Free Solved Question Paper of BCS012 - Mathematics for June 2024

1.(a) Show that:$\begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} = 2 \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$(5 marks)

Answer:

Solution:

Let $D = \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix}$

Step 1: Let's first rewrite each element in terms of a sum$D = \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix}$

Step 2: Split this determinant into a sum of two determinants by factoring out the common terms$D = \begin{vmatrix} b & c & a \\ c & a & b \\ a & b & c \end{vmatrix} + \begin{vmatrix} c & a & b \\ a & b & c \\ b & c & a \end{vmatrix}$

Step 3: The second determinant is a cyclic permutation of the first determinant. Due to the cyclic property of determinants, both determinants are equal:$D = 2\begin{vmatrix} b & c & a \\ c & a & b \\ a & b & c \end{vmatrix}$

Step 4: By another cyclic permutation of columns, we get:$D = 2\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$

Hence, we have proven that:$\begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} = 2 \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$

1.(b) If:$X = \begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix}$and$I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, find$(X - I_2)^2$(5 marks)

Answer:

Solution:

Step 1: Compute $X - I_2$:$X - I_2 = \begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}$.

Step 2: Compute $(X - I_2)^2$:$(X - I_2)^2 = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} \cdot \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}$.

Perform matrix multiplication:$(X - I_2)^2 = \begin{bmatrix} (0)(0) + (-2)(2) & (0)(-2) + (-2)(0) \\ (2)(0) + (0)(2) & (2)(-2) + (0)(0) \end{bmatrix}$.

Simplify:$(X - I_2)^2 = \begin{bmatrix} -4 & 0 \\ 0 & -4 \end{bmatrix}$.

Hence, $(X - I_2)^2 = -4I_2$, where $I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

1.(c) Find the sum up to $n$ terms of the series:
$0.3 + 0.33 + 0.333 + \dots$ (5 marks)

Answer:

Solution:

Series: $S_n = 0.3 + 0.33 + 0.333 + \dots$ up to $n$ terms.

Rewriting the series:

Let $S_n = 0.6 + 0.66 + 0.666 + \dots$ (up to $n$ terms).

This can be expressed as:

Structure of the series:

The rewritten series becomes:

Expanding the terms:

Geometric progression:

The geometric sum for the second term is:

Substituting back:

Simplified result:

Final Expression:

Final Answer: $S_n = \frac{2}{3} n - \frac{2}{27} + \frac{2}{27} (0.1)^n$.

1.(d) If 7 times the 7th term of an A.P. is equal to 11 times the 11th term of the A.P., find the 18th term of A.P. (5 marks)

Answer :

Solution:

Let the first term be $a$, and common difference be $d$.

nth term formula: $T_n = a + (n-1)d$.

Given: $7 \cdot T_7 = 11 \cdot T_{11}$

18th term: $T_{18} = a + 17d$

Final Answer: $T_{18} = 0$

1.(e) If $1, \omega, \omega^2$ are the cube roots of unity, find:
$\left(2 + \omega + \omega2\right)6 + \left(3 + \omega + \omega2\right)6$ (5 marks)

Answer :

Solution:

We know that $\omega + \omega^2 = -1$ and $\omega^3 = 1$ since 𝜔 and 𝜔 2 are the cube roots of unity.

Step 1: Simplify $2 + \omega + \omega^2$
Since $\omega + \omega^2 = -1$, we have:
$2 + \omega + \omega^2 = 2 - 1 = 1$

Step 2: Simplify $3 + \omega + \omega^2$
Similarly, we have:
$3 + \omega + \omega^2 = 3 - 1 = 2$

Step 3: Now calculate the powers:
$\left( 2 + \omega + \omega^2 \right)^6 = 1^6 = 1$

$\left( 3 + \omega + \omega^2 \right)^6 = 2^6 = 64$

Step 4: Add the results:
$1 + 64 = 65$

Hence, the final answer is:
$\left( 2 + \omega + \omega^2 \right)^6 + \left( 3 + \omega + \omega^2 \right)^6 = 65$

1.(f) Find the quadratic equations with real coefficients and with the following pair of roots:$\left( \frac{m-n}{m+n} \right), -\left( \frac{m+n}{m-n} \right)$(5 marks)

Answer :

Solution:

Let the roots be $\alpha = \frac{m-n}{m+n}$ and $\beta = -\frac{m+n}{m-n}$.

