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Explain how -2 and 3/2 are the roots of the equation 2x² + x − 6 = 0.

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4 parts
a

cos⁡A=1−sin⁡2A\cos A = \sqrt{1 - \sin^2 A}

b

tan⁡A=sin⁡A1−sin⁡2A\tan A = \frac{\sin A}{\sqrt{1 - \sin^2 A}}

c

csc⁡A=1sin⁡A\csc A = \frac{1}{\sin A}

d

sec⁡A=11−sin⁡2A\sec A = \frac{1}{\sqrt{1 - \sin^2 A}}

214.5 cm2214.5\, \text{cm}^2

536\frac{5}{36}

Step-by-step solution

4 steps

Step 1: Verify the roots by substitution (Question 21).
The given quadratic equation is 2x2+x−6=02x^2 + x - 6 = 0.
We need to explain how −2-2 and 32\frac{3}{2} are its roots. A number is a root of an equation if substituting it into the equation makes the equation true (i.e., the expression equals zero).

For x=−2x = -2:
Substitute x=−2x = -2 into the equation:

2(−2)2+(−2)−62(-2)^2 + (-2) - 6 2(4)−2−62(4) - 2 - 6 8−2−68 - 2 - 6 6−6=06 - 6 = 0

Since the expression equals 0, x=−2x = -2 is a root of the equation.

For x=32x = \frac{3}{2}:
Substitute x=32x = \frac{3}{2} into the equation:

2(32)2+(32)−62\left(\frac{3}{2}\right)^2 + \left(\frac{3}{2}\right) - 6 2(94)+32−62\left(\frac{9}{4}\right) + \frac{3}{2} - 6 184+32−6\frac{18}{4} + \frac{3}{2} - 6 92+32−6\frac{9}{2} + \frac{3}{2} - 6 122−6\frac{12}{2} - 6 6−6=06 - 6 = 0

Since the expression equals 0, x=32x = \frac{3}{2} is a root of the equation.
Thus, both −2-2 and 32\frac{3}{2} satisfy the equation, confirming they are its roots.

Step 2: Write the formula for the nthn^{th} term of an AP and find the 10th10^{th} term (Question 22).
The formula to calculate the nthn^{th} term of an Arithmetic Progression (AP) is:

an=a+(n−1)da_n = a + (n-1)d

where:

  • ana_n is the nthn^{th} term.
  • aa is the first term.
  • nn is the term number.
  • dd is the common difference.

Given AP: 3,7,11,…3, 7, 11, \dots
Here, the first term a=3a = 3.
The common difference d=7−3=4d = 7 - 3 = 4.
We need to find the 10th10^{th} term, so n=10n = 10.
Substitute these values into the formula:

a10=3+(10−1)4a_{10} = 3 + (10-1)4 a10=3+(9)4a_{10} = 3 + (9)4 a10=3+36a_{10} = 3 + 36 a10=39a_{10} = 39

The 10th10^{th} term of the AP is:

39\boxed{39}

Step 3: Express trigonometric ratios in terms of sin⁡A\sin A (Question 23).
We use the fundamental trigonometric identities. Assume AA is an acute angle, so cos⁡A>0\cos A > 0.

Part a

Express cos⁡A\cos A in terms of sin⁡A\sin A

We know the identity sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1.
Rearrange to solve for cos⁡A\cos A:

cos⁡2A=1−sin⁡2A\cos^2 A = 1 - \sin^2 A cos⁡A=1−sin⁡2A\cos A = \sqrt{1 - \sin^2 A} cos⁡A=1−sin⁡2A\boxed{\cos A = \sqrt{1 - \sin^2 A}}

Part b

tan⁡A=sin⁡A1−sin⁡2A\tan A = \frac{\sin A}{\sqrt{1 - \sin^2 A}}

Express tan⁡A\tan A in terms of sin⁡A\sin A

We know tan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A}.
Substitute the expression for cos⁡A\cos A from part (a):

tan⁡A=sin⁡A1−sin⁡2A\tan A = \frac{\sin A}{\sqrt{1 - \sin^2 A}} tan⁡A=sin⁡A1−sin⁡2A\boxed{\tan A = \frac{\sin A}{\sqrt{1 - \sin^2 A}}}

