BE Civil Engineering (IOE, TU) Engineering Mathematics II (IOE, SH 451) Question Paper 2077 Nepal

Setting up the matrix equation.

Write the system as $A\mathbf{x} = \mathbf{b}$ where

$$A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 6 \\ 11 \\ 14 \end{pmatrix}.$$

Step 1 — Determinant of $A$. Expanding along the first row:

$$|A| = 1(4\cdot6 - 5\cdot5) - 2(2\cdot6 - 5\cdot3) + 3(2\cdot5 - 4\cdot3)$$ $$= 1(24-25) - 2(12-15) + 3(10-12) = 1(-1) - 2(-3) + 3(-2) = -1 + 6 - 6 = -1.$$

Since $|A| = -1 \neq 0$, $A^{-1}$ exists and the system has a unique solution.

Step 2 — Cofactor matrix. Compute each cofactor $C_{ij}$:

$$C_{11} = +(24-25) = -1, \quad C_{12} = -(12-15) = 3, \quad C_{13} = +(10-12) = -2,$$ $$C_{21} = -(12-15) = 3, \quad C_{22} = +(6-9) = -3, \quad C_{23} = -(5-6) = 1,$$ $$C_{31} = +(10-12) = -2, \quad C_{32} = -(5-6) = 1, \quad C_{33} = +(4-4) = 0.$$

Step 3 — Adjoint (transpose of the cofactor matrix). Since the cofactor matrix is symmetric here:

$$\operatorname{adj}(A) = \begin{pmatrix} -1 & 3 & -2 \\ 3 & -3 & 1 \\ -2 & 1 & 0 \end{pmatrix}.$$

Step 4 — Inverse. $A^{-1} = \dfrac{1}{|A|}\operatorname{adj}(A) = \dfrac{1}{-1}\operatorname{adj}(A)$:

$$A^{-1} = \begin{pmatrix} 1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0 \end{pmatrix}.$$

Step 5 — Solve. $\mathbf{x} = A^{-1}\mathbf{b}$:

$$x = 1(6) + (-3)(11) + 2(14) = 6 - 33 + 28 = 1,$$ $$y = -3(6) + 3(11) + (-1)(14) = -18 + 33 - 14 = 1,$$ $$z = 2(6) + (-1)(11) + 0(14) = 12 - 11 + 0 = 1.$$

Verification. $1 + 2 + 3 = 6$ ✓, $2 + 4 + 5 = 11$ ✓, $3 + 5 + 6 = 14$ ✓.

Final answer: $\boxed{x = 1,\; y = 1,\; z = 1.}$

Step 1 — Two direction vectors in the plane.

$$\vec{AB} = B - A = (1-2,\; 1-(-1),\; 0-1) = (-1, 2, -1),$$ $$\vec{AC} = C - A = (3-2,\; 0-(-1),\; 2-1) = (1, 1, 1).$$

Step 2 — Normal vector $\mathbf{n} = \vec{AB} \times \vec{AC}$:

$$\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 2 & -1 \\ 1 & 1 & 1 \end{vmatrix} = \mathbf{i}(2\cdot1 - (-1)\cdot1) - \mathbf{j}((-1)\cdot1 - (-1)\cdot1) + \mathbf{k}((-1)\cdot1 - 2\cdot1).$$ $$\mathbf{n} = \mathbf{i}(2+1) - \mathbf{j}(-1+1) + \mathbf{k}(-1-2) = (3, 0, -3).$$

Divide by 3: take $\mathbf{n} = (1, 0, -1)$.

Step 3 — Equation of the plane. Using point $A(2,-1,1)$:

$$1(x-2) + 0(y+1) - 1(z-1) = 0 \;\Rightarrow\; x - 2 - z + 1 = 0 \;\Rightarrow\; \boxed{x - z - 1 = 0}.$$

Check with $B$ and $C$: $B$: $1 - 0 - 1 = 0$ ✓; $C$: $3 - 2 - 1 = 0$ ✓.

Step 4 — Perpendicular distance of $P(4,3,5)$. For plane $x - z - 1 = 0$:

$$d = \frac{|x_0 - z_0 - 1|}{\sqrt{1^2 + 0^2 + (-1)^2}} = \frac{|4 - 5 - 1|}{\sqrt{2}} = \frac{|-2|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}.$$

Final answers: Plane: $\boxed{x - z - 1 = 0}$; distance $= \boxed{\sqrt{2} \approx 1.414 \text{ units}}$.