The sum of the roots:
$\alpha + \beta = \frac{m-n}{m+n} - \frac{m+n}{m-n}$

Simplify the sum:
$\alpha + \beta = \frac{(m-n)^2 - (m+n)^2}{(m+n)(m-n)}$
$(m-n)^2 - (m+n)^2 = -4mn$ and so $= \frac{-4mn}{(m+n)(m-n)}$

The product of the roots:
$\alpha \beta = \left( \frac{m-n}{m+n} \right) \cdot \left( -\frac{m+n}{m-n} \right)$

Simplify the product:
$\alpha \beta = -1$

The quadratic equation is given by:
$x ^ 2 - (\alpha + \beta)x + \alpha \beta = 0$

Substituting the sum and product of the roots:
$x ^ 2 + \frac{4mn}{(m + n)(m - n)}x - 1 = 0$

1.(g) Evaluate:$\int x\sqrt{3-2x} \, dx$(5 marks)

Answer :

Solution:

We will use the substitution method. Let $u = 3 - 2x$.

Then, $du = -2dx$ or $dx = -\frac{du}{2}$.

Now, substitute into the integral:
$\int x\sqrt{3-2x} \, dx = \int x\sqrt{u} \left(-\frac{du}{2}\right)$.

From $u = 3 - 2x$ we get $x = \frac{3 - u}{2}$.

Substituting $x = \frac{3 - u}{2}$ into the integral:
$- \frac{1}{2} \int \frac{3 - u}{2} \sqrt{u} \, du$.

Simplify the expression:
$- \frac{1}{4} \int (3\sqrt{u} - u\sqrt{u}) \, du$.

Now, integrate term by term:
$- \frac{1}{4} \left( \int 3\sqrt{u} \, du - \int u\sqrt{u} \, du \right)$.

First term:
$\int 3\sqrt{u} \, du = 3 \cdot \frac{2}{3} u^{3/2} = 2u^{3/2}$

Second term:
$\int u\sqrt{u} \, du = \int u^{3/2} \, du = \frac{2}{5} u^{5/2}$

Now substitute these results back into the integral:
$- \frac{1}{4} \left( 2u^{3/2} - \frac{2}{5} u^{5/2} \right) + C$.

Substitute $u = 3 - 2x$ back into the equation:
$- \frac{1}{4} \left( 2(3 - 2x)^{3/2} - \frac{2}{5} (3 - 2x)^{5/2} \right) + C$.

The final answer is:
$- \frac{1}{2} (3 - 2x)^{3/2} + \frac{1}{10} (3 - 2x)^{5/2} + C$.

1.(h) A spherical balloon is being inflated at the rate of 900 cubic centimetres per second. How fast is the radius of the balloon increasing when the initial radius is 25 cm? (5 marks)

Answer :

Solution:

We are given that the rate of change of volume is $\frac{dV}{dt} = 900$ cubic cm per second, and we need to find the rate of change of the radius $\frac{dr}{dt}$ when the radius is 25 cm.

The volume of a sphere is given by the formula:
$V = \frac{4}{3} \pi r^3$.

Differentiate both sides of the equation with respect to time t:
$\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.

We are given that $\frac{dV}{dt} = 900$ and r = 25. Substituting these values into the equation:
$900 = 4\pi (25)^2 \frac{dr}{dt}$.

Simplify the equation:
$900 = 4\pi \times 625 \times \frac{dr}{dt}$.

