BE Civil Engineering (IOE, TU) Engineering Mathematics II (IOE, SH 451) Question Paper 2078 Nepal

Step 1 — Write in matrix form $AX=B$.

$$A=\begin{pmatrix}1&2&3\\2&4&5\\3&5&6\end{pmatrix},\quad X=\begin{pmatrix}x\\y\\z\end{pmatrix},\quad B=\begin{pmatrix}6\\13\\18\end{pmatrix}.$$

Step 2 — Condition for applicability. The inverse method requires $\det A \ne 0$ (i.e. $A$ is non-singular), so that a unique solution exists.

Step 3 — Compute $\det A$. Expand along the first row:

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

Since $\det A = -1 \ne 0$, the method applies.

Step 4 — Cofactor matrix.

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

Step 5 — Adjoint (transpose of cofactor matrix).

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

Step 6 — Inverse.

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

Step 7 — Solve $X=A^{-1}B$.

$$\begin{aligned} x&=1(6)+(-3)(13)+2(18)=6-39+36=3,\\ y&=(-3)(6)+3(13)+(-1)(18)=-18+39-18=3,\\ z&=2(6)+(-1)(13)+0(18)=12-13+0=-1. \end{aligned}$$

Step 8 — Verify in the original equations.

• Eq.1: $3+2(3)+3(-1)=3+6-3=6$ ✓
• Eq.2: $2(3)+4(3)+5(-1)=6+12-5=13$ ✓
• Eq.3: $3(3)+5(3)+6(-1)=9+15-6=18$ ✓

Final answer: $\boxed{x=3,\; y=3,\; z=-1}$.

Step 1 — Family of planes through the line of intersection. Any plane through the intersection of the two given planes is

$$(2x - y + 3z - 1) + \lambda (x + 2y - z + 2) = 0.$$

Group terms:

$$(2+\lambda)x + (-1+2\lambda)y + (3-\lambda)z + (-1+2\lambda) = 0.$$

Its normal vector is $\vec{n}=\big(2+\lambda,\; -1+2\lambda,\; 3-\lambda\big).$

Step 2 — Perpendicularity condition. Two planes are perpendicular when their normals are orthogonal. The normal of $3x-4y-z+6=0$ is $\vec{m}=(3,-4,-1)$. Require $\vec{n}\cdot\vec{m}=0$:

$$3(2+\lambda) - 4(-1+2\lambda) - 1(3-\lambda) = 0.$$ $$6+3\lambda + 4 - 8\lambda - 3 + \lambda = 0 \;\Rightarrow\; (6+4-3) + (3-8+1)\lambda = 0 \;\Rightarrow\; 7 - 4\lambda = 0.$$

Hence $\lambda = \dfrac{7}{4}.$

Step 3 — Substitute $\lambda=7/4$. Multiply the family equation through by $4$ for convenience, with $4\lambda = 7$:

$$(8+7)x + (-4+14)y + (12-7)z + (-4+14) = 0$$ $$15x + 10y + 5z + 10 = 0.$$

Divide by $5$:

$$\boxed{3x + 2y + z + 2 = 0.}$$

Step 4 — Verification of perpendicularity. Normal of the answer is $(3,2,1)$; dot with $(3,-4,-1)$ gives $9-8-1=0$ ✓. The plane also belongs to the family, so it contains the line of intersection.

Step 1 — Complementary function (CF). Auxiliary equation: $m^2 - 3m + 2 = 0 \Rightarrow (m-1)(m-2)=0 \Rightarrow m=1,2.$

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

Step 2 — Particular integral for $e^{3x}$. Using $\dfrac{1}{f(D)}e^{ax}=\dfrac{1}{f(a)}e^{ax}$ with $f(D)=D^2-3D+2$ and $a=3$:

$$f(3)=9-9+2=2 \ne 0 \Rightarrow y_{p1}=\frac{1}{2}e^{3x}.$$

Step 3 — Particular integral for $\sin x$. Replace $D^2\to -1$ in $f(D)$:

$$y_{p2}=\frac{1}{D^2-3D+2}\sin x = \frac{1}{(-1)-3D+2}\sin x = \frac{1}{1-3D}\sin x.$$