Step 1 — Complementary function (CF). The auxiliary equation is

$$m^2 - 5m + 6 = 0 \;\Rightarrow\; (m-2)(m-3) = 0 \;\Rightarrow\; m = 2,\, 3.$$

Real distinct roots give

$$y_c = C_1 e^{2x} + C_2 e^{3x}.$$

Step 2 — Particular integral (PI). For RHS $e^{4x}$, the operator method gives

$$y_p = \frac{1}{D^2 - 5D + 6}\, e^{4x}.$$

Substitute $D = 4$ (since $4$ is not a root of the auxiliary equation):

$$y_p = \frac{1}{4^2 - 5(4) + 6}\, e^{4x} = \frac{1}{16 - 20 + 6}\, e^{4x} = \frac{1}{2}\, e^{4x}.$$

Verification. Let $y_p = \tfrac12 e^{4x}$. Then $y_p' = 2e^{4x}$, $y_p'' = 8e^{4x}$, so

$$y_p'' - 5y_p' + 6y_p = 8e^{4x} - 10e^{4x} + 3e^{4x} = e^{4x}.$$

✓

Step 3 — General solution.

$$\boxed{y = C_1 e^{2x} + C_2 e^{3x} + \tfrac{1}{2} e^{4x}}$$

where $C_1, C_2$ are arbitrary constants.

Setup. Period $2\pi$, so $f(x) = \dfrac{a_0}{2} + \sum_{n=1}^{\infty}\big(a_n\cos nx + b_n\sin nx\big)$.

Since $f(x)=x^2$ is an even function, $b_n = 0$ for all $n$.

Step 1 — Constant term $a_0$.

$$a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi} x^2\,dx = \frac{2}{\pi}\int_{0}^{\pi} x^2\,dx = \frac{2}{\pi}\cdot\frac{\pi^3}{3} = \frac{2\pi^2}{3}.$$

Step 2 — Cosine coefficients $a_n$.

$$a_n = \frac{2}{\pi}\int_{0}^{\pi} x^2 \cos nx\, dx.$$

Integrate by parts twice:

$$\int x^2\cos nx\,dx = \frac{x^2\sin nx}{n} + \frac{2x\cos nx}{n^2} - \frac{2\sin nx}{n^3}.$$

Evaluate from $0$ to $\pi$. At $x=\pi$: $\sin n\pi = 0$, $\cos n\pi = (-1)^n$; at $x=0$ all terms vanish. So

$$\int_{0}^{\pi} x^2\cos nx\,dx = 0 + \frac{2\pi(-1)^n}{n^2} - 0 = \frac{2\pi(-1)^n}{n^2}.$$

Therefore

$$a_n = \frac{2}{\pi}\cdot\frac{2\pi(-1)^n}{n^2} = \frac{4(-1)^n}{n^2}.$$

Step 3 — Fourier series.

$$\boxed{x^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos nx}, \qquad -\pi < x < \pi.$$

Step 4 — Deduction. Put $x = \pi$, where $f(\pi)=\pi^2$ and $\cos n\pi = (-1)^n$, so $(-1)^n\cos n\pi = (-1)^{2n}=1$:

$$\pi^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{1}{n^2}.$$ $$4\sum_{n=1}^{\infty}\frac{1}{n^2} = \pi^2 - \frac{\pi^2}{3} = \frac{2\pi^2}{3} \;\Rightarrow\; \sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{2\pi^2}{12} = \boxed{\frac{\pi^2}{6}}.$$

✓

(a) $\sum \dfrac{n}{2^n}$ — use the Ratio Test.

Let $u_n = \dfrac{n}{2^n}$. Then

$$\frac{u_{n+1}}{u_n} = \frac{(n+1)/2^{n+1}}{n/2^n} = \frac{n+1}{2n} = \frac{1}{2}\Big(1 + \frac{1}{n}\Big).$$

Taking the limit:

$$L = \lim_{n\to\infty} \frac{u_{n+1}}{u_n} = \frac{1}{2}\,(1+0) = \frac{1}{2} < 1.$$

Since $L < 1$, by the Ratio Test the series converges (absolutely).

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(b) $\sum \dfrac{n+1}{n^3+2}$ — use the Limit Comparison Test.