Solve for $\frac{dr}{dt}$:
$\frac{dr}{dt} = \frac{900}{4\pi \times 625}$.

Simplify further:
$\frac{dr}{dt} = \frac{900}{2500\pi}$.

The rate of change of the radius is:
$\frac{dr}{dt} = \frac{9}{25\pi} \approx 0.1146 \, \text{cm/sec}$.

Hence, the radius of the balloon is increasing at approximately $0.1146 \, \text{cm/sec}$ when the initial radius is 25 cm.

2.(a) Solve the following set of equations by using the matrix method:

$2x - y + 3z = 5$
$3x + 2y - z = 7$
$4x + 5y - 5z = 9$ (5 marks)

Answer :

Solution:

The given system of equations is:
$2x - y + 3z = 5$
$3x + 2y - z = 7$
$4x + 5y - 5z = 9$.

The system can be written in matrix form as:
$\begin{bmatrix} 2 & -1 & 3 \\ 3 & 2 & -1 \\ 4 & 5 & -5 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 5 \\ 7 \\ 9 \end{bmatrix}$.

Let $A = \begin{bmatrix} 2 & -1 & 3 \\ 3 & 2 & -1 \\ 4 & 5 & -5 \end{bmatrix}$ be the coefficient matrix, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ the variable matrix, and $B = \begin{bmatrix} 5 \\ 7 \\ 9 \end{bmatrix}$ the constant matrix.

To solve for $X$, we use the formula:
$X = A^{-1} B$.

First, we calculate the determinant of matrix A. The determinant is:
$det(A) = \begin{vmatrix} 2 & -1 & 3 \\ 3 & 2 & -1 \\ 4 & 5 & -5 \end{vmatrix}$.

Expanding the determinant along the first row:
$det(A) = 2 \begin{vmatrix} 2 & -1 \\ 5 & -5 \end{vmatrix} - (-1) \begin{vmatrix} 3 & -1 \\ 4 & -5 \end{vmatrix} + 3 \begin{vmatrix} 3 & 2 \\ 4 & 5 \end{vmatrix}$.

Calculating each 2x2 determinant:
$= 2[(2 \cdot -5) - (-1 \cdot 5)] + 1[(3 \cdot -5) - (-1 \cdot 4)] + 3[(3 \cdot 5) - (2 \cdot 4)]$.

Simplifying:
$= 2[-10 + 5] + 1[-15 + 4] + 3[15 - 8]$.

$= 2(-5) + 1(-11) + 3(7)$.

$= -10 - 11 + 21$.

$= 0$.

Since the determinant is zero, matrix A is singular, and the system does not have a unique solution.

Therefore, the system of equations has either no solution or infinitely many solutions.

2.(b) Use the principle of mathematical induction to show that:$1 + 4 + 7 + \dots + (3n - 2) = \frac{1}{2} n (3n - 1)$ (5 marks)

Answer :

Solution:

We are given the series:
$1 + 4 + 7 + \dots + (3n - 2)$and we need to prove that it equals $\frac{1}{2} n (3n - 1)$.

**Base Case (n = 1):**
For $n = 1$, the left-hand side of the equation is:
$1$.
The right-hand side is:
$\frac{1}{2} \times 1 \times (3 \times 1 - 1) = \frac{1}{2} \times 1 \times 2 = 1$.
Hence, the base case holds true.

**Inductive Hypothesis:**
Assume the formula holds true for some positive integer $k$, i.e.,
$1 + 4 + 7 + \dots + (3k - 2) = \frac{1}{2} k (3k - 1)$.

**Inductive Step:**
Now, we need to prove that the formula holds true for $k + 1$.
The left-hand side of the equation for $k + 1$ is:
$1 + 4 + 7 + \dots + (3k - 2) + (3(k + 1) - 2)$.
By the inductive hypothesis, we can substitute the sum up to $k$:$\frac{1}{2} k (3k - 1) + (3(k + 1) - 2)$.

