INTRODUCTION TO MATHEMATICAL METHODS TEST ONE JULY INTAKE 2020

DATE : 9TH OCTOBER, 2020.

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Question 1

(a) Write the power set of $A=\{a,b,\{4,5\left\}\right\}$.
Answer:
$P\left(A\right)=\{\varnothing ,\{a\},\{b\},\{\{4,5\}\},\{a,b\},\{a,\{4,5\}\},\{b,\{4,5\}\},\{a,b,\{4,5\}\left\}\right\}$
Explanation:
The set $A$ has 3 elements: $a$, $b$, and the set $\{4,5\}$. The power set has $2^3=8$ subsets:

• Empty set: $\emptyset$
• Singletons: $\{a\}$, $\{b\}$, $\{\{4,5\}\}$ (since $\{4,5\}$ is a single element)
• Doubletons: $\{a, b\}$, $\{a, \{4,5\}\}$, $\{b, \{4,5\}\}$
• The full set: $\{a, b, \{4,5\}\}$

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(b) Show that $(A \cap B) \cup (A - B) = A$.
Answer:
The equality holds by set theory.
Explanation:

• Step 1: Partition $A$ into elements in $B$ and not in $B$:
  • $A \cap B = \{ x \mid x \in A \text{ and } x \in B \}$
  • $A - B = \{ x \mid x \in A \text{ and } x \notin B \}$
• Step 2: $(A \cap B) \cup (A - B)$ includes all elements of $A$, as every $x \in A$ is either in $B$ (so in $A \cap B$) or not in $B$ (so in $A - B$).
• Step 3: Conversely, if $x \in (A \cap B) \cup (A - B)$, then $x \in A$ by definition.
  Thus, $(A \cap B) \cup (A - B) = A$.

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(c) Prove that $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.
Answer:
The equality holds by set theory.
Explanation:
Let $x$ be an arbitrary element.

• Case 1: $x \in A$
  • Left side: $x \in A \cup (B \cap C)$ since $A \subseteq A \cup (B \cap C)$.
  • Right side: $x \in A \cup B$ and $x \in A \cup C$, so $x \in (A \cup B) \cap (A \cup C)$.
• Case 2: $x \notin A$
  • Left side: $x \in A \cup (B \cap C)$ iff $x \in B \cap C$ (i.e., $x \in B$ and $x \in C$).
  • Right side: $x \in (A \cup B) \cap (A \cup C)$ iff $x \in A \cup B$ and $x \in A \cup C$. Since $x \notin A$, this requires $x \in B$ and $x \in C$.
    Thus, both sides are equal.

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(e) If $A \times B = \{(x, 3), (x, 4), (x, 5), (y, 5), (y, 4), (y, 3)\}$, find $B \times A$.
Answer:
$B \times A = \{(3, x), (3, y), (4, x), (4, y), (5, x), (5, y)\}$
Explanation:

• From $A \times B$:
  • First elements: $x, y$ → $A = \{x, y\}$
  • Second elements: $3, 4, 5$ → $B = \{3, 4, 5\}$
• $B \times A$ is the set of ordered pairs $(b, a)$ where $b \in B$, $a \in A$:
  $B \times A = \{ (3, x), (3, y), (4, x), (4, y), (5, x), (5, y) \}$

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Question 2
(a) Express in the form $\frac{a}{b}$ where $b \neq 0$, and $a, b \in \mathbb{Z}$ in lowest terms.
(i) $-2.6$ (ii) $4.913$
Answer:
(i) $-2.6 = \frac{-13}{5}$
(ii) $4.913 = \frac{4913}{1000}$
Explanation:

• (i) Express $-2.6$ as a fraction in lowest terms

  The decimal $-2.6$ can be written as $-\frac{26}{10}$. Simplifying this fraction by dividing both the numerator and denominator by their greatest common divisor (GCD), which is 2, gives:

  $-\frac{26÷2}{10÷2}=-\frac{13}{5}.$

  The fraction $-\frac{13}{5}$ is in lowest terms because 13 and 5 are coprime (GCD is 1).

  (ii) Express $4.913$ as a fraction in lowest terms

  The decimal $4.913$ has three decimal places, so it can be written as $\frac{4913}{1000}$. To check if this fraction is in lowest terms, find the GCD of 4913 and 1000.

  • Factorize 1000: $1000=2^3\times 5^3$.
  • Factorize 4913: $4913=17\times 17\times 17=17^3$ (since $17^2=289$ and $17\times 289=4913$).

  The prime factors of 4913 are $17^3$, and the prime factors of 1000 are $2^3\times 5^3$. Since there are no common prime factors, the GCD is 1. Thus, $\frac{4913}{1000}$ is already in lowest terms.

