BT-102 Mathematics-I - Jun 2025 Question Paper

Under each question, How to answer lists the points an answer should cover, outlined by CampusPrep from the syllabus. Check the details against your textbook.

1. 7 Marks
   aProve that a rectangular solid of maximum volume within a sphere is a cube.
   How to answer

   • Formulate the volume of the rectangular solid inscribed in a sphere of radius $R$ as $V = 8xyz$ where $(x,y,z)$ is a vertex in the first octant.
   • Set up the constraint $g(x,y,z) = x^2 + y^2 + z^2 - R^2 = 0$.
   • Use Lagrange multipliers: $\nabla V = \lambda \nabla g$, yielding $8yz = 2\lambda x$, $8xz = 2\lambda y$, $8xy = 2\lambda z$.
   • Solve for $x, y, z$ to show $x^2 = y^2 = z^2 = R^2/3$, hence $x = y = z$.
   • Conclude that the dimensions are equal, so the solid of maximum volume is a cube.
2. 7 Marks
   bIf $u = \tan^{-1}\frac{x^3+y^3}{x-y}$, then prove that $x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \sin 2u$.
   How to answer

   • Let $z = \tan u = \frac{x^3+y^3}{x-y}$.
   • Verify that $z$ is a homogeneous function of degree $n = 3 - 1 = 2$.
   • Apply Euler's theorem to $z$: $x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = 2z$.
   • Substitute $\frac{\partial z}{\partial x} = \sec^2 u \frac{\partial u}{\partial x}$ and $\frac{\partial z}{\partial y} = \sec^2 u \frac{\partial u}{\partial y}$.
   • Obtain $x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \frac{2\tan u}{\sec^2 u} = 2\sin u \cos u = \sin 2u$.
3. 7 Marks
   aShow that the surface area of solid generated by revolution of the loop of curve $x = t^2, y = t - t^3/3$ about the x-axis is $3\pi$.
   How to answer

   • Find limits of parameter $t$ for the loop by setting $y = 0 \implies t(1 - t^2/3) = 0 \implies t = 0, \pm\sqrt{3}$.
   • Compute derivatives: $\frac{dx}{dt} = 2t$, $\frac{dy}{dt} = 1 - t^2$.
   • Compute arc length element $ds = \sqrt{(dx/dt)^2 + (dy/dt)^2}dt = \sqrt{4t^2 + (1-t^2)^2}dt = (1+t^2)dt$.
   • Use surface area formula $S = 2\pi \int_0^{\sqrt{3}} y \, ds = 2\pi \int_0^{\sqrt{3}} (t - t^3/3)(1+t^2)dt$.
   • Integrate and evaluate the limits to arrive at $S = 3\pi$.
4. 7 Marks
   bChange the order of integration in the following integral and then evaluate $$\int_0^1 \int_{e^x}^e \frac{dx dy}{\log y}$$
   How to answer

   • Identify given region of integration: $x$ goes from $0$ to $1$, $y$ goes from $e^x$ to $e$.
   • Sketch or describe region: bounded by $y = e^x$, $y = e$, $x = 0$, and $x = 1$.
   • Reverse order: $y$ ranges from $1$ to $e$, and for a fixed $y$, $x$ ranges from $0$ to $\ln y$.
   • Set up integral: $\int_1^e \int_0^{\ln y} \frac{1}{\ln y} dx dy$.
   • Integrate with respect to $x$: $\int_0^{\ln y} dx = \ln y$.
   • Integrate with respect to $y$: $\int_1^e \frac{\ln y}{\ln y} dy = \int_1^e 1 dy = e - 1$.
5. 7 Marks
   aFind the Fourier Series for $f(x) = x + x^2$ in $(-\pi, \pi)$
   How to answer