For large $n$, $\dfrac{n+1}{n^3+2} \sim \dfrac{n}{n^3} = \dfrac{1}{n^2}$. Compare with $v_n = \dfrac{1}{n^2}$ (a convergent $p$-series, $p=2>1$).

$$\lim_{n\to\infty}\frac{u_n}{v_n} = \lim_{n\to\infty}\frac{(n+1)/(n^3+2)}{1/n^2} = \lim_{n\to\infty}\frac{n^2(n+1)}{n^3+2} = \lim_{n\to\infty}\frac{n^3+n^2}{n^3+2}.$$

Divide numerator and denominator by $n^3$:

$$= \lim_{n\to\infty}\frac{1 + 1/n}{1 + 2/n^3} = \frac{1+0}{1+0} = 1.$$

The limit is a finite non-zero number ($=1$), and $\sum \dfrac{1}{n^2}$ converges, so by the Limit Comparison Test the given series converges.

Conclusion: Both series (a) and (b) converge.

Step 1 — Eliminate below the first pivot.

$R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 3R_1$:

$$\begin{pmatrix} 1 & 2 & 3 \\ 2-2 & 3-4 & 4-6 \\ 3-3 & 5-6 & 7-9 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 3 \\ 0 & -1 & -2 \\ 0 & -1 & -2 \end{pmatrix}.$$

Step 2 — Eliminate below the second pivot.

$R_3 \to R_3 - R_2$:

$$\begin{pmatrix} 1 & 2 & 3 \\ 0 & -1 & -2 \\ 0 & 0 & 0 \end{pmatrix}.$$

Step 3 — Count non-zero rows. The echelon form has 2 non-zero rows.

Final answer: $\operatorname{rank}(A) = \boxed{2}$.

(Consistent with $|A| = 1(21-20) - 2(14-12) + 3(10-9) = 1 - 4 + 3 = 0$, so the rank is less than 3.)

Step 1 — Compare with the general sphere $x^2+y^2+z^2 + 2ux + 2vy + 2wz + d = 0$, whose centre is $(-u,-v,-w)$ and radius $\sqrt{u^2+v^2+w^2-d}$.

Here $2u = -2 \Rightarrow u = -1$; $2v = 4 \Rightarrow v = 2$; $2w = -6 \Rightarrow w = -3$; $d = -11$.

Step 2 — Centre.

$$C = (-u, -v, -w) = (1, -2, 3).$$

Step 3 — Radius.

$$r = \sqrt{u^2 + v^2 + w^2 - d} = \sqrt{1 + 4 + 9 - (-11)} = \sqrt{14 + 11} = \sqrt{25} = 5.$$

Step 4 — Position of $P(1,-2,7)$. Distance from centre $C(1,-2,3)$:

$$CP = \sqrt{(1-1)^2 + (-2+2)^2 + (7-3)^2} = \sqrt{0 + 0 + 16} = 4.$$

Since $CP = 4 < r = 5$, the point $P$ lies inside the sphere.

Final answers: Centre $\boxed{(1,-2,3)}$, radius $\boxed{5}$ units; $P(1,-2,7)$ lies inside the sphere ($CP=4<5$).

Step 1 — Auxiliary equation.

$$m^2 + 4m + 13 = 0 \;\Rightarrow\; m = \frac{-4 \pm \sqrt{16 - 52}}{2} = \frac{-4 \pm \sqrt{-36}}{2} = \frac{-4 \pm 6i}{2} = -2 \pm 3i.$$

Complex conjugate roots $\alpha \pm i\beta$ with $\alpha = -2$, $\beta = 3$.

Step 2 — General solution.

$$y = e^{-2x}\big(A\cos 3x + B\sin 3x\big).$$

Step 3 — Apply $y(0)=1$.

$$y(0) = e^{0}(A\cos 0 + B\sin 0) = A = 1 \;\Rightarrow\; A = 1.$$

Step 4 — Differentiate.

$$y' = -2e^{-2x}(A\cos 3x + B\sin 3x) + e^{-2x}(-3A\sin 3x + 3B\cos 3x).$$

At $x=0$:

$$y'(0) = -2(A) + 3B = -2(1) + 3B = -2 + 3B.$$

Set equal to $2$: $-2 + 3B = 2 \Rightarrow 3B = 4 \Rightarrow B = \dfrac{4}{3}.$

Step 5 — Solution.

$$\boxed{y = e^{-2x}\Big(\cos 3x + \tfrac{4}{3}\sin 3x\Big).}$$

Check: $y(0) = 1$ ✓; $y'(0) = -2 + 3\cdot\tfrac43 = -2 + 4 = 2$ ✓.