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(b) Given operation $*$ on integers: $a * b = a^2 + 2b$.
(i) Is $*$ a binary operation?
Answer:
Yes.
Explanation:

• For integers $a, b$, $a^2$ is an integer (closed under multiplication), $2b$ is an integer, and their sum is an integer. Thus, $*$ is a binary operation.

(ii) Is $*$ commutative?
Answer:
No.
Explanation:

• Test with $a = 1, b = 2$:
  • $1 * 2 = 1^2 + 2 \times 2 = 1 + 4 = 5$
  • $2 * 1 = 2^2 + 2 \times 1 = 4 + 2 = 6$
    Since $5 \neq 6$, not commutative.

(iii) Evaluate:
(a) $(-3) * 5$ (b) $4 * (-2 * 3)$
Answer:
(a) $19$ (b) $36$
Explanation:

• (a) $(-3) * 5 = (-3)^2 + 2 \times 5 = 9 + 10 = 19$
• (b) First, $-2 * 3 = (-2)^2 + 2 \times 3 = 4 + 6 = 10$
  Then, $4 * 10 = 4^2 + 2 \times 10 = 16 + 20 = 36$

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(c) Given $A = (-1, 4]$, $B = [0, 7]$. Find:
(i) $A^c$ (ii) $A \cap B$ (iii) $B - A$
Answer:
(i) $A^c = (-\infty, -1] \cup (4, \infty)$
(ii) $A \cap B = [0, 4]$
(iii) $B - A = (4, 7]$
Explanation:

• Universal set: $\mathbb{R}$.
• (i) $A = (-1, 4] = \{ x \mid -1 < x \leq 4 \}$
  $A^c = \{ x \mid x \leq -1 \text{ or } x > 4 \} = (-\infty, -1] \cup (4, \infty)$
• (ii) $A \cap B = \{ x \mid -1 < x \leq 4 \text{ and } 0 \leq x \leq 7 \} = \{ x \mid 0 \leq x \leq 4 \} = [0, 4]$
• (iii) $B - A = \{ x \in B \mid x \notin A \} = \{ x \mid 0 \leq x \leq 7 \text{ and } (x \leq -1 \text{ or } x > 4) \}$. Since $x \geq 0 > -1$, this simplifies to $\{ x \mid 4 < x \leq 7 \} = (4, 7]$.

Number Line Representation:

• $A^c$: Shaded from $-\infty$ to $-1$ (inclusive) and from $4$ (exclusive) to $\infty$.
• $A \cap B$: Shaded from $0$ (inclusive) to $4$ (inclusive).
• $B - A$: Shaded from $4$ (exclusive) to $7$ (inclusive).
• 

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Question 3
(a) Solve:
(i) $\left| \frac{|2x-3|}{2} \right| = 1$
Answer:
$x = \frac{1}{2}, \frac{5}{2}$
Explanation:

Simplify: $\frac{|2x-3|}{2} = 1$ (since absolute value ≥ 0, the case = -1 is impossible).

So $|2x-3| = 2$. ( WHEN YOU CROSS MULTIPLY)

Cases:

$2x-3 = 2$

→ $2x = 5$

→ $x = \frac{5}{2}$

$2x-3 = -2$

→ $2x = 1$

→ $x = \frac{1}{2}$

(ii) $|-2x - 3| = |x + 1|$
Answer:
$x = -2, -\frac{4}{3}$
Explanation:

Cases:

$-2x - 3 = x + 1$

→ $-3x = 4$

→ $x = -\frac{4}{3}$

$-2x - 3 = -(x + 1)$

→ $-2x - 3 = -x - 1$

→ $-x = 2$

→ $x = -2$

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(b) Solve:
(i) $|2x - 1| < 9$
Answer:
$-4 < x < 5$ or $x \in (-4, 5)$
Explanation:

• The inequality can be rewritten as a compound inequality:

  Add 1 to all parts:

  $$

Divide all parts by 2:

$$

The solution is all $x$ such that .

To verify:

• At $x=0$ (within the interval), .
• At $x=5$ (endpoint), $\mid 2\left(5\right)-1\mid =\mid 10-1\mid =9$, which is not less than 9, so excluded.
• At $x=-4$ (endpoint), $\mid 2(-4)-1\mid =\mid -8-1\mid =\mid -9\mid =9$, which is not less than 9, so excluded.
• At $x=6$ (outside), $\mid 2\left(6\right)-1\mid =\mid 12-1\mid =11>9$.
• At $x=-5$ (outside), $\mid 2(-5)-1\mid =\mid -10-1\mid =\mid -11\mid =11>9$.

Thus, the solution is .

