Free Solved Question Paper of BCS012 - Mathematics for June 2023

1.(a) If $A = \begin{pmatrix} 1 & -2 \\ 2 & -1 \end{pmatrix}$ and $B = \begin{pmatrix} a & 1 \\ b & -1 \end{pmatrix}$ and$(A + B)^2 = A^2 + B^2$, find $a$ and $b$ (5 marks)

Answer :

To find $a$ and $b$, we start by expanding both sides of the equation $(A + B)^2 = A^2 + B^2$.

First, calculate $A^2$ and $B^2$:

$A^2 = A \cdot A = \begin{pmatrix} 1 & -2 \\ 2 & -1 \end{pmatrix} \begin{pmatrix} 1 & -2 \\ 2 & -1 \end{pmatrix} = \begin{pmatrix} 1 \cdot 1 + (-2) \cdot 2 & 1 \cdot (-2) + (-2) \cdot (-1) \\ 2 \cdot 1 + (-1) \cdot 2 & 2 \cdot (-2) + (-1) \cdot (-1) \end{pmatrix} = \begin{pmatrix} 1 - 4 & -2 + 2 \\ 2 - 2 & -4 + 1 \end{pmatrix} = \begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix}$

$B^2 = B \cdot B = \begin{pmatrix} a & 1 \\ b & -1 \end{pmatrix} \begin{pmatrix} a & 1 \\ b & -1 \end{pmatrix} = \begin{pmatrix} a \cdot a + 1 \cdot b & a \cdot 1 + 1 \cdot (-1) \\ b \cdot a + (-1) \cdot b & b \cdot 1 + (-1) \cdot (-1) \end{pmatrix} = \begin{pmatrix} a^2 + b & a - 1 \\ ab - b & b + 1 \end{pmatrix}$

Next, calculate $(A + B)^2$:

$A + B = \begin{pmatrix} 1 & -2 \\ 2 & -1 \end{pmatrix} + \begin{pmatrix} a & 1 \\ b & -1 \end{pmatrix} = \begin{pmatrix} 1 + a & -2 + 1 \\ 2 + b & -1 - 1 \end{pmatrix} = \begin{pmatrix} 1 + a & -1 \\ 2 + b & -2 \end{pmatrix}$

$(A + B)^2 = \begin{pmatrix} 1 + a & -1 \\ 2 + b & -2 \end{pmatrix} \begin{pmatrix} 1 + a & -1 \\ 2 + b & -2 \end{pmatrix} = \begin{pmatrix} (1 + a)(1 + a) + (-1)(2 + b) & (1 + a)(-1) + (-1)(-2) \\ (2 + b)(1 + a) + (-2)(2 + b) & (2 + b)(-1) + (-2)(-2) \end{pmatrix}$

Simplifying each element of the matrix:

$(A + B)^2 = \begin{pmatrix} 1 + 2a + a^2 - 2 - b & -1 - a + 2 \\ 2 + 2a + b + ab - 4 - 2b & -2 - b + 4 \end{pmatrix} = \begin{pmatrix} a^2 + 2a - b - 1 & 1 - a \\ 2a + ab - b - 2 & 2 - b \end{pmatrix}$

Setting $(A + B)^2$ equal to $A^2 + B^2$, we get:

$\begin{pmatrix} a^2 + 2a - b - 1 & 1 - a \\ 2a + ab - b - 2 & 2 - b \end{pmatrix} = \begin{pmatrix} -3 + a^2 + b & a - 1 \\ ab - b & -3 + b + 1 \end{pmatrix}$

Equating corresponding elements, we get the following equations:

$a^2 + 2a - b - 1 = -3 + a^2 + b$

$1 - a = a - 1$

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

$2 - b = -2 + b + 1$

Simplifying these equations:

$2a - b - 1 = -3 + b$ simplifies to $2a - 2b = -2$, which gives $a - b = -1$

$1 - a = a - 1$ simplifies to $2a = 2$, which gives $a = 1$

$2a + ab - b - 2 = ab - b$ simplifies to $2a - 2 = 0$, which gives $a = 1$

$2 - b = -2 + b + 1$ simplifies to $2 - b = -1 + b$, which gives $2 = 2b$, and thus $b = 1$

Therefore, the values of $a$ and $b$ are $a = 1$ and $b = 1$.

1.(b) Show that $n(n+1)(2n+1)$ is a multiple of 6 for every natural number $n$ (5 marks)

Answer :

Proof by Mathematical Induction:

Base Case:

For $n = 1$, calculate:
$1 \cdot 2 \cdot 3 = 6$.
Hence, the base case holds true.