Part c

csc⁡A=1sin⁡A\csc A = \frac{1}{\sin A}

Express csc⁡A\csc A in terms of sin⁡A\sin A

We know csc⁡A=1sin⁡A\csc A = \frac{1}{\sin A}.

csc⁡A=1sin⁡A\boxed{\csc A = \frac{1}{\sin A}}

Part d

sec⁡A=11−sin⁡2A\sec A = \frac{1}{\sqrt{1 - \sin^2 A}}

Express sec⁡A\sec A in terms of sin⁡A\sin A

We know sec⁡A=1cos⁡A\sec A = \frac{1}{\cos A}.
Substitute the expression for cos⁡A\cos A from part (a):

sec⁡A=11−sin⁡2A\sec A = \frac{1}{\sqrt{1 - \sin^2 A}} sec⁡A=11−sin⁡2A\boxed{\sec A = \frac{1}{\sqrt{1 - \sin^2 A}}}

  1. Prove that a parallelogram circumscribing a circle is a rhombus (Question 24).

    Let ABCD be a parallelogram circumscribing a circle with center O.
    Let the sides AB, BC, CD, and DA touch the circle at points P, Q, R, and S respectively.

    According to the property that tangents drawn from an external point to a circle are equal in length:

    • From vertex A: AP=ASAP = AS (1)
    • From vertex B: BP=BQBP = BQ (2)
    • From vertex C: CQ=CRCQ = CR (3)
    • From vertex D: DR=DSDR = DS (4)

    Add equations (1), (2), (3), and (4):
    AP+BP+CR+DR=AS+BQ+CQ+DSAP + BP + CR + DR = AS + BQ + CQ + DS
    Rearrange the terms:
    (AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
    From the figure, we can see that:
    AP+BP=ABAP + BP = AB
    CR+DR=CDCR + DR = CD
    AS+DS=ADAS + DS = AD
    BQ+CQ=BCBQ + CQ = BC
    So, the equation becomes:
    AB+CD=AD+BCAB + CD = AD + BC (5)

    Since ABCD is a parallelogram, we know that opposite sides are equal in length:
    AB=CDAB = CD (6)
    BC=ADBC = AD (7)

    Substitute (6) and (7) into (5):
    AB+AB=BC+BCAB + AB = BC + BC
    2AB=2BC2AB = 2BC
    AB=BCAB = BC

    Since AB=BCAB = BC, and we know that in a parallelogram, opposite sides are equal (AB=CDAB=CD and BC=ADBC=AD), this implies that all four sides are equal:
    AB=BC=CD=DAAB = BC = CD = DA.
    A parallelogram with all four sides equal is a rhombus.
    Therefore, a parallelogram circumscribing a circle is a rhombus.

  2. Find the total surface area of the toy (Question 25).

    The toy is a cone mounted on a hemisphere of the same radius.
    Given:
    Radius of hemisphere (rr) = 3.5 cm
    Radius of cone (rr) = 3.5 cm
    Total height of the toy (HH) = 15.5 cm

    The height of the hemispherical part is equal to its radius, r=3.5r = 3.5 cm.
    The height of the conical part (hch_c) is the total height minus the height of the hemisphere:

    hc=H−rh_c = H - r hc=15.5 cm−3.5 cmh_c = 15.5\, \text{cm} - 3.5\, \text{cm} hc=12 cmh_c = 12\, \text{cm}

    Now, calculate the slant height (ll) of the cone using the Pythagorean theorem:

    l=r2+hc2l = \sqrt{r^2 + h_c^2} l=(3.5 cm)2+(12 cm)2l = \sqrt{(3.5\, \text{cm})^2 + (12\, \text{cm})^2} l=12.25 cm2+144 cm2l = \sqrt{12.25\, \text{cm}^2 + 144\, \text{cm}^2} l=156.25 cm2l = \sqrt{156.25\, \text{cm}^2} l=12.5 cml = 12.5\, \text{cm}