$-4 < x < 5$

(ii) $|3x + 1| \geq |2x + 3|$
Answer:
$x \leq -\frac{4}{5}$ or $x \geq 2$ or $x \in (-\infty, -\frac{4}{5}] \cup [2, \infty)$
Explanation:

• Square both sides (valid as both ≥ 0):
  $(3x + 1)^2 \geq (2x + 3)^2$
  $9x^2 + 6x + 1 \geq 4x^2 + 12x + 9$
  $5x^2 - 6x - 8 \geq 0$
• Solve $5x^2 - 6x - 8 = 0$:
  $$
• $x=\frac{6\pm \sqrt{(-6{)}^2-4\cdot 5\cdot (-8)}}{10}$
• $=\frac{6\pm \sqrt{196}}{10}$
• $x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 5 \cdot (-8)}}{10} = \frac{6 \pm \sqrt{196}}{10} = \frac{6 \pm 14}{10}$
  $x = 2, -\frac{8}{10} = -\frac{4}{5}$
• Parabola $5x^2 - 6x - 8$ opens upwards, so ≥ 0 when $x \leq -\frac{4}{5}$ or $x \geq 2$.

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Question 4
(a) Domain and Range:
(i) $y = 5x - 7$
Answer:

Explanation

The function $y=5x-7$ is a linear function with no restrictions (such as division by zero or square roots of negative numbers). Therefore:

• Domain: The set of all real numbers for which the function is defined. Since there are no restrictions, the domain is all real numbers, denoted in interval notation as $(-\infty ,\infty )$.

• Range: The set of all possible output values. As a linear function with a non-zero slope (slope = 5), the function can produce any real number as output. Thus, the range is also all real numbers, denoted as $(-\infty ,\infty )$.

$\overline{)\text{Domain: }(-\infty ,\infty )\text{Range: }(-\infty ,\infty )}$

• Polynomial → domain all reals. Linear function → range all reals.

(ii) $y = x^2 - 3x + 2$

Answer:
Domain: $\mathbb{R}$, Range: $[-\frac{1}{4}, \infty)$
Explanation:

• Domain: all reals (polynomial).
• Vertex at $x = \frac{3}{2}$:
  $y = \left(\frac{3}{2}\right)^2 - 3 \cdot \frac{3}{2} + 2 = \frac{9}{4} - \frac{9}{2} + 2 = \frac{9 - 18 + 8}{4} = -\frac{1}{4}$
• Parabola opens upwards → range $[-\frac{1}{4}, \infty)$.

OR SAY

The function $y=x^2-3x+2$ is a quadratic polynomial.

Domain:

Polynomial functions are defined for all real numbers. Therefore, the domain is $(-\infty ,\infty )$.

Range:

The quadratic function $y=x^2-3x+2$ has a positive leading coefficient (coefficient of $x^2$ is 1), so the parabola opens upwards. The vertex represents the minimum point.

The x-coordinate of the vertex is given by $x=-\frac{b}{2a}$, where $a=1$ and $b=-3$:

$x=-\frac{-3}{2\cdot 1}=\frac{3}{2}=\mathrm{1.5.}$

Substitute $x=1.5$ into the function to find the y-coordinate:

$y=(1.5{)}^2-3(1.5)+2=2.25-4.5+2=-0.25=-\frac{1}{4}.$

Since the parabola opens upwards, the minimum value of $y$ is $-\frac{1}{4}$, and

$y$ increases without bound as $x$ approaches $\pm \infty$. Therefore, the range is $[-\frac{1}{4},\infty )$.

$\overline{)\text{Domain: }(-\infty ,\infty )\text{Range: }[-\frac{1}{4},\infty )}$

(iii) $f(x) = \sqrt{16 - x^2}$
Answer:
Domain: $[-4, 4]$, Range: $[0, 4]$
Explanation:

$16 - x^2 \geq 0$

→ $x^2 \leq 16$

→ $-4 \leq x \leq 4$.

$\sqrt{\cdot} \geq 0$; max at $x = 0$: $f(0) = 4$; min at $x = \pm 4$: $f(\pm 4) = 0$.

OR USE THIS BELOW

Domain

The function is $f\left(x\right)=\sqrt{16-x^2}$. The expression under the square root, $16-x^2$, must be non-negative for the function to be defined. Therefore:

$16-x^2\ge 0$

Solving the inequality:

$x^2\le 16$ $$

$\mid x\mid \le 4$ $$

$-4\le x\le 4$

Thus, the domain is the closed interval $[-4,4]$.