Step 1 — Leibnitz (alternating series) Test. Write $u_n = \dfrac{1}{2n-1} > 0$.

(i) Monotonic decrease: Since $2(n+1)-1 = 2n+1 > 2n-1$, we have

$$u_{n+1} = \frac{1}{2n+1} < \frac{1}{2n-1} = u_n \quad\text{for all } n.$$

So $\{u_n\}$ is monotonically decreasing.

(ii) Limit:

$$\lim_{n\to\infty} u_n = \lim_{n\to\infty}\frac{1}{2n-1} = 0.$$

Both Leibnitz conditions hold, so the series converges.

Step 2 — Test for absolute convergence. Consider

$$\sum_{n=1}^{\infty} |u_n| = \sum_{n=1}^{\infty}\frac{1}{2n-1} = 1 + \frac{1}{3} + \frac{1}{5} + \cdots$$

Compare with the divergent harmonic-type series. By the Limit Comparison Test with $v_n = \dfrac{1}{n}$:

$$\lim_{n\to\infty}\frac{1/(2n-1)}{1/n} = \lim_{n\to\infty}\frac{n}{2n-1} = \frac{1}{2} \neq 0.$$

Since $\sum \dfrac1n$ diverges and the limit is finite non-zero, $\sum |u_n|$ diverges.

Conclusion. The series converges (by Leibnitz) but the series of absolute values diverges. Hence the series is conditionally convergent.

(In fact its sum is $\dfrac{\pi}{4}$.)

Step 1 — Trial solution. For a Cauchy–Euler equation, try $y = x^m$. Then

$$y' = m x^{m-1}, \qquad y'' = m(m-1)x^{m-2}.$$

Substitute:

$$x^2\,m(m-1)x^{m-2} - 3x\,m x^{m-1} + 4x^m = 0$$ $$\Rightarrow \big[m(m-1) - 3m + 4\big]x^m = 0.$$

Step 2 — Indicial (auxiliary) equation.

$$m(m-1) - 3m + 4 = m^2 - m - 3m + 4 = m^2 - 4m + 4 = 0.$$ $$(m-2)^2 = 0 \;\Rightarrow\; m = 2,\, 2 \quad(\text{repeated root}).$$

Step 3 — General solution. For a repeated root $m$ in a Cauchy–Euler equation, the second independent solution carries a factor $\ln x$:

$$\boxed{y = \big(C_1 + C_2 \ln x\big)x^2}, \qquad x > 0,$$

where $C_1, C_2$ are arbitrary constants.

Setup. For a half-range sine series on $(0, L)$ with $L = 2$:

$$f(x) = \sum_{n=1}^{\infty} b_n \sin\frac{n\pi x}{L}, \qquad b_n = \frac{2}{L}\int_0^L f(x)\sin\frac{n\pi x}{L}\,dx.$$

Here $L = 2$, so $b_n = \dfrac{2}{2}\displaystyle\int_0^2 x\sin\frac{n\pi x}{2}\,dx = \int_0^2 x\sin\frac{n\pi x}{2}\,dx.$

Step 1 — Integrate by parts. With $u = x$, $dv = \sin\frac{n\pi x}{2}dx$:

$$\int x\sin\frac{n\pi x}{2}\,dx = -\frac{2x}{n\pi}\cos\frac{n\pi x}{2} + \frac{2}{n\pi}\int \cos\frac{n\pi x}{2}\,dx$$ $$= -\frac{2x}{n\pi}\cos\frac{n\pi x}{2} + \frac{4}{n^2\pi^2}\sin\frac{n\pi x}{2}.$$

Step 2 — Evaluate from $0$ to $2$. At $x=2$: $\cos\frac{n\pi\cdot2}{2} = \cos n\pi = (-1)^n$ and $\sin n\pi = 0$. At $x=0$ both terms vanish.

$$b_n = \left[-\frac{2(2)}{n\pi}(-1)^n + 0\right] - 0 = -\frac{4(-1)^n}{n\pi} = \frac{4(-1)^{n+1}}{n\pi}.$$

Step 3 — Series.

$$\boxed{x = \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin\frac{n\pi x}{2}}, \qquad 0 < x < 2.$$

Written out: $x = \dfrac{4}{\pi}\Big(\sin\dfrac{\pi x}{2} - \dfrac{1}{2}\sin\pi x + \dfrac{1}{3}\sin\dfrac{3\pi x}{2} - \cdots\Big).$