   • Write Fourier series formula: $f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty (a_n \cos nx + b_n \sin nx)$.
   • Calculate $a_0 = \frac{1}{\pi} \int_{-\pi}^\pi (x + x^2)dx = \frac{2\pi^2}{3}$.
   • Calculate $a_n = \frac{1}{\pi} \int_{-\pi}^\pi (x + x^2) \cos nx dx = \frac{4(-1)^n}{n^2}$.
   • Calculate $b_n = \frac{1}{\pi} \int_{-\pi}^\pi (x + x^2) \sin nx dx = \frac{2(-1)^{n+1}}{n}$.
   • Write final Fourier series combining all coefficients.
6. 7 Marks
   bFind the half range sine series for $f(x) = x(\pi - x)$ in $(0, \pi)$. Hence Deduce that $$\frac{1}{1^3} - \frac{1}{3^3} + \frac{1}{5^3} \cdots\dots = \frac{\pi^3}{32}$$
   How to answer

   • State half-range Fourier sine series formula: $f(x) = \sum_{n=1}^\infty b_n \sin nx$.
   • Compute $b_n = \frac{2}{\pi}\int_0^\pi (\pi x - x^2)\sin nx \, dx$ using integration by parts.
   • Show $b_n = \frac{4}{\pi n^3}[1 - (-1)^n]$, so $b_n = \frac{8}{\pi n^3}$ for odd $n$ and $0$ for even $n$.
   • Form the series: $f(x) = \frac{8}{\pi} \sum_{m=1}^\infty \frac{\sin((2m-1)x)}{(2m-1)^3}$.
   • Substitute $x = \pi/2$: $f(\pi/2) = \frac{\pi^2}{4} = \frac{8}{\pi} [\frac{1}{1^3} - \frac{1}{3^3} + \frac{1}{5^3} - \dots]$.
   • Conclude $\frac{1}{1^3} - \frac{1}{3^3} + \frac{1}{5^3} - \dots = \frac{\pi^3}{32}$.
7. 7 Marks
   aLet $W_1$ and $W_2$ be subspaces of a vector space $V$ and assume that $W_1 \cap W_2 = \{0\}$. Let $w_1 \in W_1$ and $w_2 \in W_2$ be such that $w_1 \neq 0$ and $w_2 \neq 0$. Prove that $\{w_1, w_2\}$ is linearly independent.
   How to answer

   • Set up the linear combination: $c_1 w_1 + c_2 w_2 = 0$ for scalars $c_1, c_2$.
   • Rewrite as $c_1 w_1 = -c_2 w_2$.
   • Note that $c_1 w_1 \in W_1$ and since $-c_2 w_2 \in W_2$, $c_1 w_1 \in W_1 \cap W_2$.
   • Since $W_1 \cap W_2 = \{0\}$, we have $c_1 w_1 = 0$ and $-c_2 w_2 = 0$.
   • Since $w_1 \neq 0$ and $w_2 \neq 0$, it follows that $c_1 = 0$ and $c_2 = 0$.
   • Conclude that $\{w_1, w_2\}$ is linearly independent.
8. 7 Marks
   bLet $W_1$ and $W_2$ be subspaces of a vector space $V$. i) Prove that $W_1 \cap W_2$ is a subspace of $V$. ii) Give an example to show that $W_1 \cup W_2$ need not be a subspace of $V$. iii) Is $W_1 \setminus W_2$ a subspace of $V$?
   How to answer

   • For (i): Show $0 \in W_1 \cap W_2$; take $u, v \in W_1 \cap W_2$ and scalars $\alpha, \beta$, prove $\alpha u + \beta v \in W_1$ and $\alpha u + \beta v \in W_2$, thus in $W_1 \cap W_2$.
   • For (ii): Let $V = \mathbb{R}^2$, $W_1 = \{(x,0) : x \in \mathbb{R}\}$ (x-axis), $W_2 = \{(0,y) : y \in \mathbb{R}\}$ (y-axis). Show $(1,0) + (0,1) = (1,1) \notin W_1 \cup W_2$.
   • For (iii): State 'No'. Explain that the zero vector $0 \in W_2$, so $0 \notin W_1 \setminus W_2$, meaning it can never contain the zero vector (or if $W_1 \setminus W_2$ is empty/non-empty, it lacks the zero element).
9. 7 Marks
   aFind Eigen Value and Eigen vectors of $\begin{bmatrix} 2 & -2 & 2 \\ 1 & 1 & 1 \\ 1 & 3 & -1 \end{bmatrix}$
   How to answer