Range

The square root function outputs non-negative values, so $f\left(x\right)\ge 0$. The expression $16-x^2$ ranges from 0 to 16 as $x$ varies over $[-4,4]$:

• When $x=\pm 4$, $f(\pm 4)=\sqrt{16-16}=\sqrt{0}=0$.
• When $x=0$, $f\left(0\right)=\sqrt{16-0}=\sqrt{16}=4$.

Since $f\left(x\right)$ is continuous, and it achieves all values between 0 and 4

(e.g., for $f\left(x\right)=2$, solve $\sqrt{16-x^2}=2$, which gives $16-x^2=4$,

so $x^2=12$, and $x=\pm 2\sqrt{3}$, which are in $[-4,4]$), the range is the closed interval $[0,4]$.

Final Answer

$\overline{)\text{Domain: }[-4,4]\text{Range: }[0,4]}$

(iv) $f(x) = \frac{3}{x-3}$
Answer:
Domain: $\mathbb{R} \setminus \{3\}$, Range: $\mathbb{R} \setminus \{0\}$
Explanation:

• Denominator ≠ 0 → domain $x \neq 3$.
• Solve for $x$: $y = \frac{3}{x-3}$ → $x = 3 + \frac{3}{y}$, defined for all $y \neq 0$.

OR USE THIS BELOW

Domain

The function is $f\left(x\right)=\frac{3}{x-3}$. The denominator $x-3$ cannot be zero, as division by zero is undefined. Solving $x-3=0$ gives $x=3$. Thus, the domain excludes $x=3$ and includes all other real numbers. In interval notation, the domain is $(-\infty ,3)\cup (3,\infty )$.

Range

To find the range, solve for $x$ in terms of $y$, where $y=f\left(x\right)$:

$y=\frac{3}{x-3}$

Rearranging gives:

$y(x-3)=3$

$\mathrm{}$ $yx-3y=3$

$\mathrm{}$ $yx=3+3y$

$$ $x=\frac{3+3y}{y}=3(1+\frac{1}{y}),y\ne 0$

The value $y=0$ is not possible because $\frac{3}{x-3}=0$ implies $3=0$, which is a contradiction. For any $y\ne 0$, the corresponding $x$ is defined and $x\ne 3$ (since $x=3+\frac{3}{y}$ and $\frac{3}{y}\ne 0$). Thus, the range is all real numbers except zero. In interval notation, the range is $(-\infty ,0)\cup (0,\infty )$.

Verification

• As $x$ approaches 3 from the right, $f\left(x\right)$ approaches $\infty$.
• As $x$ approaches 3 from the left, $f\left(x\right)$ approaches $-\infty$.
• As $x$ approaches $\infty$ or $-\infty$, $f\left(x\right)$ approaches 0 but never reaches it.
• The function takes all other real values (e.g., $f\left(6\right)=1$, $f\left(0\right)=-1$, $f\left(9\right)=0.5$).

$\overline{)\text{Domain: }(-\infty ,3)\cup (3,\infty )\text{Range: }(-\infty ,0)\cup (0,\infty )}$

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(b) (i) Show $f(x) = \frac{x+3}{x-2}$ is one-to-one.

Answer:
Assume $f(a) = f(b)$:
$\frac{a+3}{a-2} = \frac{b+3}{b-2}$
Cross-multiply:
$(a+3)(b-2) = (b+3)(a-2)$
$ab - 2a + 3b - 6 = ab - 2b + 3a - 6$
$-2a + 3b = 3a - 2b$
$5b = 5a$
$a = b$
Thus, injective (one-to-one).

OR USE THIS BELOW

To show that the function $f\left(x\right)=\frac{x+3}{x-2}$ is one-to-one, it must be proven that if $f\left(a\right)=f\left(b\right)$, then $a=b$, for all $a$ and $b$ in the domain of $f$.

The domain of $f$ is all real numbers except $x=2$, since the denominator cannot be zero. Thus, $a\ne 2$ and $b\ne 2$.

Assume $f\left(a\right)=f\left(b\right)$. Then:

$\frac{a+3}{a-2}=\frac{b+3}{b-2}$

Cross-multiplying gives:

$(a+3)(b-2)=(b+3)(a-2)$

Expanding both sides:

• Left side: $(a+3)(b-2)=ab-2a+3b-6$
• Right side: $(b+3)(a-2)=ab-2b+3a-6$

So the equation is:

$ab-2a+3b-6=ab-2b+3a-6$

Subtract $ab$ and add 6 to both sides:

$-2a+3b=-2b+3a$

Bring all terms to one side:

$-2a+3b+2b-3a=0$ $-5a+5b=0$

Factor out 5:

$5(-a+b)=0$ $-a+b=0$ $b=a$

Thus, $f\left(a\right)=f\left(b\right)$ implies $a=b$, which proves that $f$ is one-to-one.