Rationalise by multiplying numerator and denominator by $(1+3D)$:

$$y_{p2}=\frac{1+3D}{1-9D^2}\sin x = \frac{1+3D}{1-9(-1)}\sin x = \frac{1+3D}{10}\sin x.$$

Now $D\sin x=\cos x$, so

$$y_{p2}=\frac{\sin x + 3\cos x}{10}.$$

Step 4 — General solution.

$$\boxed{y = C_1 e^{x} + C_2 e^{2x} + \tfrac{1}{2}e^{3x} + \tfrac{1}{10}\big(\sin x + 3\cos x\big).}$$

Check (PI for $e^{3x}$): if $y=\tfrac12 e^{3x}$ then $y''-3y'+2y=\tfrac12(9-9+2)e^{3x}=e^{3x}$ ✓.

Step 1 — Note symmetry. $f(x)=x^2$ is even on $(-\pi,\pi)$, so all sine coefficients $b_n=0$.

Step 2 — 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 3 — Cosine coefficients $a_n$.

$$a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}x^2\cos nx\,dx=\frac{2}{\pi}\int_{0}^{\pi}x^2\cos nx\,dx.$$

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.

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

Therefore

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

Step 4 — Fourier series. With $f(x)=\dfrac{a_0}{2}+\sum a_n\cos nx$:

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

Step 5 — Deduction. Put $x=\pi$ (a point where the periodic extension is continuous, value $\pi^2$; here $\cos n\pi=(-1)^n$ so $(-1)^n\cos n\pi=1$):

$$\pi^2 = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{(-1)^n(-1)^n}{n^2} = \frac{\pi^2}{3} + 4\sum_{n=1}^{\infty}\frac{1}{n^2}.$$

So $4\sum \frac{1}{n^2} = \pi^2 - \frac{\pi^2}{3} = \frac{2\pi^2}{3}$, giving

$$\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{2\pi^2}{3}\cdot\frac{1}{4} = \boxed{\frac{\pi^2}{6}.}$$

(a) $\sum \dfrac{n!}{n^n}$ — Ratio test. Let $u_n=\dfrac{n!}{n^n}$. Then

$$\frac{u_{n+1}}{u_n}=\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\frac{(n+1)\,n^n}{(n+1)^{n+1}}=\frac{n^n}{(n+1)^n}=\left(\frac{n}{n+1}\right)^n=\frac{1}{\left(1+\frac1n\right)^n}.$$ $$\lim_{n\to\infty}\frac{u_{n+1}}{u_n}=\frac{1}{e}\approx 0.368 < 1.$$

By the ratio test the series converges.

(b) $\sum \dfrac{n^2}{2^n}$ — Ratio test.

$$\frac{u_{n+1}}{u_n}=\frac{(n+1)^2}{2^{n+1}}\cdot\frac{2^n}{n^2}=\frac{1}{2}\left(\frac{n+1}{n}\right)^2.$$ $$\lim_{n\to\infty}\frac{u_{n+1}}{u_n}=\frac12 \cdot 1 = \frac12 < 1.$$

Hence the series converges.

(c) $\sum \dfrac{1}{n^2+1}$ — Comparison test. For all $n\ge 1$, $\;n^2+1 > n^2$, so $\dfrac{1}{n^2+1} < \dfrac{1}{n^2}$. The $p$-series $\sum \dfrac{1}{n^2}$ converges ($p=2>1$). By the comparison test, the smaller-term series $\sum \dfrac{1}{n^2+1}$ also converges.

Summary: all three series converge.