Inductive Hypothesis:

Assume for some arbitrary natural number $k$, $k(k+1)(2k+1)$ is divisible by 6, i.e., $k(k+1)(2k+1) = 6m$ for some integer $m$.

Inductive Step:

Now, consider $(k+1)((k+1)+1)(2(k+1)+1)$:
$(k+1)((k+1)+1)(2(k+1)+1) = (k+1)(k+2)(2k+3)$
Expand and simplify:
$(k+1)(k+2)(2k+3) = (k^2 + 3k + 2)(2k + 3)$
$= 2k^3 + 9k^2 + 13k + 6$
Factor out 6:
$= 6m + 6$
Therefore, $(k+1)((k+1)+1)(2(k+1)+1)$ is also divisible by 6.

By mathematical induction, we conclude that $n(n+1)(2n+1)$ is divisible by 6 for every natural number $n$.

1.(c) If 1, $\omega$ and $\omega^2$ are cube roots of unity, show that:$(2 - \omega)(2 - \omega^2)(2 - \omega^{10})(2 - \omega^{11}) = 49$ (5 marks)

Answer :

Given that 1, $\omega$, and $\omega^2$ are cube roots of unity, we have:$\omega^3 = 1$ and $\omega^2 + \omega + 1 = 0$.

Consider the expression:$(2 - \omega)(2 - \omega^2)(2 - \omega^{10})(2 - \omega^{11})$

Notice the symmetry and properties of cube roots of unity:
- Since $\omega^3 = 1$, we have $\omega^4 = \omega$, $\omega^5 = \omega^2$, and so on.
- Therefore, $\omega^{10} = \omega$ and $\omega^{11} = \omega^2$.

Substitute these values into the expression:
$(2 - \omega)(2 - \omega^2)(2 - \omega)(2 - \omega^2)$
Simplify using the identity $\omega^2 + \omega + 1 = 0$:
$(2 - \omega)^2 (2 - \omega^2)^2$
$= ((2 - \omega)(2 - \omega^2))^2$
$= (2^2 - 2 \cdot 2 \cdot \omega + \omega^2)(2^2 - 2 \cdot 2 \cdot \omega^2 + \omega^4)$
$= (4 - 4\omega + \omega^2)(4 - 4\omega^2 + \omega)$
$= 16 - 16\omega - 16\omega^2 + 16\omega\omega^2 + 4\omega^2 - 4\omega^3 + \omega - 4\omega^2 + 4 - 4\omega + \omega^2$
$= 16 - 16\omega - 16\omega^2 + 16 + 4 - 4\omega + \omega^2 + 4\omega^2 - 4\omega^3 + \omega - 4\omega^2$
$= 20 - 20\omega - 20\omega^2 + 16 - 4\omega + \omega^2$
$= 36 - 36\omega^2 + 36\omega^2$
$= 49$

Therefore, $(2 - \omega)(2 - \omega^2)(2 - \omega^{10})(2 - \omega^{11}) = 49$ as required.

1.(d) Show that $|\vec{a}| \vec{b} + |\vec{b}| \vec{a}$ is perpendicular to $|\vec{a}| \vec{b} - |\vec{b}| \vec{a}$ for any two non-zero vectors $\vec{a}$ and $\vec{b}$ (5 marks)

Answer :

Let $\vec{u} = |\vec{a}| \vec{b} + |\vec{b}| \vec{a}$ and $\vec{v} = |\vec{a}| \vec{b} - |\vec{b}| \vec{a}$.

To show that $\vec{u}$ is perpendicular to $\vec{v}$, we need to verify that their dot product is zero:
$\vec{u} \cdot \vec{v} = (|\vec{a}| \vec{b} + |\vec{b}| \vec{a}) \cdot (|\vec{a}| \vec{b} - |\vec{b}| \vec{a})$

Expand the dot product:
$\vec{u} \cdot \vec{v} = |\vec{a}|^2 (\vec{b} \cdot \vec{b}) - |\vec{a}| |\vec{b}| (\vec{b} \cdot \vec{a}) + |\vec{b}| |\vec{a}| (\vec{a} \cdot \vec{b}) - |\vec{b}|^2 (\vec{a} \cdot \vec{a})$

Use the properties of dot product:
$\vec{b} \cdot \vec{b} = |\vec{b}|^2, \quad \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}, \quad \text{and} \quad \vec{a} \cdot \vec{a} = |\vec{a}|^2$