(ii) Sketch $y = 2 - \sqrt{x - 1}$
Answer:

• Domain: $x \geq 1$
• Key Points:
  • $x = 1$: $y = 2 - \sqrt{0} = 2$ → point (1,2)
  • $x = 2$: $y = 2 - \sqrt{1} = 1$ → point (2,1)
  • $x = 5$: $y = 2 - \sqrt{4} = 0$ → point (5,0)
  • $x = 10$: $y = 2 - \sqrt{9} = -1$ → point (10,-1)
• Shape: Starts at (1,2), decreases slowly then rapidly. Reflection of $y = \sqrt{x}$ over x-axis, shifted right 1, up 2.

Graph Sketch:

• Axes: x from 0 to 11, y from -2 to 3.
• Curve begins at (1,2), passes through (2,1), (5,0), (10,-1), decreasing and concave up.
• 

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Question 5
(a) (i) Given $f(x) = 4x - 3$, $g(x) = \frac{x+3}{4}$, solve $f \circ g(x) = g \circ f(x) = x$.
Answer:
Holds for all $x \in \mathbb{R}$.
Explanation:

$f\circ g(x)=f\left(\frac{x+3}{4}\right)=4X\frac{x+3}{4}-3$

$=x+3-3$

f∘g(x)= x$f \circ g(x) = f\left( \frac{x+3}{4} \right) = 4 \cdot \frac{x+3}{4} - 3 = x + 3 - 3 = x$

$g\circ f(x)=g(4x-3)=\frac{(4x-3)+3}{4}$

$=\frac{4x}{4}$

$g \circ f(x) = g(4x - 3) = \frac{(4x - 3) + 3}{4} = \frac{4x}{4} = x$
Thus, $f \circ g(x) = g \circ f(x) = x$ for all real $x$.

(ii) Given $f(x) = 3x^2 + x - 2$, $g(x) = x + 1$, solve $f \circ g(x) = 0$.
Answer:
$x = -2, -\frac{1}{3}$
Explanation:

$f\circ g(x)=f(x+1)=3(x+1{)}^2+(x+1)-2$

$=3(x^2+2x+1)+x+1-2$

$=3x^2+6x+3+x-1$

$f \circ g(x) = f(x + 1) = 3(x + 1)^2 + (x + 1) - 2 = 3(x^2 + 2x + 1) + x + 1 - 2 = 3x^2 + 6x + 3 + x - 1 = 3x^2 + 7x + 2$

Set equal to 0:
$3x^2 + 7x + 2 = 0$

$x=\frac{-7\pm \sqrt{49-24}}{6}$

$x = \frac{ -7 \pm \sqrt{49 - 24} }{6} = \frac{ -7 \pm 5 }{6}$
$x = \frac{-2}{6} = -\frac{1}{3}, \quad x = \frac{-12}{6} = -2$

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(b) Find inverse of $f(x) = \sqrt{x}$ ($x \geq 0$). Sketch $f$ and $f^{-1}$.
Answer:
Inverse: $f^{-1}(x) = x^2$ ($x \geq 0$)
Explanation:

• Let $y = \sqrt{x}$ → $y \geq 0$, $x = y^2$.
• Thus, $f^{-1}(x) = x^2$ for $x \geq 0$.
• Graphs:
  • $f(x) = \sqrt{x}$: Starts at (0,0), passes through (1,1), (4,2). Concave down.
  • $f^{-1}(x) = x^2$: Starts at (0,0), passes through (1,1), (2,4). Concave up.
• Symmetric about line $y = x$.

Graph Sketch:

• Axes: x and y from 0 to 5.
• $y = \sqrt{x}$: Curve from (0,0) to (4,2).
• $y = x^2$: Curve from (0,0) to (2,4).
• Line $y = x$: Diagonal.
• 

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Question 6
(a) Find $k$ such that $kx^2 + 4x + (3 + k) = 0$ has no real roots.
Answer:
$k < -4$ or $k > 1$ or $k \in (-\infty, -4) \cup (1, \infty)$
Explanation:

• Discriminant $D < 0$:
  $D = 4^2 - 4 \cdot k \cdot (k + 3) = 16 - 4k^2 - 12k$
• Set $D < 0$:
  $16 - 4k^2 - 12k < 0$
  $-4k^2 - 12k + 16 < 0$
  Multiply by -1 (reverse inequality):
  $4k^2 + 12k - 16 > 0$
  Divide by 4:
  $k^2 + 3k - 4 > 0$
• Roots: $k = \frac{ -3 \pm \sqrt{9 + 16} }{2} = \frac{ -3 \pm 5 }{2}$ → $k = 1, -4$
• Parabola opens upwards → positive when $k < -4$ or $k > 1$.