Step 1 — Row operations. $R_2\to R_2-2R_1,\; R_3\to R_3-3R_1,\; R_4\to R_4-6R_1$:

$$\begin{pmatrix}1&2&3&0\\0&0&-3&2\\0&-4&-8&3\\0&-4&-11&5\end{pmatrix}.$$

Step 2 — Swap $R_2\leftrightarrow R_3$ to bring a pivot into column 2:

$$\begin{pmatrix}1&2&3&0\\0&-4&-8&3\\0&0&-3&2\\0&-4&-11&5\end{pmatrix}.$$

Step 3 — $R_4\to R_4-R_2$:

$$\begin{pmatrix}1&2&3&0\\0&-4&-8&3\\0&0&-3&2\\0&0&-3&2\end{pmatrix}.$$

Step 4 — $R_4\to R_4-R_3$:

$$\begin{pmatrix}1&2&3&0\\0&-4&-8&3\\0&0&-3&2\\0&0&0&0\end{pmatrix}.$$

Step 5 — Count non-zero rows. There are 3 non-zero rows.

$$\boxed{\operatorname{rank}(A)=3.}$$

Step 1 — Identify points and direction vectors. Line 1: point $A=(1,2,3)$, direction $\vec{b_1}=(2,3,4)$. Line 2: point $C=(2,4,5)$, direction $\vec{b_2}=(3,4,5)$.

$$\vec{AC}=C-A=(1,2,2).$$

Step 2 — Cross product $\vec{b_1}\times\vec{b_2}$.

$$\vec{b_1}\times\vec{b_2}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&3&4\\3&4&5\end{vmatrix}=\mathbf{i}(3\cdot5-4\cdot4)-\mathbf{j}(2\cdot5-4\cdot3)+\mathbf{k}(2\cdot4-3\cdot3).$$ $$=\mathbf{i}(15-16)-\mathbf{j}(10-12)+\mathbf{k}(8-9)=(-1,\,2,\,-1).$$

Magnitude: $|\vec{b_1}\times\vec{b_2}|=\sqrt{(-1)^2+2^2+(-1)^2}=\sqrt{1+4+1}=\sqrt6.$

Step 3 — Scalar triple product $\vec{AC}\cdot(\vec{b_1}\times\vec{b_2})$.

$$(1,2,2)\cdot(-1,2,-1)=1(-1)+2(2)+2(-1)=-1+4-2=1.$$

Step 4 — Shortest distance.

$$d=\frac{|\vec{AC}\cdot(\vec{b_1}\times\vec{b_2})|}{|\vec{b_1}\times\vec{b_2}|}=\frac{|1|}{\sqrt6}=\frac{1}{\sqrt6}.$$ $$\boxed{d=\dfrac{1}{\sqrt6}\approx 0.408\text{ units}.}$$

Step 1 — Substitute $x=e^t$ (so $t=\ln x$). With $\theta=\dfrac{d}{dt}$: $x\dfrac{dy}{dx}=\theta y$ and $x^2\dfrac{d^2y}{dx^2}=\theta(\theta-1)y$. The equation becomes

$$\theta(\theta-1)y - 2\theta y - 4y = e^{4t},$$ $$\big[\theta^2 - \theta - 2\theta - 4\big]y = e^{4t} \;\Rightarrow\; (\theta^2 - 3\theta - 4)y = e^{4t}.$$

Step 2 — Complementary function. Auxiliary: $m^2-3m-4=0 \Rightarrow (m-4)(m+1)=0 \Rightarrow m=4,-1.$

$$y_c = C_1 e^{4t} + C_2 e^{-t} = C_1 x^4 + C_2 x^{-1}.$$

Step 3 — Particular integral.

$$y_p=\frac{1}{\theta^2-3\theta-4}e^{4t}.$$

Since $\theta=4$ makes the denominator zero (it is a root), use the standard rule $\dfrac{1}{f(\theta)}e^{at}=\dfrac{t}{f'(a)}e^{at}$ when $f(a)=0$. Here $f'(\theta)=2\theta-3$, so $f'(4)=8-3=5$:

$$y_p=\frac{t}{5}e^{4t}=\frac{\ln x}{5}x^4.$$

Step 4 — General solution.