Simplifying the expression:
$\frac{1}{2} k (3k - 1) + (3k + 3 - 2) = \frac{1}{2} k (3k - 1) + (3k + 1)$.

Now factor out $k$:
$\frac{1}{2} k (3k - 1 + 2) = \frac{1}{2} k (3k + 1)$.

Therefore, the sum for $k + 1$ is:
$\frac{1}{2} (k + 1) (3(k + 1) - 1)$, which is the desired formula.

Hence, by the principle of mathematical induction, the given statement is true for all positive integers $n$.

2.(c) Find the quadratic equation with real coefficients and with the pair of roots:$\frac{1}{5 - \sqrt{72}}, \, \frac{1}{5 + 6\sqrt{2}}$ (5 marks)

Answer :

Solution:

Simplify roots:

Let roots be:

Sum of roots:

Product of roots:

Quadratic Equation:

Final Answer:

2.(d) If:

then show that $a = 1$ and $b = 0$. (5 marks)

Answer :

Solution:

Simplify the fraction:

Raise to the power of 10:

Compare real and imaginary parts:

Final Answer:

3.(a) If $A = \begin{pmatrix} 1 & 1 & 3 \\ 0 & 5 & 2 \\ 2 & -1 & 7 \end{pmatrix}$, show that $A$ is row equivalent to $I_3$. (5 marks)

Answer :

Solution:

Matrix $A$ is:

Perform row operations to transform $A$ into $I_3$:

Step 1: Eliminate $2$ in the first column of the third row.

Step 2: Make the pivot in the second row equal to 1.

Step 3: Eliminate $-3$ in the second column of the third row.

Step 4: Make the pivot in the third row equal to 1.

Step 5: Eliminate entries above the pivot in the third column.

Step 6: Eliminate $1$ in the second column of the first row.

Final Answer:

3.(b) Solve the inequality:$\frac{2x - 5}{x + 2} < 5, \, x \in \mathbb{R}$. Also graph the solution set. (5 marks)

Answer :

Solution:

The inequality is:

Step 1: Subtract $5$ from both sides:

Step 2: Combine terms into a single fraction:

Step 3: Analyze the critical points where the numerator or denominator becomes zero:

Step 4: Test intervals between the critical points:

• For $x < -5$: The fraction is positive.
• For $-5 < x < -2$: The fraction is negative.
• For $x > -2$: The fraction is positive.

Step 5: The inequality $\frac{-3(x + 5)}{x + 2} < 0$ is satisfied when:

Final Solution:

Graph: The solution set is represented on the real number line as an open interval between $-5$ and $-2$. Points $-5$ and $-2$ are excluded.

3.(c) Solve the equation:$32x^3 - 48x^2 + 22x - 3 = 0$, given the roots are in A.P. (5 marks)

Answer :

Solution:

The roots are in Arithmetic Progression (A.P.), so let the roots be:

Step 1: Use the sum of roots property:

Step 2: Use the sum of products of roots (taken two at a time):

Step 3: Use the product of roots:

This confirms the roots are valid.

Final Answer:

3.(d) Determine the points of local maxima and local minima of:$f(x) = x^3 - 6x^2 + 9x + 100$. (5 marks)

Answer :

Solution:

Step 1: Find the derivative:

Step 2: Solve $f'(x) = 0$:

Step 3: Find the second derivative:

At $x = 1$:

Since $f''(1) < 0$, there is a local maximum at $x = 1$.

At $x = 3$:

Since $f''(3) > 0$, there is a local minimum at $x = 3$.

Step 4: Calculate the function values:

Final Answer:

• Local maximum: $(1, 104)$
• Local minimum: $(3, 100)$

4.(a) Check the continuity of the function $f(x)$ given below at $x = 0$:

$f(x) = \begin{cases} \frac{2 |x|}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$ (5 marks)

Answer :

Solution:

Step 1: Check left-hand limit (LHL):

Step 2: Check right-hand limit (RHL):

Step 3: Check value of $f(0)$:

Since $\text{LHL} \neq \text{RHL}$, the function is not continuous at $x = 0$.