   • Set up the characteristic equation $\det(A - \lambda I) = 0$.
   • Expand determinant: $-\lambda^3 + \text{tr}(A)\lambda^2 - M\lambda + \det(A) = 0$.
   • Solve characteristic polynomial to find eigenvalues $\lambda_1, \lambda_2, \lambda_3$.
   • For each eigenvalue $\lambda$, solve the system $(A - \lambda I)X = 0$ using row reduction.
   • State the corresponding linearly independent eigenvectors.
10. 7 Marks
    bDiagonalizable the matrix $\begin{bmatrix} 3 & -1 & 1 \\ -1 & 5 & -1 \\ 1 & -1 & 3 \end{bmatrix}$
    How to answer

    • Find the characteristic polynomial of $A$: $|A - \lambda I| = 0$.
    • Find eigenvalues (e.g., $\lambda = 2, 3, 6$).
    • Find orthogonal/independent eigenvectors corresponding to each eigenvalue.
    • Construct the modal matrix $P = [X_1 \quad X_2 \quad X_3]$.
    • Compute or state the diagonal matrix $D = P^{-1}AP = \operatorname{diag}(\lambda_1, \lambda_2, \lambda_3)$.
11. 7 Marks
    aReduce the matrix $$\begin{bmatrix} 3 & 2 & -1 \\ 4 & 2 & 6 \\ 7 & 4 & 5 \end{bmatrix}$$ to the normal form, hence find its rank.
    How to answer

    • Apply elementary row and column operations to transform the matrix to the normal form $\begin{bmatrix} I_r & 0 \\ 0 & 0 \end{bmatrix}$.
    • Notice $R_3 = R_1 + R_2$, indicating row 3 becomes zero.
    • Perform column transformations to clear off non-diagonal elements.
    • Arrive at the canonical/normal form $\begin{bmatrix} I_2 & 0 \\ 0 & 0 \end{bmatrix}$.
    • State that the rank is $r = 2$.
12. 7 Marks
    bFind the first 3 terms in the Maclaurin series for i) $\sin^2 x$ ii) $x e^{-x}$
    How to answer

    • For (i) $\sin^2 x$: Rewrite as $\frac{1 - \cos 2x}{2} = \frac{1}{2}\left(1 - \left(1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \frac{(2x)^6}{6!} + \dots\right)\right) = x^2 - \frac{x^4}{3} + \frac{2x^6}{45}$.
    • For (ii) $x e^{-x}$: Use expansion $e^{-x} = 1 - x + \frac{x^2}{2!} - \dots$, so $x e^{-x} = x - x^2 + \frac{x^3}{2} - \dots$
    • Explicitly list the first three non-zero terms for each function.
13. 7 Marks
    aInvestigate for what values of $\lambda$ and $\mu$, the simultaneous equation $$\begin{aligned} x + y + z &= 6 \\ x + 2y + 3z &= 10 \\ x + 2y + \lambda z &= \mu \end{aligned}$$ Have i) no solution ii) a unique solution iii) an infinite number of solutions
    How to answer

