BE Civil Engineering (IOE, TU) Engineering Mathematics II (IOE, SH 451) Question Paper 2079 Nepal

Step 1 — Write the system in matrix form $AX = B$.

$$A=\begin{pmatrix}2&1&1\\1&3&2\\3&2&1\end{pmatrix},\quad X=\begin{pmatrix}x\\y\\z\end{pmatrix},\quad B=\begin{pmatrix}7\\13\\10\end{pmatrix}.$$

Step 2 — Determinant of $A$.

$$\det A = 2(3\cdot1-2\cdot2) - 1(1\cdot1-2\cdot3) + 1(1\cdot2-3\cdot3)$$ $$= 2(3-4) - 1(1-6) + 1(2-9) = 2(-1) -1(-5) + 1(-7) = -2+5-7 = -4.$$

Since $\det A = -4 \neq 0$, $A^{-1}$ exists and the solution is unique.

Step 3 — Cofactor matrix.

$$C_{11}=+(3-4)=-1,\; C_{12}=-(1-6)=5,\; C_{13}=+(2-9)=-7,$$ $$C_{21}=-(1-2)=1,\; C_{22}=+(2-3)=-1,\; C_{23}=-(4-3)=-1,$$ $$C_{31}=+(2-3)=-1,\; C_{32}=-(4-1)=-3,\; C_{33}=+(6-1)=5.$$

Step 4 — Adjoint (transpose of cofactor matrix).

$$\operatorname{adj}A = \begin{pmatrix}-1&1&-1\\5&-1&-3\\-7&-1&5\end{pmatrix}.$$

Step 5 — Inverse.

$$A^{-1}=\frac{1}{\det A}\operatorname{adj}A = -\frac{1}{4}\begin{pmatrix}-1&1&-1\\5&-1&-3\\-7&-1&5\end{pmatrix}.$$

Step 6 — Solve $X = A^{-1}B$.

Compute $\operatorname{adj}A \cdot B$:

$$\begin{pmatrix}-1&1&-1\\5&-1&-3\\-7&-1&5\end{pmatrix}\begin{pmatrix}7\\13\\10\end{pmatrix}=\begin{pmatrix}-7+13-10\\35-13-30\\-49-13+50\end{pmatrix}=\begin{pmatrix}-4\\-8\\-12\end{pmatrix}.$$

Then

$$X = -\frac{1}{4}\begin{pmatrix}-4\\-8\\-12\end{pmatrix}=\begin{pmatrix}1\\2\\3\end{pmatrix}.$$

Hence $\boxed{x=1,\; y=2,\; z=3}$.

Step 7 — Verification (third equation): $3(1)+2(2)+1(3)=3+4+3=10.$ ✓ (Also: $2(1)+2+3=7$ ✓, $1+6+6=13$ ✓.)

Step 1 — Two direction vectors in the plane.

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

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

$$\vec n = \begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\1&-1&1\\-1&1&3\end{vmatrix}.$$ $$n_x = (-1)(3)-(1)(1) = -3-1 = -4,$$ $$n_y = -\big[(1)(3)-(1)(-1)\big] = -(3+1) = -4,$$ $$n_z = (1)(1)-(-1)(-1) = 1-1 = 0.$$

So $\vec n = (-4,-4,0)$, or dividing by $-4$, $\vec n \parallel (1,1,0)$.

Step 3 — Equation of the plane. Using point $A(1,1,0)$ and normal $(1,1,0)$:

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

Plane: $\boxed{x+y=2}$.

Check the three points: $A:1+1=2$ ✓, $B:2+0=2$ ✓, $C:0+2=2$ ✓.

Step 4 — Perpendicular distance from $P(3,4,5)$.

$$d=\frac{|x_0+y_0-2|}{\sqrt{1^2+1^2+0^2}}=\frac{|3+4-2|}{\sqrt2}=\frac{5}{\sqrt2}=\frac{5\sqrt2}{2}.$$

Distance $= \dfrac{5}{\sqrt2}=\dfrac{5\sqrt2}{2}\approx 3.54$ units.