$$\boxed{y = C_1 x^4 + \frac{C_2}{x} + \frac{x^4\ln x}{5}.}$$

Step 1 — Apply the ratio test. Let $u_n=\dfrac{(x-2)^n}{n\,3^n}$. Then

$$\left|\frac{u_{n+1}}{u_n}\right|=\left|\frac{(x-2)^{n+1}}{(n+1)3^{n+1}}\cdot\frac{n\,3^n}{(x-2)^n}\right|=\frac{n}{n+1}\cdot\frac{|x-2|}{3}.$$ $$\lim_{n\to\infty}\left|\frac{u_{n+1}}{u_n}\right|=\frac{|x-2|}{3}.$$

For convergence this limit must be $<1$:

$$\frac{|x-2|}{3}<1 \;\Rightarrow\; |x-2|<3 \;\Rightarrow\; -1 < x < 5.$$

Radius of convergence $R=3$, centred at $x=2$.

Step 2 — Test endpoints.

At $x=5$: series becomes $\displaystyle\sum \frac{3^n}{n\,3^n}=\sum\frac{1}{n}$, the harmonic series, which diverges.

At $x=-1$: series becomes $\displaystyle\sum \frac{(-3)^n}{n\,3^n}=\sum\frac{(-1)^n}{n}$, the alternating harmonic series, which converges (by the Leibniz test).

Step 3 — Interval of convergence.

$$\boxed{-1 \le x < 5,\quad\text{i.e. } [-1,\,5).}$$

Step 1 — Form of half-range sine series. On $(0,\pi)$ the half-range sine series is

$$f(x)=\sum_{n=1}^{\infty}b_n\sin nx,\qquad b_n=\frac{2}{\pi}\int_{0}^{\pi}f(x)\sin nx\,dx.$$

Step 2 — Compute $b_n$ for $f(x)=x$.

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

Integration by parts ($u=x,\;dv=\sin nx\,dx$):

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

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

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

Therefore

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

Step 3 — Series.

$$\boxed{x = 2\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin nx = 2\left(\sin x - \frac{\sin 2x}{2} + \frac{\sin 3x}{3} - \cdots\right),\quad 0<x<\pi.}$$

Step 1 — Characteristic equation $\det(A-\lambda I)=0$.

$$\det\begin{pmatrix}4-\lambda & 1\\ 2 & 3-\lambda\end{pmatrix}=(4-\lambda)(3-\lambda)-(1)(2)=0.$$ $$\lambda^2 - 7\lambda + 12 - 2 = 0 \;\Rightarrow\; \lambda^2 - 7\lambda + 10 = 0.$$ $$(\lambda-5)(\lambda-2)=0 \;\Rightarrow\; \lambda_1=5,\;\lambda_2=2.$$

Step 2 — Eigenvector for $\lambda_1=5$. Solve $(A-5I)v=0$:

$$\begin{pmatrix}-1 & 1\\ 2 & -2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=0 \;\Rightarrow\; -x+y=0 \;\Rightarrow\; y=x.$$

Taking $x=1$: $v_1=\begin{pmatrix}1\\1\end{pmatrix}.$

Step 3 — Eigenvector for $\lambda_2=2$. Solve $(A-2I)v=0$:

$$\begin{pmatrix}2 & 1\\ 2 & 1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=0 \;\Rightarrow\; 2x+y=0 \;\Rightarrow\; y=-2x.$$

Taking $x=1$: $v_2=\begin{pmatrix}1\\-2\end{pmatrix}.$

Step 4 — Result.

$$\boxed{\lambda_1=5,\;v_1=\begin{pmatrix}1\\1\end{pmatrix};\qquad \lambda_2=2,\;v_2=\begin{pmatrix}1\\-2\end{pmatrix}.}$$

Check ($\lambda=5$): $Av_1=\begin{pmatrix}4+1\\2+3\end{pmatrix}=\begin{pmatrix}5\\5\end{pmatrix}=5v_1$ ✓.