INTRODUCTION TO MATHEMATICAL METHODS QUIZ TWO JULY INTAKE 2024

DATE : 18th OCTOBER, 2024

QUESTION ONE

Which of the following is true about $R = \{ (1, 2), (2, 3), (3, 4), (4, 5) \}$
a) $R$ is a relation but it’s not a function
b) $R$ is not both a relation and a function
c) $R$ is both a relation and a function
d) $R$ is not a relation but it’s a function

Answer: c) $R$ is both a relation and a function

Explanation:

A relation is any set of ordered pairs, so $R$ is a relation.
A function requires that each input (first element) maps to exactly one output (second element).
Inputs in $R$ : 1 maps to 2, 2 maps to 3, 3 maps to 4, 4 maps to 5. All inputs are unique and map to one output, so $R$ is a function.
Thus, $R$ is both a relation and a function.

Question 2:

According to the method of partial fractions, there is an equation of the form
$\frac{x}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$
For some numbers $A, B$ and $C$ . What is the number $C$ ?
a) $\frac{1}{2}$
b) $\frac{1}{2}$
c) -1
d) $\frac{-1}{2}$

Answer: a) $\frac{1}{2}$

Explanation:

Note: The numerator in the problem is $x$ , but the options suggest a possible typo (as the correct $C = \frac{3}{2}$ is not listed). Assuming the intended numerator is 1 for consistency with options.
Given: $\frac{1}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x-3}$ .
Multiply both sides by $(x-1)(x-2)(x-3)$ :
$1 = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$
Solve for $C$ by substituting $x = 3$ (to eliminate other terms):
$1 = C(3-1)(3-2) \implies 1 = C \cdot 2 \cdot 1 \implies C = \frac{1}{2}$
Thus, $C = \frac{1}{2}$ .

Question 3:

Find the solution set of $|2x - 4| > 2$
a) (1,3)
b) [1,3]
c) $(-\infty, 1) \cup (3, \infty)$
d) $(-\infty, 1] \cup [3, \infty)$

Answer: c) $(-\infty, 1) \cup (3, \infty)$

Explanation:

Simplify: $|2x - 4| > 2 \implies |x - 2| > 1$ (divide by 2).
Absolute inequality $|A| > B$ (with $B > 0$ ) means $A < -B$ or $A > B$ .
So, $|x - 2| > 1 \implies x - 2 < -1$ or $x - 2 > 1 \implies x < 1$ or $x > 3$ .
Interval notation: $(-\infty, 1) \cup (3, \infty)$ .

Question 4:

Which of the following functions is NOT one to one?
a) $F = \{(0,1),(1,2),(2,3),(3,4)\}$
b) $F = \{(0,0),(1,1),(2,2),(3,3)\}$
c) $F = \{(0,1),(2,4),(3,1),(4,2)\}$
d) $F = \{(0,4),(1,3),(2,2),(3,1)\}$

Answer: c) $F = \{(0,1),(2,4),(3,1),(4,2)\}$

Explanation:

A function is one-to-one (injective) if each output is paired with at most one input.
Check outputs:
- a) Outputs: 1, 2, 3, 4 (all distinct) → injective.
- b) Outputs: 0, 1, 2, 3 (all distinct) → injective.
- c) Outputs: 1, 4, 1, 2 (output 1 repeated for inputs 0 and 3) → not injective.
- d) Outputs: 4, 3, 2, 1 (all distinct) → injective.
Thus, c) is not one-to-one.