Step 1 — Complementary function (CF). Auxiliary equation:

$$m^2 - 5m + 6 = 0 \Rightarrow (m-2)(m-3)=0 \Rightarrow m=2,\,3.$$ $$y_c = C_1 e^{2x} + C_2 e^{3x}.$$

Step 2 — Particular integral for $e^{4x}$. With operator $f(D)=D^2-5D+6$, put $D=4$:

$$f(4)=16-20+6=2 \neq 0.$$ $$y_{p1}=\frac{1}{f(4)}e^{4x}=\frac{1}{2}e^{4x}.$$

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

$$y_{p2}=\frac{1}{D^2-5D+6}\sin x = \frac{1}{-1-5D+6}\sin x=\frac{1}{5-5D}\sin x=\frac{1}{5}\cdot\frac{1}{1-D}\sin x.$$

Multiply numerator and denominator by $(1+D)$:

$$=\frac{1}{5}\cdot\frac{1+D}{1-D^2}\sin x = \frac{1}{5}\cdot\frac{1+D}{1-(-1)}\sin x = \frac{1}{10}(1+D)\sin x.$$ $$=\frac{1}{10}(\sin x + \cos x).$$

Step 4 — General solution.

$$\boxed{y = C_1 e^{2x} + C_2 e^{3x} + \tfrac{1}{2}e^{4x} + \tfrac{1}{10}(\sin x + \cos x).}$$

Verification of the $\sin x$ part: let $y_{p2}=\frac{1}{10}(\sin x+\cos x)$, then $y'=\frac{1}{10}(\cos x-\sin x)$, $y''=\frac{1}{10}(-\sin x-\cos x)$. Substituting: $y''-5y'+6y=\frac{1}{10}[(-\sin x-\cos x)-5(\cos x-\sin x)+6(\sin x+\cos x)] = \frac{1}{10}[(-1+5+6)\sin x +(-1-5+6)\cos x] = \frac{1}{10}(10\sin x) = \sin x.$ ✓

Part (a) — Test $\sum n^2/2^n$ by the Ratio Test.

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

$$\frac{a_{n+1}}{a_n} = \frac{(n+1)^2}{2^{n+1}}\cdot\frac{2^n}{n^2}=\frac{1}{2}\left(\frac{n+1}{n}\right)^2 = \frac{1}{2}\left(1+\frac{1}{n}\right)^2.$$

Taking the limit:

$$L=\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = \frac{1}{2}\cdot 1 = \frac{1}{2}.$$

Since $L = \tfrac12 < 1$, by the Ratio Test the series converges.

Part (b) — General $\sum n^2/x^n$ for $x>0$.

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

$$\frac{b_{n+1}}{b_n}=\frac{(n+1)^2}{x^{n+1}}\cdot\frac{x^n}{n^2}=\frac{1}{x}\left(1+\frac1n\right)^2 \to \frac{1}{x}.$$

• If $\dfrac1x < 1$, i.e. $x>1$: series converges.
• If $\dfrac1x > 1$, i.e. $0<x<1$: series diverges.
• If $x=1$: the series becomes $\sum n^2$, whose terms $\to\infty$, so it diverges (term test fails).

Conclusion: the series $\displaystyle\sum \frac{n^2}{x^n}$ converges precisely for $\boxed{x>1}$.

Step 1 — Symmetry. $f(x)=x^2$ is an even function on $(-\pi,\pi)$, so all sine coefficients vanish: $b_n=0$.

Step 2 — Constant term $a_0$.

$$a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}x^2\,dx=\frac{2}{\pi}\int_0^{\pi}x^2\,dx=\frac{2}{\pi}\cdot\frac{\pi^3}{3}=\frac{2\pi^2}{3}.$$

(The series uses $\tfrac{a_0}{2}=\tfrac{\pi^2}{3}$ as the mean term.)

Step 3 — Cosine coefficients $a_n$.

$$a_n=\frac{2}{\pi}\int_0^{\pi}x^2\cos(nx)\,dx.$$

Integrate by parts twice. The standard result is

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

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

$$\int_0^{\pi}x^2\cos(nx)\,dx = \frac{2\pi(-1)^n}{n^2}.$$

Therefore

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

Step 4 — Fourier series.

$$\boxed{x^2 = \frac{\pi^2}{3} + \sum_{n=1}^{\infty}\frac{4(-1)^n}{n^2}\cos(nx) = \frac{\pi^2}{3} - 4\left(\cos x - \frac{\cos 2x}{4} + \frac{\cos 3x}{9} - \cdots\right).}$$

Step 5 — Deduce $\sum 1/n^2$. Put $x=\pi$ (where $f$ is continuous in the periodic extension and equals $\pi^2$):

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

Since $\cos(n\pi)=(-1)^n$, we have $(-1)^n(-1)^n=1$, so

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

Rearrange:

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

Step 1 — Eliminate below the first pivot.

