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Given the relation f: x (2x-1)/(x2+3x+2).

Mathematics

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Question 1:
Given the relation f:x→2x−1x2+3x+2f: x \to \frac{2x-1}{x^2+3x+2}.

i) Find the zero of the relation.

  1. The relation is zero when the numerator is equal to zero.

    2x−1=02x - 1 = 0
  2. Solve for xx.

    2x=12x = 1 x=12x = \frac{1}{2}

    The zero of the relation is 12\boxed{\frac{1}{2}}.

    ii) Find the value of xx which makes the relation undefined.

  3. The relation is undefined when the denominator is equal to zero.

    x2+3x+2=0x^2 + 3x + 2 = 0
  4. Factor the quadratic equation.

    (x+1)(x+2)=0(x+1)(x+2) = 0
  5. Solve for xx.

    x+1=0orx+2=0x+1 = 0 \quad \text{or} \quad x+2 = 0 x=−1orx=−2x = -1 \quad \text{or} \quad x = -2

    The values of xx which make the relation undefined are −1,−2\boxed{-1, -2}.

    Question 2:
    Calculate the (i) Area (ii) Perimeter of the given foundation drawing.

    i) Area

  6. Divide the L-shaped figure into two rectangles.

    We can split it into a large bottom rectangle and a smaller top-left rectangle.
    The bottom rectangle has dimensions 22 cm×8 cm22 \text{ cm} \times 8 \text{ cm}.
    The top-left rectangle has a width of 22 cm−14 cm=8 cm22 \text{ cm} - 14 \text{ cm} = 8 \text{ cm} and a height of 19 cm−8 cm=11 cm19 \text{ cm} - 8 \text{ cm} = 11 \text{ cm}.

  7. Calculate the area of each rectangle.

    Area of bottom rectangle =22 cm×8 cm=176 cm2= 22 \text{ cm} \times 8 \text{ cm} = 176 \text{ cm}^2.
    Area of top-left rectangle =8 cm×11 cm=88 cm2= 8 \text{ cm} \times 11 \text{ cm} = 88 \text{ cm}^2.

  8. Add the areas to find the total area.

    Total Area =176 cm2+88 cm2=264 cm2= 176 \text{ cm}^2 + 88 \text{ cm}^2 = 264 \text{ cm}^2.
    The area is 264 cm2\boxed{264 \text{ cm}^2}.

    ii) Perimeter

  9. Identify all the outer side lengths of the figure.

    The given lengths are 19 cm19 \text{ cm}, 8 cm8 \text{ cm}, 11 cm11 \text{ cm}, 14 cm14 \text{ cm}, 8 cm8 \text{ cm}, and 22 cm22 \text{ cm}.

  10. Sum all the outer side lengths.

    Perimeter =19 cm+8 cm+11 cm+14 cm+8 cm+22 cm= 19 \text{ cm} + 8 \text{ cm} + 11 \text{ cm} + 14 \text{ cm} + 8 \text{ cm} + 22 \text{ cm}
    Perimeter =82 cm= 82 \text{ cm}.
    The perimeter is 82 cm\boxed{82 \text{ cm}}.

    Question 3:
    Find the value of xx and yy in the diagram below if O is the centre of the circle.

  11. Use the property of angles in a cyclic quadrilateral.

    The sum of opposite angles in a cyclic quadrilateral is 180∘180^\circ.
    The angles xx and 3x3x are opposite angles in the cyclic quadrilateral.

    x+3x=180∘x + 3x = 180^\circ 4x=180∘4x = 180^\circ
  12. Solve for xx.

    x=180∘4x = \frac{180^\circ}{4} x=45∘x = 45^\circ
  13. Use the property that the angle subtended by an arc at the center is twice the angle subtended by the same arc at any point on the remaining part of the circle.

    The angle at the center is yy, and the angle at the circumference subtended by the same arc is xx.

    y=2xy = 2x
  14. Substitute the value of xx to find yy.

    y=2(45∘)y = 2(45^\circ) y=90∘y = 90^\circ

    The value of xx is 45∘\boxed{45^\circ} and the value of yy is 90∘\boxed{90^\circ}.

    Question 4:

    a) y varies directly as x. If y=18y = 18 when x=6x = 6, find:

    i) the equation connecting y and x

    ii) the value of y when x=15x = 15

    i) The equation connecting y and x

  15. For direct variation, the relationship is y=kxy = kx, where kk is the constant of proportionality.

  16. Use the given values y=18y=18 and x=6x=6 to find kk.

    18=k(6)18 = k(6) k=186k = \frac{18}{6} k=3k = 3

    The equation connecting yy and xx is y=3x\boxed{y = 3x}.

    ii) The value of y when x=15x = 15

  17. Use the equation y=3xy = 3x found in part (i).

  18. Substitute x=15x=15 into the equation.

    y=3(15)y = 3(15) y=45y = 45

    The value of yy when x=15x=15 is 45\boxed{45}.

    b) y varies inversely as x. If y=4y = 4 when x=3x = 3, find y when x=12x = 12.

  19. For inverse variation, the relationship is y=kxy = \frac{k}{x}, where kk is the constant of proportionality.

  20. Use the given values y=4y=4 and x=3x=3 to find kk.

    4=k34 = \frac{k}{3} k=4×3k = 4 \times 3 k=12k = 12

    The equation connecting yy and xx is y=12xy = \frac{12}{x}.

  21. Find yy when x=12x=12.

    y=1212y = \frac{12}{12} y=1y = 1

    The value of yy when x=12x=12 is 1\boxed{1}.

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