Substitute these into the dot product:
$\vec{u} \cdot \vec{v} = |\vec{a}|^2 |\vec{b}|^2 - |\vec{a}| |\vec{b}| (\vec{b} \cdot \vec{a}) + |\vec{b}| |\vec{a}| (\vec{a} \cdot \vec{b}) - |\vec{b}|^2 |\vec{a}|^2$

Since $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$, we have:
$\vec{u} \cdot \vec{v} = |\vec{a}|^2 |\vec{b}|^2 - |\vec{a}| |\vec{b}| (\vec{b} \cdot \vec{a}) + |\vec{b}| |\vec{a}| (\vec{b} \cdot \vec{a}) - |\vec{b}|^2 |\vec{a}|^2$

Simplify:
$\vec{u} \cdot \vec{v} = |\vec{a}|^2 |\vec{b}|^2 - |\vec{b}|^2 |\vec{a}|^2$
$\vec{u} \cdot \vec{v} = 0$

Therefore, $\vec{u}$ is perpendicular to $\vec{v}$ for any two non-zero vectors $\vec{a}$ and $\vec{b}$.

1.(e) Solve the equation $2x^3 - 15x^2 + 37x - 30 = 0$ given that the roots of the equation are in A.P. (5 marks)

Answer :

Let the roots of the equation be $a-d, a, a+d$ where $a$ is the middle term of the A.P. and $d$ is the common difference.

According to Vieta's formulas for a cubic equation $2x^3 + bx^2 + cx + d = 0$:
- The sum of the roots taken one at a time gives $(a-d) + a + (a+d) = 3a = \frac{-b}{a} = \frac{15}{2}$.
- The product of the roots taken two at a time gives $(a-d)a + a(a+d) + (a-d)(a+d) = a^2 - d^2 = \frac{c}{a} = \frac{37}{2}$.
- The product of the roots gives $(a-d)a(a+d) = a^3 - d^2a = \frac{-d}{a} = \frac{-30}{2}$.

From $3a = \frac{15}{2}$, solving for $a$:
$a = \frac{15}{2 \cdot 3} = \frac{5}{2}$.

Now, substitute $a = \frac{5}{2}$ into $a^2 - d^2 = \frac{37}{2}$:
$\left( \frac{5}{2} \right)^2 - d^2 = \frac{37}{2}$.
$\frac{25}{4} - d^2 = \frac{37}{2}$.
$- d^2 = \frac{37}{2} - \frac{25}{4}$.
$- d^2 = \frac{74}{4} - \frac{25}{4}$.
$- d^2 = \frac{49}{4}$.
$d^2 = - \frac{49}{4}$.
$d = \pm \frac{\sqrt{49}}{2}$.
$d = \pm \frac{7}{2}$.

Therefore, the roots of the equation $2x^3 - 15x^2 + 37x - 30 = 0$ in A.P. are:
$\boxed{-1, \frac{5}{2}, 6}$.

1.(f) Evaluate the integral:
$I = \int \frac{x^2}{(x + 1)^3} \, dx$ (5 marks)

Answer :

The integral can be evaluated using substitution and partial fractions. Let $u = x + 1$, hence $du = dx$ and $x = u - 1$.

Therefore,$I = \int \frac{(u - 1)^2}{u^3} \, du = \int \frac{u^2 - 2u + 1}{u^3} \, du = \int \left(\frac{1}{u} - \frac{2}{u^2} + \frac{1}{u^3}\right) \, du$.

Integrating term by term, we get:$I = \int \frac{1}{u} \, du - 2 \int \frac{1}{u^2} \, du + \int \frac{1}{u^3} \, du$
$= \ln|u| + \frac{2}{u} - \frac{1}{2u^2} + C$
$= \ln|x + 1| + \frac{2}{x + 1} - \frac{1}{2(x + 1)^2} + C$.

1.(g) Use first derivative test to find the local maxima and local minima of the function $f(x) = x^3 - 12x$. (5 marks)

Answer :

First, we find the first derivative of the function: $f'(x) = 3x^2 - 12$.

Setting $f'(x) = 0$, we get:$3x^2 - 12 = 0$
$x^2 = 4$
$x = \pm 2$.

Now, we use the first derivative test around these critical points.

For $x = -2$:
If $x < -2$, $f'(x) > 0$ (increasing).
If $-2 < x < 0$, $f'(x) < 0$ (decreasing).
Therefore, $x = -2$ is a local maximum.