Question 5:

Given that $\alpha$ and $\beta$ are the roots of the equation $x^2 - 3x + 5 = 0$ . Find the equation whose roots are $\frac{\beta}{\alpha}$ and $\frac{\alpha}{\beta}$ .
a) $5x^2 + 5x - 1 = 0$
b) $5x^2 - x - 5 = 0$
c) $5x^2 - x + 5 = 0$
d) $5x^2 + x + 5 = 0$

Answer: d) $5x^2 + x + 5 = 0$

Explanation:

For $x^2 - 3x + 5 = 0$ , sum of roots $\alpha + \beta = 3$ , product $\alpha \beta = 5$ .
New roots: $r_1 = \frac{\beta}{\alpha}$ , $r_2 = \frac{\alpha}{\beta}$ .
Sum of new roots:
$r_1 + r_2 = \frac{\beta}{\alpha} + \frac{\alpha}{\beta} = \frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{(\alpha + \beta)^2 - 2\alpha \beta}{\alpha \beta} = \frac{3^2 - 2 \cdot 5}{5} = \frac{9 - 10}{5} = -\frac{1}{5}$
Product of new roots:
$r_1 r_2 = \left( \frac{\beta}{\alpha} \right) \left( \frac{\alpha}{\beta} \right) = 1$
Quadratic equation: $x^2 - (\text{sum})x + (\text{product}) = 0 \implies x^2 - \left( -\frac{1}{5} \right)x + 1 = 0 \implies x^2 + \frac{1}{5}x + 1 = 0$ .
Multiply by 5 to clear fraction: $5x^2 + x + 5 = 0$ .

Question 6:

The expression $p(x) = x^3 - \alpha x^2 - x + \beta$ is exactly divisible by $x - 1$ and on division by $x - 2$ gives a remainder of 21. Calculate the value of $\alpha$ and $\beta$

$\begin{array}{cc} \alpha & \beta \\ a) & -5 & -5 \\ b) & 5 & 5 \\ c) & -5 & 5 \\ d) & 5 & -5 \end{array}$

Answer: a) $\alpha = -5$ , $\beta = -5$

Explanation:

Divisible by $x - 1$ implies $p(1) = 0$ :
$p(1) = 1^3 - \alpha(1)^2 - 1 + \beta = 1 - \alpha - 1 + \beta = -\alpha + \beta = 0 \implies \beta = \alpha$
Remainder 21 when divided by $x - 2$ implies $p(2) = 21$ :
$p(2) = 2^3 - \alpha(2)^2 - 2 + \beta = 8 - 4\alpha - 2 + \beta = 6 - 4\alpha + \beta = 21$
Substitute $\beta = \alpha$ :
$6 - 4\alpha + \alpha = 21 \implies 6 - 3\alpha = 21 \implies -3\alpha = 15 \implies \alpha = -5$
Then $\beta = -5$ .
Verify: $p(x) = x^3 + 5x^2 - x - 5$ , $p(1) = 1 + 5 - 1 - 5 = 0$ , $p(2) = 8 + 20 - 2 - 5 = 21$ .

Question 7:

A function f: R→R is defined by $f(x)=5x^3-8$ . The type of function is
a) one-one
b) onto
c) many-one
d) both one-one and onto

Answer: d) both one-one and onto

Explanation:

One-one (injective): If $f(a) = f(b)$ , then $5a^3 - 8 = 5b^3 - 8 \implies 5a^3 = 5b^3 \implies a^3 = b^3 \implies a = b$ . So injective.
Onto (surjective): For any $y \in \mathbb{R}$ , solve $5x^3 - 8 = y \implies x^3 = \frac{y + 8}{5} \implies x = \sqrt[3]{\frac{y + 8}{5}}$ . Since cube root is defined for all reals, there is always a real $x$ . So surjective.
Thus, bijective.

Question 8:

The function $f(x) = x^4 - x^3$ is......
a) Even
b) Odd
c) Neither odd nor even
d) Both odd and even

Answer: c) Neither odd nor even

Explanation:

Even if $f(-x) = f(x)$ :
$f(-x) = (-x)^4 - (-x)^3 = x^4 - (-x^3) = x^4 + x^3 \neq f(x) = x^4 - x^3$
Odd if $f(-x) = -f(x)$ :
$f(-x) = x^4 + x^3 \neq -f(x) = -x^4 + x^3$
Thus, neither.

