Sample Mathematics Solutions#
Q.1 [14 marks]#
Fill in the blanks using appropriate choice from the given options
Q1.1 [1 mark]#
If $A_{2×3}$ and $B_{3×4}$ are two matrices then find order of $AB$ = ______
a. $4×2$ b. $2×4$ c. $3×3$ d. AB is not possible
Answer: a. $2×4$
Solution: For matrix multiplication $AB$ to be possible, the number of columns in matrix $A$ must equal the number of rows in matrix $B$.
Given:
- Matrix $A$ has order $2×3$ (2 rows, 3 columns)
- Matrix $B$ has order $3×4$ (3 rows, 4 columns)
Since the number of columns in $A$ (3) equals the number of rows in $B$ (3), multiplication is possible.
The order of the resultant matrix $AB$ will be: $$AB_{(2×3)} \times B_{(3×4)} = (AB)_{(2×4)}$$
Therefore, the order of $AB$ is $2×4$.
Q1.2 [1 mark]#
If $A = [1\quad 3\quad 2]$ and $B = \begin{bmatrix} 1 \ 2 \ 1 \end{bmatrix}$ then find $AB$ = ______
a. Not possible b. $9$ c. $[1\quad 1]$ d. $[1\quad 6\quad 2]$
Answer: b. $9$
Solution: Given: $$A = [1\quad 3\quad 2] \text{ (order: } 1×3\text{)}$$ $$B = \begin{bmatrix} 1 \ 2 \ 1 \end{bmatrix} \text{ (order: } 3×1\text{)}$$
Since $A$ is $1×3$ and $B$ is $3×1$, multiplication is possible and the result will be $1×1$ (a scalar).
$$AB = [1\quad 3\quad 2] \begin{bmatrix} 1 \ 2 \ 1 \end{bmatrix}$$
$$AB = (1)(1) + (3)(2) + (2)(1)$$ $$AB = 1 + 6 + 2 = 9$$
Q1.3 [1 mark]#
If $A \cdot I_2 = A$ then $I_2$ = ______
a. $\begin{bmatrix} 0 & 1 \ 1 & 0 \end{bmatrix}$ b. $\begin{bmatrix} 1 & 0 \ 1 & 0 \end{bmatrix}$ c. $\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$ d. $\begin{bmatrix} 0 & 0 \ 0 & 1 \end{bmatrix}$
Answer: c. $\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$
Solution: The condition $A \cdot I_2 = A$ means that $I_2$ is the identity matrix of order $2×2$.
The identity matrix has the property that when any matrix is multiplied by it, the original matrix remains unchanged.
The $2×2$ identity matrix is: $$I_2 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$$
Verification: For any $2×2$ matrix $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$:
$$A \cdot I_2 = \begin{bmatrix} a & b \ c & d \end{bmatrix} \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} a & b \ c & d \end{bmatrix} = A$$
Calculus Problems#
Q1.4 [1 mark]#
If $\frac{d}{dx}(\sin^2 x + \cos^2 x) = $ ______
a. $1$ b. $0$ c. $-1$ d. $x$
Answer: b. $0$
Solution: We know that $\sin^2 x + \cos^2 x = 1$ (fundamental trigonometric identity).
Since the derivative of a constant is zero: $$\frac{d}{dx}(\sin^2 x + \cos^2 x) = \frac{d}{dx}(1) = 0$$
Alternative approach (step by step): $$\frac{d}{dx}(\sin^2 x + \cos^2 x)$$ $$= \frac{d}{dx}(\sin^2 x) + \frac{d}{dx}(\cos^2 x)$$ $$= 2\sin x \cos x + 2\cos x(-\sin x)$$ $$= 2\sin x \cos x - 2\sin x \cos x = 0$$
Integration Example#
Evaluate: $\int x^5 dx$
Solution: Using the power rule for integration: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ where $n \neq -1$
$$\int x^5 dx = \frac{x^{5+1}}{5+1} + C = \frac{x^6}{6} + C$$
Complex Numbers#
Sample Problem#
Simplify: $(3+4i)(4-5i)$
Solution: $$(3+4i)(4-5i)$$ $$= 3(4) + 3(-5i) + 4i(4) + 4i(-5i)$$ $$= 12 - 15i + 16i - 20i^2$$
Since $i^2 = -1$: $$= 12 - 15i + 16i - 20(-1)$$ $$= 12 - 15i + 16i + 20$$ $$= 32 + i$$
Differential Equations#
Sample Problem#
Solve: $x\frac{dy}{dx} + y = 0$
Solution: This is a separable differential equation.
Rearranging: $x\frac{dy}{dx} = -y$
Separating variables: $\frac{dy}{y} = -\frac{dx}{x}$
Integrating both sides: $$\int \frac{dy}{y} = \int -\frac{dx}{x}$$ $$\ln|y| = -\ln|x| + C_1$$ $$\ln|y| = \ln|x^{-1}| + C_1$$ $$|y| = e^{\ln|x^{-1}| + C_1} = e^{C_1} \cdot |x^{-1}|$$
Let $C = \pm e^{C_1}$, then: $$y = \frac{C}{x}$$
This is the general solution.