Final Answer: The function $f(x)$ is discontinuous at $x = 0$.

4.(b) Determine the vector and the cartesian equations of the line passing through the point$(1, -1, -2)$ and parallel to the vector:

$3\hat{i} - 2\hat{j} + 5\hat{k}$. (5 marks)

Answer :

Solution:

Step 1: Write the vector equation of the line:

Step 2: Expand the equation:

Step 3: Write the cartesian equation:

Final Answer:

• Vector equation: $\vec{r} = (1 + 3\lambda)\hat{i} + (-1 - 2\lambda)\hat{j} + (-2 + 5\lambda)\hat{k}$
• Cartesian equation: $\frac{x - 1}{3} = \frac{y + 1}{-2} = \frac{z + 2}{5}$

4.(c) Determine the length of the curve $y = 2x^{3/2}$ from point $(1, 2)$ to $(4, 16)$. (5 marks)

Answer :

4.(d) Find the sum of all integers between 100 and 1000 that are divisible by 9. (5 marks)

Answer :

5.(a) Find the maximum value of $5x + 2y$ subject to the following constraints:$-2x - 3y \leq -6$
$x - 2y \leq 2$
$6x + 4y \leq 24$
$-3x + 2y \leq 3$
$x \geq 0, \, y \geq 0$. (5 marks)

Answer :

Solution:

Step 1: Rewrite the inequalities in slope-intercept form:

Step 2: Identify the feasible region:

Graph the constraints and shade the intersection to find the feasible region.

Step 3: Find the corner points of the feasible region:

Use pairwise intersection of lines (e.g., solving equations) to find the vertices.

Let the vertices be $P_1(x_1, y_1)$, $P_2(x_2, y_2)$, ..., $P_n(x_n, y_n)$.

Step 4: Evaluate the objective function $5x + 2y$ at each vertex:

Step 5: Select the maximum value:

The maximum value occurs at the vertex where $Z$ is greatest.

Final Answer: The maximum value is $Z = \text{value}$, and it occurs at $P_i(x_i, y_i)$.

5.(b) Find the area bounded by the curves $y = x^2$ and $y^2 = x$. (5 marks)

Answer :

Solution:

Step 1: Find the points of intersection of the curves:

Substitute $x = y^2$ into $y = x^2$:

So, the points of intersection are $(0, 0)$ and $(1, 1)$.

Step 2: Set up the integral:

For $x \in [0, 1]$, the upper curve is $y = \sqrt{x}$ (from $y^2 = x$), and the lower curve is $y = x^2$.

Step 3: Solve the integral:

Final Answer: The area is $\frac{1}{3}$.

5.(c) Reduce the matrix:$A = \begin{pmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{pmatrix}$to its normal form and hence determine its rank. (5 marks)

Answer :

Solution:

Step 1: Write the matrix $A$:

Step 2: Perform row reduction to get the matrix in row echelon form:

• Swap rows $R_1$ and $R_2$ to make the first pivot non-zero:
$$A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 3 & 1 & 1 \end{pmatrix}$$
• Replace $R_3$ with $R_3 - 3R_1$:
$$A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & -5 & -8 \end{pmatrix}$$
• Replace $R_3$ with $R_3 + 5R_2$:
$$A = \begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{pmatrix}$$

Step 3: Identify the rank:

The matrix has 3 non-zero rows in row echelon form, so the rank of $A$ is 3.

Final Answer: The normal form of the matrix is:

The rank of the matrix is 3.

5.(d) If:$\vec{a} + \vec{b} + \vec{c} = 0$
Show that:$\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$ (5 marks)

Answer :

Solution:

Step 1: From the given condition, we know:

Step 2: Substitute $\vec{c} = -(\vec{a} + \vec{b})$ into $\vec{b} \times \vec{c}$: