)=2x+11f(x) = 2x + 11
  19. Find f(−2)f(-2).

    Substitute x=−2x = -2 into the expression for f(x)f(x):

    f(−2)=2(−2)+11f(-2) = 2(-2) + 11 f(−2)=−4+11f(-2) = -4 + 11 f(−2)=7f(-2) = 7

    So, f(x)=2x+11f(x) = \boxed{2x + 11} and f(−2)=7f(-2) = \boxed{7}.

    12)

  20. Perform polynomial long division.

    We need to divide 4x3+2x2−4x+34x^3 + 2x^2 - 4x + 3 by 2x+32x + 3.

            2x^2  - 2x   + 1
          _________________
    2x + 3 | 4x^3 + 2x^2 - 4x + 3
            -(4x^3 + 6x^2)
            _________________
                  -4x^2 - 4x
                -(-4x^2 - 6x)
                _________________
                        2x + 3
                      -(2x + 3)
                      _________
                              0
    
  21. Identify the quotient and remainder.

    The quotient is 2x2−2x+12x^2 - 2x + 1.
    The remainder is 0.
    The quotient is 2x2−2x+1\boxed{2x^2 - 2x + 1} and the remainder is 0\boxed{0}.

    13)

  22. Substitute the coordinates of the point into the equation of the locus.

    The point is (4,5)(4, 5), so x=4x = 4 and y=5y = 5.
    The equation of the locus is x2+y2+kx−8y−25=0x^2 + y^2 + kx - 8y - 25 = 0.
    Substitute x=4x=4 and y=5y=5:

    (4)2+(5)2+k(4)−8(5)−25=0(4)^2 + (5)^2 + k(4) - 8(5) - 25 = 0
  23. Simplify the equation.

    16+25+4k−40−25=016 + 25 + 4k - 40 - 25 = 0
  24. Solve for kk.

    41+4k−65=041 + 4k - 65 = 0 4k−24=04k - 24 = 0 4k=244k = 24 k=244k = \frac{24}{4} k=6k = 6

    The value of kk is 6\boxed{6}.

    14)

  25. Understand the matrix construction rule.

    We need to construct a 2×22 \times 2 matrix, so it will have elements a11,a12,a21,a22a_{11}, a_{12}, a_{21}, a_{22}.
    The rule for elements is aij=2i−3ja_{ij} = 2i - 3j.

  26. Calculate each element.

    For a11a_{11} (i=1,j=1i=1, j=1):

    a11=2(1)−3(1)=2−3=−1a_{11} = 2(1) - 3(1) = 2 - 3 = -1

    For a12a_{12} (i=1,j=2i=1, j=2):

    a12=2(1)−3(2)=2−6=−4a_{12} = 2(1) - 3(2) = 2 - 6 = -4

    For a21a_{21} (i=2,j=1i=2, j=1):

    a21=2(2)−3(1)=4−3=1a_{21} = 2(2) - 3(1) = 4 - 3 = 1

    For a22a_{22} (i=2,j=2i=2, j=2):

    a22=2(2)−3(2)=4−6=−2a_{22} = 2(2) - 3(2) = 4 - 6 = -2
  27. Form the matrix.

    The matrix is:

    [a11a12a21a22]=[−1−41−2]\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} = \begin{bmatrix} -1 & -4 \\ 1 & -2 \end{bmatrix}

    The constructed matrix is [−1−41−2]\boxed{\begin{bmatrix} -1 & -4 \\ 1 & -2 \end{bmatrix}}.

    15)
    This question was solved in the previous turn. The values are a=1a=1 and b=4b=4.

    16)

  28. Rewrite the equation using trigonometric identities.

    Given sin⁡2θ=cos⁡θ\sin 2\theta = \cos \theta.
    Use the double angle identity for sine: sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2 \sin \theta \cos \theta.

    2sin⁡θcos⁡θ=cos⁡θ2 \sin \theta \cos \theta = \cos \theta
  29. Rearrange the equation to solve for θ\theta.

    2sin⁡θcos⁡θ−cos⁡θ=02 \sin \theta \cos \theta - \cos \theta = 0

    Factor out cos⁡θ\cos \theta:

    cos⁡θ(2sin⁡θ−1)=0\cos \theta (2 \sin \theta - 1) = 0
  30. Set each factor to zero and solve for θ\theta.

    Case 1: cos⁡θ=0\cos \theta = 0
    For 0∘≤θ≤90∘0^\circ \le \theta \le 90^\circ, the value of θ\theta for which cos⁡θ=0\cos \theta = 0 is θ=90∘\theta = 90^\circ.

    Case 2: 2sin⁡θ−1=02 \sin \theta - 1 = 0

    2sin⁡θ=12 \sin \theta = 1 sin⁡θ=12\sin \theta = \frac{1}{2}

    For 0∘≤θ≤90∘0^\circ \le \theta \le 90^\circ, the value of θ\theta for which sin⁡θ=12\sin \theta = \frac{1}{2} is θ=30∘\theta = 30^\circ.

  31. List all solutions within the given range.

    The solutions are θ=30∘\theta = 30^\circ and θ=90∘\theta = 90^\circ.
    The solutions are 30∘,90∘\boxed{30^\circ, 90^\circ}.

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