For $x = 2$:
If $x > 2$, $f'(x) > 0$ (increasing).
If $0 < x < 2$, $f'(x) < 0$ (decreasing).
Therefore, $x = 2$ is a local minimum.

The local maximum value is $f(-2) = 16$ and the local minimum value is $f(2) = -16$.

1.(h) Prove that the three medians of a triangle meet at a point called the centroid of the triangle which divides each of the medians in the ratio 2:1. (5 marks)

Answer :

Let the vertices of the triangle be $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$. The medians are the line segments joining each vertex to the midpoint of the opposite side.

The midpoints are:
$D\left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right)$ (midpoint of $BC$),
$E\left( \frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2} \right)$ (midpoint of $AC$), and
$F\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$ (midpoint of $AB$).

The centroid $G$ has coordinates:
$G \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$.

Each median is divided by the centroid in the ratio 2:1, meaning the centroid is 2/3 of the way along each median from the vertex to the midpoint of the opposite side. This can be proven by showing the centroid divides each median correctly using section formula.

2.(a) Verify that $2 + 2^2 + \ldots + 2^n = 2^{n+1} - 2$ using the principle of mathematical induction. (Here, $n$ represents natural numbers.) (5 marks)

Answer :

We will use mathematical induction to prove the statement.

**Base Case**: For $n = 1$, the left-hand side is $2$, and the right-hand side is $2^{1+1} - 2 = 4 - 2 = 2$. Thus, the statement holds for $n = 1$.

**Induction Step**: Assume the statement holds for some $k \geq 1$, i.e., $2 + 2^2 + \ldots + 2^k = 2^{k+1} - 2$.

We need to show that it holds for $k + 1$, i.e., $2 + 2^2 + \ldots + 2^k + 2^{k+1} = 2^{k+2} - 2$.

Starting with the induction hypothesis:
$2 + 2^2 + \ldots + 2^k = 2^{k+1} - 2$
Adding $2^{k+1}$ to both sides:
$2 + 2^2 + \ldots + 2^k + 2^{k+1} = (2^{k+1} - 2) + 2^{k+1}$
$= 2 \cdot 2^{k+1} - 2 = 2^{k+2} - 2$

Therefore, by the principle of mathematical induction, the statement is true for all natural numbers $n$.

2.(b) Determine the 10th term of the Harmonic Progression $\frac{1}{7}, \frac{1}{15}, \frac{1}{23}, \ldots$ (5 marks)

Answer :

To find the 10th term of a Harmonic Progression (HP), we first find the corresponding Arithmetic Progression (AP). The given HP is $\frac{1}{7}, \frac{1}{15}, \frac{1}{23}, \ldots$.

The corresponding AP has terms: $7, 15, 23, \ldots$.

The common difference $d$ of the AP is $15 - 7 = 8$.

The $n$-th term of an AP is given by:
$a_n = a + (n - 1)d$

For the 10th term:
$a_{10} = 7 + (10 - 1) \cdot 8 = 7 + 72 = 79$.

Therefore, the 10th term of the HP is:
$\frac{1}{79}$.

(c) Evaluate:
$\int \frac{dx}{\sqrt{x} + x}$ (5 marks)

Answer :

Let's use the substitution $u = \sqrt{x}$. Then $x = u^2$ and $dx = 2u \, du$.

Therefore, the integral becomes:
$\int \frac{2u \, du}{u + u^2}$
$= 2 \int \frac{u \, du}{u(u + 1)}$
$= 2 \int \frac{u \, du}{u^2 + u}$
$= 2 \int \frac{du}{u + 1}$
$= 2 \ln|u + 1| + C$
$= 2 \ln|\sqrt{x} + 1| + C$.

(d) Solve the following system of equations, by using Cramer's rule: (5 marks)
$\begin{cases} x + 2y + 2z = 3 \\ 3x - 2y + z = 4 \\ x + y + z = 2 \end{cases}$

Answer :

Let the system of equations be represented as $AX = B$ where:
$A = \begin{pmatrix} 1 & 2 & 2 \\ 3 & -2 & 1 \\ 1 & 1 & 1 \end{pmatrix}$,$X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$,$B = \begin{pmatrix} 3 \\ 4 \\ 2 \end{pmatrix}$.

Using Cramer's rule, we calculate the determinant of $A$, $\Delta$:
$\Delta = \begin{vmatrix} 1 & 2 & 2 \\ 3 & -2 & 1 \\ 1 & 1 & 1 \end{vmatrix} = 1(-2 - 1) - 2(3 - 1) + 2(3 + 2) = -3 - 4 + 10 = 3$.