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(b) If $\alpha, \beta$ are roots of $x^2 + 2x - 3 = 0$, find $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$.
Answer:
$-\frac{10}{3}$
Explanation:

Sum $\alpha + \beta = -2$, product $\alpha \beta = -3$.

$\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=\frac{{\alpha }^2+{\beta }^2}{\alpha \beta }$

$=\frac{(\alpha +\beta {)}^2-2\alpha \beta }{\alpha \beta }$

$=\frac{(-2{)}^2-2(-3)}{-3}$

$=\frac{4+6}{-3}$

$=\frac{10}{-3}$

$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \beta} = \frac{ (\alpha + \beta)^2 - 2\alpha \beta }{ \alpha \beta } = \frac{ (-2)^2 - 2(-3) }{ -3 } = \frac{4 + 6}{-3} = \frac{10}{-3} = -\frac{10}{3}$

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Question 7
(a) Expression $3x^3 + 2x^2 - b x + a$ divisible by $(x-1)$, remainder 10 when divided by $(x+1)$. Find $a, b$.
Answer:
$a = 3$, $b = 8$
Explanation:

• Divisible by $(x-1)$ → at $x = 1$:
  $3(1)^3 + 2(1)^2 - b(1) + a = 3 + 2 - b + a = 5 - b + a = 0 \quad \text{(1)}$
• Remainder 10 when divided by $(x+1)$ → at $x = -1$:
  $3(-1)^3 + 2(-1)^2 - b(-1) + a = -3 + 2 + b + a = -1 + a + b = 10 \quad \text{(2)}$
• Solve system:
  From (1): $a - b = -5$
  From (2): $a + b = 11$
  Add: $2a = 6$ → $a = 3$
  Substitute: $3 + b = 11$ → $b=8$
• $\mathrm{}$

$\mathrm{OR\; USE\; THIS\; BELOW}$

Let’s solve this step by step.

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🧮 Given:

The polynomial:

$p\left(x\right)=3x^3+2x^2-bx+a$

• It is divisible by $(x-1{)}^2$, so:

  • $p\left(1\right)=0$
  • $p^{\prime }\left(1\right)=0$

• It leaves remainder 10 when divided by $(x+1{)}^2$, so:

  • $p(-1)=10$

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✏️ Step 1: Use $p\left(1\right)=0$

Substitute $x=1$:

$p\left(1\right)=3(1{)}^3+2(1{)}^2-b\left(1\right)+a=3+2-b+a=5-b+a$

Set equal to 0:

$5-b+a=0\Rightarrow a=b-5\text{(Equation ①)}$

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✏️ Step 2: Use $p^{\prime }\left(1\right)=0$

Differentiate:

$p^{\prime }\left(x\right)=9x^2+4x-b$

Substitute $x=1$:

$p^{\prime }\left(1\right)=9(1{)}^2+4(1)-b=9+4-b=13-b$

Set equal to 0:

$13-b=0\Rightarrow b=13\text{(Equation ②)}$

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✏️ Step 3: Use $p(-1)=10$

Substitute $x=-1$:

$p(-1)=3(-1{)}^3+2(-1{)}^2-b(-1)+a=-3+2+b+a=-1+b+a$

Set equal to 10:

$-1+b+a=10\Rightarrow b+a=11\text{(Equation ③)}$

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✅ Final Step: Solve system

From Equation ②: $b=13$
Substitute into Equation ①:

$a=13-5=8$

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📌 Final Answer:

$\overline{)a=8,b=13}$

$b = 8$

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(b) Given $f(x) = x^3 - x^2 - 4x + 4$.
(i) Solve $f(x) < 0$.
Answer:
$x < -2$ or $1 < x < 2$ or $x \in (-\infty, -2) \cup (1, 2)$
Explanation:

• Factorize:
  • Root $x = 1$: $f(1) = 1 - 1 - 4 + 4 = 0$
  • Synthetic division:
    $\begin{array}{c|cccc} 1 & 1 & -1 & -4 & 4 \\ \hline & & 1 & 0 & -4 \\ \end{array}$
    → $f(x) = (x - 1)(x^2 - 4) = (x - 1)(x - 2)(x + 2)$
• Sign analysis:

  Interval	$(-\infty, -2)$	$(-2, 1)$	$(1, 2)$	$(2, \infty)$
  $(x + 2)$	-	+	+	+
  $(x - 1)$	-	-	+	+
  $(x - 2)$	-	-	-	+
  Product	-	+	-	+

• $f(x) < 0$ when negative: $(-\infty, -2)$ and $(1, 2)$.