    The total surface area of the toy is the sum of the curved surface area of the cone and the curved surface area of the hemisphere.
    Curved surface area of cone (CSAconeCSA_{cone}) = πrl\pi r l
    Curved surface area of hemisphere (CSAhemisphereCSA_{hemisphere}) = 2πr22\pi r^2

    Total Surface Area (TSATSA) = CSAcone+CSAhemisphereCSA_{cone} + CSA_{hemisphere}

    TSA=πrl+2πr2TSA = \pi r l + 2\pi r^2

    Factor out πr\pi r:

    TSA=πr(l+2r)TSA = \pi r (l + 2r)

    Substitute the values (r=3.5r = 3.5, l=12.5l = 12.5, π≈227\pi \approx \frac{22}{7}):

    TSA=227×3.5 cm×(12.5 cm+2×3.5 cm)TSA = \frac{22}{7} \times 3.5\, \text{cm} \times (12.5\, \text{cm} + 2 \times 3.5\, \text{cm}) TSA=227×3.5 cm×(12.5 cm+7 cm)TSA = \frac{22}{7} \times 3.5\, \text{cm} \times (12.5\, \text{cm} + 7\, \text{cm}) TSA=227×3.5 cm×19.5 cmTSA = \frac{22}{7} \times 3.5\, \text{cm} \times 19.5\, \text{cm} TSA=22×0.5 cm×19.5 cmTSA = 22 \times 0.5\, \text{cm} \times 19.5\, \text{cm} TSA=11 cm×19.5 cmTSA = 11\, \text{cm} \times 19.5\, \text{cm} TSA=214.5 cm2TSA = 214.5\, \text{cm}^2

    The total surface area of the toy is:

    214.5 cm2\boxed{214.5\, \text{cm}^2}

  3. Write the formula for mean by step deviation method and explain its terms (Question 26).

    The formula to find the mean (xˉ\bar{x}) by the step deviation method for grouped data is:

    xˉ=A+(∑fiui∑fi)h\bar{x} = A + \left(\frac{\sum f_i u_i}{\sum f_i}\right)h

    Where:

    • xˉ\bar{x} is the mean of the data.
    • AA is the assumed mean, which is a value chosen from the midpoints of the class intervals (usually the midpoint of the middle class interval).
    • fif_i is the frequency of the ithi^{th} class interval.
    • uiu_i is the step deviation for the ithi^{th} class interval, calculated as ui=xi−Ahu_i = \frac{x_i - A}{h}.
    • xix_i is the class mark (midpoint) of the ithi^{th} class interval.
    • hh is the class size (or class width), which is the difference between the upper and lower limits of a class interval.
    • ∑fiui\sum f_i u_i is the sum of the products of frequencies and corresponding step deviations.
    • ∑fi\sum f_i is the sum of all frequencies (total number of observations).

  4. Find the probability that the sum of two numbers appearing on the top of the dice is 8 (Question 27).

    When two dice are thrown, the total number of possible outcomes is 6×6=366 \times 6 = 36.
    The possible outcomes are pairs (d1,d2)(d_1, d_2) where d1d_1 is the result of the first die and d2d_2 is the result of the second die.

    We need to find the outcomes where the sum of the two numbers is 8. These are:

    • (2,6)(2, 6)
    • (3,5)(3, 5)
    • (4,4)(4, 4)
    • (5,3)(5, 3)
    • (6,2)(6, 2) There are 5 favorable outcomes.

    The probability P(sum is 8)P(\text{sum is 8}) is given by:

    P(sum is 8)=Number of favorable outcomesTotal number of outcomesP(\text{sum is 8}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} P(sum is 8)=536P(\text{sum is 8}) = \frac{5}{36}

    The probability that the sum of the two numbers appearing on the top of the dice is 8 is:

    536\boxed{\frac{5}{36}}
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