Question 9:

Which of the following functions is one to one.
a) $f(x) = \frac{x - 7}{x + 4}$
b) $f(x) = |x| - 4$
c) $f(x) = x^2 - 4$
d) $f(x) = |4x - 7|$

Answer: a) $f(x) = \frac{x - 7}{x + 4}$

Explanation:

a) Suppose $f(a) = f(b)$ :
$\frac{a-7}{a+4} = \frac{b-7}{b+4} \implies (a-7)(b+4) = (b-7)(a+4)$
$ab + 4a - 7b - 28 = ab + 4b - 7a - 28 \implies 4a - 7b = 4b - 7a \implies 11a = 11b \implies a = b$
So injective.
b) Not injective: $f(1) = |1| - 4 = -3$ , $f(-1) = |-1| - 4 = -3$ , but $1 \neq -1$ .
c) Not injective: $f(2) = 2^2 - 4 = 0$ , $f(-2) = (-2)^2 - 4 = 0$ .
d) Not injective: $f(2) = |8 - 7| = 1$ , $f(1.5) = |6 - 7| = 1$ , but $2 \neq 1.5$ .
Thus, only a) is one-to-one.

Question 10:

Which of the following can be expressed in the form $f(x) = 2(x - 3)^2 - 5$
a) $f(x) = 2x^2 - 4$
b) $f(x) = 2x^2 - 12x - 5$
c) $f(x) = 2x^2 - 12x + 13$
d) $f(x) = 2x^2 - 6x$

Answer: c) $f(x) = 2x^2 - 12x + 13$

Explanation:

Expand the given form:
$2(x - 3)^2 - 5 = 2(x^2 - 6x + 9) - 5 = 2x^2 - 12x + 18 - 5 = 2x^2 - 12x + 13$
Compare to options:
- a) $2x^2 - 4$
- b) $2x^2 - 12x - 5$
- c) $2x^2 - 12x + 13$ (matches)
- d) $2x^2 - 6x$

Question 11:

The roots of the quadratic equation $kx^2 - 2(k + 2)x + 3k = 0$ differ by 2. Find the values of k.
a) 2 or -3
b) 1 or -3
c) -2/3 or -3
d) 2 or -2/3

Answer: d) 2 or -2/3

Explanation:

Let roots be $r$ and $s$ , with $|r - s| = 2$ .
Sum $r + s = \frac{2(k+2)}{k}$ , product $r s = \frac{3k}{k} = 3$ .
Use identity $(r - s)^2 = (r + s)^2 - 4rs$ :
$(r - s)^2 = 2^2 = 4$
$(r + s)^2 - 4rs = \left( \frac{2(k+2)}{k} \right)^2 - 4 \cdot 3 = \frac{4(k+2)^2}{k^2} - 12$
Set equal:
$\frac{4(k+2)^2}{k^2} - 12 = 4$
$\frac{4(k+2)^2}{k^2} = 16$
$4(k+2)^2 = 16k^2$
$(k+2)^2 = 4k^2$
$k^2 + 4k + 4 = 4k^2$
$3k^2 - 4k - 4 = 0$
Solve quadratic: discriminant $d = (-4)^2 - 4 \cdot 3 \cdot (-4) = 16 + 48 = 64$ ,
$k = \frac{4 \pm \sqrt{64}}{6} = \frac{4 \pm 8}{6}$
$k = \frac{12}{6} = 2 \quad \text{or} \quad k = \frac{-4}{6} = -\frac{2}{3}$
Verify:
- For $k = 2$ : equation $2x^2 - 8x + 6 = 0$ , roots $x = 1, 3$ (differ by 2).
- For $k = -\frac{2}{3}$ : equation $-\frac{2}{3}x^2 - \frac{8}{3}x - 2 = 0$ (multiply by 3: $-2x^2 -8x -6=0$ , or $2x^2 +8x+6=0$ ), roots $x = -1, -3$ (differ by 2).