OR USE THIS BELOW

To solve the inequality

we’ll follow a structured approach:

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✏️ Step 1: Factor the cubic expression

We try rational root theorem or factor by grouping:

Group terms:

$f\left(x\right)=(x^3-x^2)-(4x-4)=x^2(x-1)-4(x-1)$

Factor:

$f\left(x\right)=(x-1)(x^2-4)=(x-1)(x-2)(x+2)$

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📌 Step 2: Solve the inequality

We want:

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📊 Step 3: Sign chart analysis

Critical points: $x=-2,1,2$

Test intervals:

Interval	Test Point	Sign of Expression
	$x=-3$	$(-)(-)(-)$ = –
	$x=0$	$(-)(-)(+)$ = +
	$x=1.5$	$(+)(-)(+)$ = –
$x>2$	$x=3$	$(+)(+)(+)$ = +

We want where the expression is negative:

• $1<x<2$

✅ Final Answer:

$\overline{)x\in (-\infty ,-2)\cup (1,2)}$

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(ii) Sketch graph of $f(x)$.
Explanation:

• Cubic, positive leading coefficient.
• Roots at $x = -2, 1, 2$.
• Behavior:
  • As $x \to \infty$, $f(x) \to \infty$; $x \to -\infty$, $f(x) \to -\infty$.
  • Local max/min: Use sign chart or derivative (not required), but from (i):
    • Positive in $(-2, 1) \cup (2, \infty)$
    • Negative in $(-\infty, -2) \cup (1, 2)$
• Sketch:
  • Crosses x-axis at -2, 1, 2.
  • From left: below x-axis, crosses up at x=-2, above in (-2,1), crosses down at x=1, below in (1,2), crosses up at x=2, above for x>2.

• ✅ Roots clearly marked at $x=-2,1,2$
• 📈 Positive regions shaded green: $(-2,1)\cup (2,\infty )$
• 📉 Negative regions shaded orange: $(-\infty ,-2)\cup (1,2)$
• 🔁 Cubic behavior: rises to $+\infty$ as $x\to \infty$, falls to $-\infty$ as $x\to -\infty$

You’ll see the function:

• Starts below the x-axis, crosses up at $x=-2$
• Stays positive until $x=1$, then dips negative until $x=2$
• Rises again above the axis for $x>2$

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(c) Partial fractions:
(i) $\frac{3x^2 - 3x - 1}{(x^2 - 1)(x - 1)}$
Answer:
$\frac{7}{4(x-1)} - \frac{1}{2(x-1)^2} + \frac{5}{4(x+1)}$
Explanation:

• Denominator: $(x^2 - 1)(x - 1) = (x-1)^2 (x+1)$
• Set up:
  $\frac{3x^2 - 3x - 1}{(x-1)^2 (x+1)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1}$
• Multiply by denominator:
  $3x^2 - 3x - 1 = A(x-1)(x+1) + B(x+1) + C(x-1)^2$
• Expand:
  $A(x^2 - 1) + B(x + 1) + C(x^2 - 2x + 1) = (A + C)x^2 + (B - 2C)x + (-A + B + C)$
• Equate coefficients:
  • $A + C = 3$
  • $B - 2C = -3$
  • $-A + B + C = -1$
• Solve:
  From (1): $C = 3 - A$
  From (2): $B = -3 + 2(3 - A) = 3 - 2A$
  Plug into (3): $-A + (3 - 2A) + (3 - A) = -1$ → $-4A + 6 = -1$ → $-4A = -7$ → $A = \frac{7}{4}$
  Then $C = 3 - \frac{7}{4} = \frac{5}{4}$, $B = 3 - 2 \cdot \frac{7}{4} = 3 - \frac{7}{2} = -\frac{1}{2}$
• Thus:
  $\frac{7/4}{x-1} + \frac{-1/2}{(x-1)^2} + \frac{5/4}{x+1} = \frac{7}{4(x-1)} - \frac{1}{2(x-1)^2} + \frac{5}{4(x+1)}$

OR USE THIS BELOW

Let’s break this down and express the rational function

$\frac{3x^2-3x-1}{(x^2-1)(x-1)}$

as partial fractions.