    • Write the augmented matrix $[A|B] = \begin{bmatrix} 1 & 1 & 1 & 6 \\ 1 & 2 & 3 & 10 \\ 1 & 2 & \lambda & \mu \end{bmatrix}$.
    • Apply row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$, then $R_3 \to R_3 - R_2$.
    • Obtain upper triangular form: $\begin{bmatrix} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & 4 \\ 0 & 0 & \lambda - 3 & \mu - 10 \end{bmatrix}$.
    • Analyze conditions: (i) No solution: $\operatorname{rank}(A) \neq \operatorname{rank}(A|B) \implies \lambda = 3, \mu \neq 10$.
    • (ii) Unique solution: $\operatorname{rank}(A) = \operatorname{rank}(A|B) = 3 \implies \lambda \neq 3$ (any value of $\mu$).
    • (iii) Infinite solutions: $\operatorname{rank}(A) = \operatorname{rank}(A|B) < 3 \implies \lambda = 3, \mu = 10$.
14. 7 Marks
    bFind the Taylor series for the function $x^4 + x - 2$ centered at $a = 1$.
    How to answer

    • State Taylor series formula about $a = 1$: $f(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f'''(1)}{3!}(x-1)^3 + \frac{f^{(4)}(1)}{4!}(x-1)^4$.
    • Evaluate derivatives: $f(x) = x^4 + x - 2 \implies f(1) = 0$.
    • $f'(x) = 4x^3 + 1 \implies f'(1) = 5$.
    • $f''(x) = 12x^2 \implies f''(1) = 12$.
    • $f'''(x) = 24x \implies f'''(1) = 24$.
    • $f^{(4)}(x) = 24 \implies f^{(4)}(1) = 24$. All higher derivatives are zero.
    • Write Taylor polynomial: $f(x) = 5(x-1) + 6(x-1)^2 + 4(x-1)^3 + (x-1)^4$.
15. 7 Marks
    aEvaluate $\int_0^\infty \int_x^\infty \frac{e^{-y}}{y} dydx$
    How to answer

    • Identify limits: $x$ from $0$ to $\infty$, $y$ from $x$ to $\infty$.
    • Change the order of integration: region is bounded by $y = x$, $x = 0$, $y \to \infty$, so $y$ goes from $0$ to $\infty$ and $x$ goes from $0$ to $y$.
    • Rewrite integral: $I = \int_0^\infty \int_0^y \frac{e^{-y}}{y} dx dy$.
    • Perform inner integration: $\int_0^y \frac{e^{-y}}{y} dx = \frac{e^{-y}}{y} [x]_0^y = \frac{e^{-y}}{y} \cdot y = e^{-y}$.
    • Perform outer integration: $\int_0^\infty e^{-y} dy = [-e^{-y}]_0^\infty = 0 - (-1) = 1$.
16. 7 Marks
    bIf $x^x y^y z^z = C$ then show that $\frac{\partial^2 z}{\partial x \partial y} = -(x \log ex)^{-1}$ at $x = y = z$.
    How to answer

    • Take natural log of both sides: $x\ln x + y\ln y + z\ln z = \ln C$.
    • Differentiate implicitly with respect to $x$: $(1 + \ln x) + (1 + \ln z)\frac{\partial z}{\partial x} = 0 \implies \frac{\partial z}{\partial x} = -\frac{1 + \ln x}{1 + \ln z}$.
    • Similarly differentiate with respect to $y$: $\frac{\partial z}{\partial y} = -\frac{1 + \ln y}{1 + \ln z}$.
    • Differentiate $\frac{\partial z}{\partial y}$ with respect to $x$: $\frac{\partial^2 z}{\partial x \partial y} = \frac{\partial}{\partial x}\left(-\frac{1 + \ln y}{1 + \ln z}\right) = \frac{1 + \ln y}{(1 + \ln z)^2} \frac{1}{z} \frac{\partial z}{\partial x}$.
    • Substitute $\frac{\partial z}{\partial x} = -\frac{1 + \ln x}{1 + \ln z}$, giving $\frac{\partial^2 z}{\partial x \partial y} = -\frac{(1 + \ln x)(1 + \ln y)}{z(1 + \ln z)^3}$.
    • At $x = y = z$: expression simplifies to $-\frac{(1 + \ln x)^2}{x(1 + \ln x)^3} = -\frac{1}{x(1 + \ln x)} = -(x \log ex)^{-1}$.