Question 12:

Ruth and Racheal have 45 marbles together. After losing 5 marbles each, the product of the number of marbles they have now is 124. Find out how many marbles they had to start with.
a) 31 and 4
b) 36 and 9
c) 31 and 14
d) 36 and 4

Answer: b) 36 and 9

Explanation:

Let initial marbles: Ruth $r$ , Racheal $s$ , with $r + s = 45$ .
After loss: $(r - 5)$ , $(s - 5)$ , product $(r - 5)(s - 5) = 124$ .
Expand:
$(r - 5)(s - 5) = rs - 5(r + s) + 25 = 124$
Substitute $r + s = 45$ :
$rs - 5(45) + 25 = 124 \implies rs - 225 + 25 = 124 \implies rs - 200 = 124 \implies rs = 324$
Solve $r + s = 45$ , $rs = 324$ : quadratic equation $t^2 - 45t + 324 = 0$ .
Discriminant $d = 45^2 - 4 \cdot 1 \cdot 324 = 2025 - 1296 = 729$ ,
$t = \frac{45 \pm \sqrt{729}}{2} = \frac{45 \pm 27}{2}$
$t = \frac{72}{2} = 36 \quad \text{or} \quad t = \frac{18}{2} = 9$
Thus, marbles are 36 and 9.
Verify: After loss, 31 and 4, product $31 \times 4 = 124$ .

Question 13:

Factorize $f(x) = x^3 - 2x^2 - x + 2$
a) (2x-2)(x+1)(x-1)
b) (2x-2)(x+1)(x-2)
c) (x-2)(x+1)(x-1)
d) (2x-1)(x+1)(x-1)

Answer: c) (x-2)(x+1)(x-1)

Explanation:

Possible rational roots: factors of 2 over 1, so $\pm 1, \pm 2$ .
$f(1) = 1 - 2 - 1 + 2 = 0$ , so $(x - 1)$ is a factor.

Synthetic division by

(x - 1)

1 | 1  -2  -1  2
      |     1  -1  -2
    ----------------
      1  -1  -2 | 0

Quotient:

x^2 - x - 2

Factor quotient: $x^2 - x - 2 = (x - 2)(x + 1)$ .
Thus, $f(x) = (x - 1)(x - 2)(x + 1)$ .
Option c) is $(x-2)(x+1)(x-1)$ , same as above.

Question 14:

Find the remainder when $p(x) = 6x^3 + 13x^2 - 4$ is divided by $2x - 3$
a) 40.5
b) 45
c) 45.5
d) 70

Answer: c) 45.5

Explanation:

Remainder theorem: when dividing by $2x - 3$ , root is $x = \frac{3}{2}$ , remainder is $p\left( \frac{3}{2} \right)$ .
Compute:
$p\left( \frac{3}{2} \right) = 6 \left( \frac{3}{2} \right)^3 + 13 \left( \frac{3}{2} \right)^2 - 4 = 6 \cdot \frac{27}{8} + 13 \cdot \frac{9}{4} - 4 = \frac{162}{8} + \frac{117}{4} - 4$
Convert to fractions with denominator 8:
$\frac{162}{8} + \frac{117 \cdot 2}{8} - \frac{4 \cdot 8}{8} = \frac{162}{8} + \frac{234}{8} - \frac{32}{8} = \frac{162 + 234 - 32}{8} = \frac{364}{8} = 45.5$
Thus, remainder is 45.5.

Question 15:

Determine the nature of the roots of a quadratic function $\frac{x}{2} \times \frac{1}{x} = 2$
a) Two distinct real roots
b) Two distinct complex roots
c) One real root
d) No real roots

Answer: d) No real roots

Explanation:

The equation is $\frac{x}{2} \cdot \frac{1}{x} = 2$ , which simplifies to $\frac{1}{2} = 2$ , a contradiction.
Thus, no real solutions.
Note: The equation is not quadratic, but the contradiction implies no real roots.

Question 16:

Equation of $(x+1)^2 \cdot x^2 = 0$ has number of real roots equal to:
a) 1
b) 2
c) 3
d) 4

Answer: d) 4

Explanation:

Equation: $(x+1)^2 x^2 = 0$ .
Roots when $(x+1)^2 = 0$ or $x^2 = 0$ :
- $x = -1$ (multiplicity 2),
- $x = 0$ (multiplicity 2).
Counting multiplicity, there are 4 real roots.
Note: The polynomial is degree 4, and all roots are real with multiplicities.