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✏️ Step 1: Factor the denominator

We have:

• $x^2-1=(x-1)(x+1)$

So the full denominator becomes:

$(x-1{)}^2(x+1)$

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🧩 Step 2: Set up partial fractions

Since the denominator has a repeated linear factor $(x-1{)}^2$ and a distinct linear factor $(x+1)$, we write:

$\frac{3x^2-3x-1}{(x-1{)}^2(x+1)}=\frac{A}{x-1}+\frac{B}{(x-1{)}^2}+\frac{C}{x+1}$

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🧮 Step 3: Multiply both sides by the denominator

Multiply through by $(x-1{)}^2(x+1)$:

$3x^2-3x-1=A(x-1)(x+1)+B(x+1)+C(x-1{)}^2$

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✏️ Step 4: Choose smart values for $x$

Let $x=1$:

$3(1{)}^2-3(1)-1=A(0\left)\right(2)+B(2)+C(0{)}^2$

$\Rightarrow -1=2B$

$\Rightarrow B=-\frac{1}{2}$

Let $x=-1$:

$3(-1{)}^2-3(-1)-1=A(-2\left)\right(0)+B(0)+C(-2{)}^2$

$\Rightarrow 3+3-1=4C$

$\Rightarrow C=\frac{5}{4}$

Let $x=0$:

$0-0-1=A(-1)\left(1\right)+B\left(1\right)+C(1{)}^2$

$\Rightarrow -1=-A+B+C$

Substitute known values:

$-1=-A-\frac{1}{2}+\frac{5}{4}$

$\Rightarrow -1=-A+\frac{3}{4}$

$\Rightarrow -A=-\frac{7}{4}$

$\Rightarrow A=\frac{7}{4}$

$\frac{\mathrm{}}{}$

✅ Final Answer:

$\overline{)\frac{3x^2-3x-1}{(x^2-1)(x-1)}=\frac{7}{4(x-1)}-\frac{1}{2(x-1{)}^2}+\frac{5}{4(x+1)}}$

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(ii) $\frac{x^2 + x + 2}{(x^2 + 1)(x + 1)}$
Answer:
$\frac{1}{x+1} + \frac{1}{x^2 + 1}$
Explanation:

• Set up:
  $\frac{x^2 + x + 2}{(x^2 + 1)(x + 1)} = \frac{Ax + B}{x^2 + 1} + \frac{C}{x+1}$
• Multiply by denominator:
  $x^2 + x + 2 = (Ax + B)(x + 1) + C(x^2 + 1)$
• Expand:
  $(Ax + B)(x + 1) = Ax^2 + Ax + Bx + B$
  $+ Cx^2 + C = (A + C)x^2 + (A + B)x + (B + C)$
• Equate coefficients:
  • $A + C = 1$
  • $A + B = 1$
  • $B + C = 2$
• Solve:
  From (1): $C = 1 - A$
  From (2): $B = 1 - A$
  Plug into (3): $(1 - A) + (1 - A) = 2$ → $2 - 2A = 2$ → $-2A = 0$ → $A = 0$
  Then $B = 1$, $C = 1$
• Thus:
  $\frac{0\cdot x+1}{x^2+1}+\frac{1}{x+1}=\frac{1}{x+1}+\frac{1}{x^2+1}$

OR USE THIS BELOW

$\frac{0 \cdot x + 1}{x^2 + 1} + \frac{1}{x+1} = \frac{1}{x+1} + \frac{1}{x^2 + 1}$

Let’s decompose the rational expression

$\frac{x^2+x+2}{(x^2+1)(x+1)}$

into partial fractions.

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✏️ Step 1: Denominator structure

• $x^2+1$ is irreducible over the reals.
• $x+1$ is a linear factor.

So we use the form:

$\frac{x^2+x+2}{(x^2+1)(x+1)}=\frac{Ax+B}{x^2+1}+\frac{C}{x+1}$

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🧮 Step 2: Multiply both sides by the denominator

Multiply through by $(x^2+1)(x+1)$:

$x^2+x+2=(Ax+B)(x+1)+C(x^2+1)$

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✏️ Step 3: Expand both sides

Expand RHS:

• $(Ax+B)(x+1)=A{x}^2+Ax+Bx+B=A{x}^2+(A+B)x+B$
• $C(x^2+1)=C{x}^2+C$

Add both:

$RHS=(A+C)x^2+(A+B)x+(B+C)$

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🧩 Step 4: Match coefficients

Compare:

$x^2+x+2=(A+C)x^2+(A+B)x+(B+C)$

Match terms:

• $A+C=1$   (①)
• $A+B=1$   (②)
• $B+C=2$   (③)

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🧮 Step 5: Solve the system

From (①): $C=1-A$
From (②): $B=1-A$
Substitute into (③):

$(1-A)+(1-A)=2\Rightarrow 2-2A=2\Rightarrow A=0$

Then:

• $B=1$
• $C=1$

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✅ Final Answer:

$\overline{)\frac{x^2+x+2}{(x^2+1)(x+1)}=\frac{1}{x^2+1}+\frac{1}{x+1}}$

@Dr. Microbiota

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END OF SOLUTIONS