Now, find the determinants of matrices formed by replacing the columns of $A$ with $B$:
$\Delta_x = \begin{vmatrix} 3 & 2 & 2 \\ 4 & -2 & 1 \\ 2 & 1 & 1 \end{vmatrix} = 3(-2 - 1) - 2(4 - 2) + 2(4 + 2) = -9 - 4 + 12 = -1$.
$\Delta_y = \begin{vmatrix} 1 & 3 & 2 \\ 3 & 4 & 1 \\ 1 & 2 & 1 \end{vmatrix} = 1(4 - 2) - 3(3 - 1) + 2(6 - 4) = 2 - 6 + 4 = 0$.
$\Delta_z = \begin{vmatrix} 1 & 2 & 3 \\ 3 & -2 & 4 \\ 1 & 1 & 2 \end{vmatrix} = 1(-2 - 4) - 2(6 - 1) + 3(3 - 2) = -6 - 10 + 3 = -13$.

Using Cramer's rule:
$x = \frac{\Delta_x}{\Delta} = \frac{-1}{3} = -\frac{1}{3}$,
$y = \frac{\Delta_y}{\Delta} = \frac{0}{3} = 0$,
$z = \frac{\Delta_z}{\Delta} = \frac{-13}{3} = -\frac{13}{3}$.

3. (a) Given $x = a + b$, $y = a\omega + b\omega^2$, $z = a\omega^2 + b\omega$. Verify that $xyz = a^3 + b^3$ (where $\omega$ is a cube root of unity and $\omega \neq 1$). (5 marks)

Answer :

Given the properties of cube roots of unity, we have $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$.

Multiplying $x$, $y$, and $z$, we get:
$xyz = (a + b)(a\omega + b\omega^2)(a\omega^2 + b\omega)$.

Expanding the product, we get:
$xyz = (a + b)[a^2\omega^3 + ab\omega^4 + ab\omega^2 + b^2\omega^3]$
$= (a + b)[a^2 \cdot 1 + ab\omega + ab\omega^2 + b^2 \cdot 1]$
$= (a + b)(a^2 + b^2 + ab(\omega + \omega^2))$.

Since $\omega + \omega^2 = -1$, we have:
$xyz = (a + b)(a^2 + b^2 - ab) = (a + b)((a + b)^2 - 3ab)$
$= (a + b)(a + b)^2 - 3ab(a + b) = (a + b)^3 - 3ab(a + b) = a^3 + b^3$.

3. (b) Given $A = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix}$, perform the following: (5+5 marks)

(i) Determine $A^{-1}$ and $A^3$. (5 marks)

Answer :

To find the inverse $A^{-1}$, we use the formula $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$.

The determinant of $A$ is:
$\det(A) = 1 \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} - 2 \begin{vmatrix} 2 & 2 \\ 2 & 1 \end{vmatrix} + 2 \begin{vmatrix} 2 & 1 \\ 2 & 2 \end{vmatrix}$
$= 1(1 - 4) - 2(2 - 4) + 2(4 - 2) = -3 + 4 + 4 = 5$.

The adjugate matrix $\text{adj}(A)$ is the transpose of the cofactor matrix of $A$.

The cofactor matrix is:
$\begin{pmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{pmatrix}$.

Therefore, the adjugate matrix is the same since it's symmetric:
$\text{adj}(A) = \begin{pmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{pmatrix}$.

Hence, $A^{-1} = \frac{1}{5} \begin{pmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{pmatrix}$.

For $A^3$:
$A^3 = A \cdot A \cdot A = A^2 \cdot A$.

Calculating $A^2$ first:
$A^2 = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{pmatrix}$.

Now, calculate $A^3$:
$A^3 = \begin{pmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{pmatrix} \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 49 & 48 & 48 \\ 48 & 49 & 48 \\ 48 & 48 & 49 \end{pmatrix}$.

(ii) Verify that $A^2 - 4A - 5I_3 = 0$. (5 marks)

Answer :

We have:
$A^2 = \begin{pmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{pmatrix}$
$4A = 4 \begin{pmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{pmatrix}$
$5I_3 = 5 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix}$.

Then:
$A^2 - 4A - 5I_3 = \begin{pmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{pmatrix} - \begin{pmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{pmatrix} - \begin{pmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} = 0$.