Question 17:

The sum of two numbers is 27 and product is 182. The numbers are:
a) 12 and 13
b) 13 and 14
c) 12 and 15
d) 13 and 24

Answer: b) 13 and 14

Explanation:

Let numbers be $x$ and $y$ : $x + y = 27$ , $x y = 182$ .
Quadratic equation: $t^2 - 27t + 182 = 0$ .
Discriminant $d = (-27)^2 - 4 \cdot 1 \cdot 182 = 729 - 728 = 1$ ,
$t = \frac{27 \pm \sqrt{1}}{2} = \frac{27 \pm 1}{2}$
$t = \frac{28}{2} = 14 \quad \text{or} \quad t = \frac{26}{2} = 13$
Thus, numbers are 13 and 14.

Question 18:

The roots of quadratic equation $2x^2 + x + 4 = 0$ are:
a) Positive and negative
b) Both Positive
c) Both Negative
d) No real roots

Answer: d) No real roots

Explanation:

Discriminant $d = b^2 - 4ac = 1^2 - 4 \cdot 2 \cdot 4 = 1 - 32 = -31 < 0$ .
Negative discriminant implies no real roots (complex roots).

Question 19:

If one root of equation $4x^2 - 2x + k - 4 = 0$ is reciprocal of the other. The value of $k$ is:
a) -8
b) 8
c) -4
d) 4

Answer: b) 8

Explanation:

Note: The equation is interpreted as $4x^2 - 2x + (k - 4) = 0$ (assuming typo in notation).
Let roots be $r$ and $\frac{1}{r}$ .
Product of roots: $r \cdot \frac{1}{r} = 1 = \frac{c}{a} = \frac{k - 4}{4}$ .
So:
$\frac{k - 4}{4} = 1 \implies k - 4 = 4 \implies k = 8$
Verify: For $k = 8$ , equation $4x^2 - 2x + 4 = 0$ . Roots: $d = (-2)^2 - 4 \cdot 4 \cdot 4 = 4 - 64 = -60 < 0$ (complex), but product is $\frac{4}{4} = 1$ , so if roots exist, they are reciprocals.

Question 20:

Find the turning point of the graph of a quadratic function $f(x) = 2x^4 - 6x + 10$
a) {-8,10}
b) {2,2}
c) {2,10}
d) {-8,2}

Answer: b) {2,2}

Explanation:

Note: The function is assumed to be quadratic, likely $f(x) = 2x^2 - 8x + 10$ (typo in exponent).
For $f(x) = 2x^2 - 8x + 10$ , vertex at $x = -\frac{b}{2a} = -\frac{-8}{2 \cdot 2} = \frac{8}{4} = 2$ .
Then $y = f(2) = 2(2)^2 - 8(2) + 10 = 8 - 16 + 10 = 2$ .
Turning point (vertex) is $(2, 2)$ , so {2, 2}.

Question 21:

What is the line of the symmetry of the graph of a quadratic equation
$f(x) = -x^2 - 5x - 10$
a) $x = 5/2$
b) $x = 2/5$
c) $x = -5/2$
d) $x = -2/5$

Answer: c) $x = -5/2$

Explanation:

Axis of symmetry for $ax^2 + bx + c$ is $x = -\frac{b}{2a}$ .
Here, $a = -1$ , $b = -5$ :
$x = -\frac{-5}{2 \cdot (-1)} = -\frac{5}{-2} = -\frac{5}{2}$
Thus, $x = -\frac{5}{2}$ .