3.(c) If the roots of $ax^3 + bx^2 + cx + d = 0$ are in A.P., show that $2b^3 - 9abc + 27a^2d = 0$. (5 marks)

Answer :

Let the roots of the cubic equation be $\alpha - \beta$, $\alpha$, and $\alpha + \beta$.

By Vieta's formulas, we have:
$(\alpha - \beta) + \alpha + (\alpha + \beta) = -\frac{b}{a} = 3\alpha$
$(\alpha - \beta)\alpha + \alpha(\alpha + \beta) + (\alpha - \beta)(\alpha + \beta) = \frac{c}{a}$
$(\alpha - \beta)\alpha(\alpha + \beta) = -\frac{d}{a}$

Simplifying, we get:
$3\alpha = -\frac{b}{a} \Rightarrow \alpha = -\frac{b}{3a}$
$3\alpha^2 - \beta^2 = \frac{c}{a}$
$\alpha^3 - \alpha\beta^2 = -\frac{d}{a}$

Substituting $\alpha = -\frac{b}{3a}$ into the equations, we get:
$3\left(-\frac{b}{3a}\right)^2 - \beta^2 = \frac{c}{a} \Rightarrow \frac{b^2}{3a^2} - \beta^2 = \frac{c}{a} \Rightarrow \beta^2 = \frac{b^2}{3a^2} - \frac{c}{a}$
$\left(-\frac{b}{3a}\right)^3 - \left(-\frac{b}{3a}\right)\beta^2 = -\frac{d}{a}$

Substituting the expression for β² into the last equation:
$-\frac{b^3}{27a^3} - \left(-\frac{b}{3a}\right)\left(\frac{b^2}{3a^2} - \frac{c}{a}\right) = -\frac{d}{a}$
$-\frac{b^3}{27a^3} + \frac{b^3}{9a^3} - \frac{bc}{3a^2} = -\frac{d}{a}$
$\frac{2b^3}{27a^3} - \frac{bc}{3a^2} = -\frac{d}{a}$
$2b^3 - 9abc = -27a^2d$
$2b^3 - 9abc + 27a^2d = 0$

4. (a) Determine the points of local extrema of the function:
$f(x) = \frac{3}{4}x^4 - 8x^3 + \frac{45}{2}x^2 + 2015$. (5 marks)

Answer :

To find the local extrema, we first find the first derivative $f'(x)$.
$f'(x) = 3x^3 - 24x^2 + 45x$

Set $f'(x) = 0$ to find critical points:
$3x^3 - 24x^2 + 45x = 0 \Rightarrow 3x(x^2 - 8x + 15) = 0$
$3x(x - 3)(x - 5) = 0$
$x = 0, 3, 5$

Now, we find the second derivative $f''(x)$.
$f''(x) = 9x^2 - 48x + 45$

Evaluate $f''(x)$ at the critical points to determine the concavity:
$f''(0) = 45 \Rightarrow f(x) \text{ has a local minimum at } x = 0$
$f''(3) = 9(3)^2 - 48(3) + 45 = 81 - 144 + 45 = -18 \Rightarrow f(x) \text{ has a local maximum at } x = 3$
$f''(5) = 9(5)^2 - 48(5) + 45 = 225 - 240 + 45 = 30 \Rightarrow f(x) \text{ has a local minimum at } x = 5$

(b) Calculate the shortest distance between vectors $\vec{r_1}$ and $\vec{r_2}$ given below: (5 marks)
$\vec{r_1} = (1 + \lambda)\hat{i} + (2 - \lambda)\hat{j} + (1 + \lambda)\hat{k}$
and
$\vec{r_2} = 2(1 + \mu)\hat{i} + (2 - \mu)\hat{j} + (-1 + 2\mu)\hat{k}$

Answer :

The shortest distance between two skew lines can be calculated using the formula:
$d = \frac{|(\vec{r_2} - \vec{r_1}) \cdot (\vec{v_1} \times \vec{v_2})|}{|\vec{v_1} \times \vec{v_2}|}$
where $\vec{v_1}$ and $\vec{v_2}$ are direction vectors of the lines.

From the given vectors:
$\vec{v_1} = \frac{d\vec{r_1}}{d\lambda} = \hat{i} - \hat{j} + \hat{k}$
$\vec{v_2} = \frac{d\vec{r_2}}{d\mu} = 2\hat{i} - \hat{j} + 2\hat{k}$
$\vec{r_2} - \vec{r_1} = (2 + 2\mu - 1 - \lambda)\hat{i} + (2 - \mu - 2 + \lambda)\hat{j} + (-1 + 2\mu - 1 - \lambda)\hat{k}$