Question 22:

Let $\alpha$ and $\beta$ be the roots of a quadratic equation $f(x) = x^2 + 5x + 6$ . Evaluate $\frac{4}{\alpha} - \frac{4}{\beta}$
a) -2/3 or 2/3
b) 5 or 6
c) -5 or 6
d) -2/5 or 2/5

Answer: a) -2/3 or 2/3

Explanation:

Roots of $x^2 + 5x + 6 = 0$ : factors as $(x+2)(x+3)=0$ , so $\alpha = -2, \beta = -3$ or vice versa.
Compute:
$\frac{4}{\alpha} - \frac{4}{\beta} = 4 \left( \frac{1}{\alpha} - \frac{1}{\beta} \right) = 4 \left( \frac{\beta - \alpha}{\alpha \beta} \right)$
$\alpha + \beta = -5$ , $\alpha \beta = 6$ , $(\beta - \alpha)^2 = (\alpha + \beta)^2 - 4\alpha\beta = 25 - 24 = 1$ , so $|\beta - \alpha| = 1$ .
Thus:
$\frac{\beta - \alpha}{\alpha \beta} = \frac{\pm 1}{6}$
$4 \cdot \frac{\pm 1}{6} = \pm \frac{4}{6} = \pm \frac{2}{3}$
So $-\frac{2}{3}$ or $\frac{2}{3}$ .

Question 23:

Let $\alpha$ and $\beta$ be the roots of a quadratic equation $x^2 + 7x - 3 = 0$ . Find the quadratic equation whose roots are $2\alpha - 1$ and $2\beta - 1$
a) $x^2 - 16x + 3 = 0$
b) $x^2 + 16x - 3 = 0$
c) $x^2 - 16x - 3 = 0$
d) $x^2 + 16x + 3 = 0$

Answer: d) $x^2 + 16x + 3 = 0$

Explanation:

Sum $\alpha + \beta = -7$ , product $\alpha \beta = -3$ .
New roots: $\gamma = 2\alpha - 1$ , $\delta = 2\beta - 1$ .
Sum: $\gamma + \delta = (2\alpha - 1) + (2\beta - 1) = 2(\alpha + \beta) - 2 = 2(-7) - 2 = -14 - 2 = -16$ .
Product: $\gamma \delta = (2\alpha - 1)(2\beta - 1) = 4\alpha\beta - 2\alpha - 2\beta + 1 = 4\alpha\beta - 2(\alpha + \beta) + 1 = 4(-3) - 2(-7) + 1 = -12 + 14 + 1 = 3$ .
Quadratic: $x^2 - (\gamma + \delta)x + \gamma \delta = x^2 - (-16)x + 3 = x^2 + 16x + 3$ .

Question 24:

Determine the values of p for which the quadratic $f(x) = 2x^2 + px + 8 = 0$ has equal roots
a) $(-\infty, -8) \cup (8, \infty)$
b) -8 or 8
c) {-8,8}
d) 8

Answer: c) {-8,8}

Explanation:

Equal roots when discriminant $d = 0$ :
$d = p^2 - 4 \cdot 2 \cdot 8 = p^2 - 64 = 0$
$p^2 = 64 \implies p = \pm 8$
Values are $p = -8$ or $p = 8$ , so set {-8, 8}.

Question 25:

Solve $\left| \frac{2x-3}{x+4} \right| = 3$
a) 3 or 9
b) -3 or 9
c) -3 or -9
d) 3 or -9

Answer: a) 3 or 9

Explanation:

Note: The denominator is assumed to be $x - 4$ (typo in sign) for consistency with options.
Equation: $\left| \frac{2x-3}{x-4} \right| = 3$ .
Cases:
- Case 1: $\frac{2x-3}{x-4} = 3$
  $2x - 3 = 3(x - 4) \implies 2x - 3 = 3x - 12 \implies -3 + 12 = 3x - 2x \implies x = 9$
  Verify: $\left| \frac{18-3}{9-4} \right| = \left| \frac{15}{5} \right| = 3$ .
- Case 2: $\frac{2x-3}{x-4} = -3$
  $2x - 3 = -3(x - 4) \implies 2x - 3 = -3x + 12 \implies 2x + 3x = 12 + 3 \implies 5x = 15 \implies x = 3$
  Verify: $\left| \frac{6-3}{3-4} \right| = \left| \frac{3}{-1} \right| = 3$ .
Solutions: $x = 9$ or $x = 3$ .

@Dr. Microbiota

INTRODUCTION TO MATHEMATICAL METHODS QUIZ TWO JULY INTAKE 2024

DATE : 18th OCTOBER, 2024

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