$R_2 \to R_2 - 2R_1$: $(2-2,\;4-4,\;7-6)=(0,0,1)$. $R_3 \to R_3 - 3R_1$: $(3-3,\;6-6,\;10-9)=(0,0,1)$.

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

Step 2 — Eliminate the third row.

$R_3 \to R_3 - R_2$: $(0,0,0)$.

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

Step 3 — Count non-zero rows. There are 2 non-zero rows.

Rank$(A) = \boxed{2}$.

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 $\vec b_2=(3,4,5)$.

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

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

So $\vec b_1\times\vec b_2=(-1,2,-1)$, with $|\vec b_1\times\vec b_2|=\sqrt{1+4+1}=\sqrt6.$

Step 3 — Vector joining the points. $\vec{AC}=C-A=(1,2,2).$

Step 4 — Scalar triple product.

$$\vec{AC}\cdot(\vec b_1\times\vec b_2)=(1)(-1)+(2)(2)+(2)(-1)=-1+4-2=1.$$

Step 5 — Shortest distance.

$$d=\frac{|\vec{AC}\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}=\frac{|1|}{\sqrt6}=\frac{1}{\sqrt6}.$$

Shortest distance $=\dfrac{1}{\sqrt6}=\dfrac{\sqrt6}{6}\approx 0.408$ units.

Step 1 — Trial solution. For a Cauchy–Euler equation try $y=x^m$. Then $y'=mx^{m-1}$ and $y''=m(m-1)x^{m-2}$.

Step 2 — Substitute.

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

Step 3 — Indicial (auxiliary) equation.

$$m^2 - m - 3m + 4 = 0 \;\Rightarrow\; m^2 - 4m + 4 = 0 \;\Rightarrow\; (m-2)^2 = 0.$$

This gives a repeated root $m=2$.

Step 4 — General solution. For a repeated root $m$, the solution is $y=(C_1 + C_2\ln x)x^m$:

$$\boxed{y = (C_1 + C_2\ln x)\,x^2.}$$

Check: $y_2=x^2\ln x$ gives $y_2'=2x\ln x + x$, $y_2''=2\ln x+3$. Then $x^2(2\ln x+3)-3x(2x\ln x+x)+4x^2\ln x = 2x^2\ln x+3x^2-6x^2\ln x-3x^2+4x^2\ln x = 0.$ ✓

Step 1 — Choose a comparison series. For large $n$, $\sqrt{n(n+1)}\approx n$, so compare with the $p$-series $\sum 1/n$ (the harmonic series, $p=1$), which diverges.

Step 2 — Limit comparison test. Let $a_n=\dfrac{1}{\sqrt{n(n+1)}}$ and $b_n=\dfrac{1}{n}$. Then

$$\frac{a_n}{b_n}=\frac{1/\sqrt{n(n+1)}}{1/n}=\frac{n}{\sqrt{n(n+1)}}=\frac{n}{\sqrt{n^2+n}}=\frac{1}{\sqrt{1+\frac1n}}.$$

Step 3 — Take the limit.

$$L=\lim_{n\to\infty}\frac{a_n}{b_n}=\frac{1}{\sqrt{1+0}}=1.$$

Since $0<L=1<\infty$ (finite and non-zero), $\sum a_n$ and $\sum b_n$ behave alike.

Step 4 — Conclusion. Because $\sum \dfrac1n$ diverges, the given series

$$\sum_{n=1}^{\infty}\frac{1}{\sqrt{n(n+1)}}\;\textbf{diverges}.$$

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

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

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

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

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

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

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

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

Result: $\lambda_1=5$ with $v_1=(1,1)^T$; $\lambda_2=2$ with $v_2=(1,-2)^T$.

Check: $Av_1=\binom{4+1}{2+3}=\binom{5}{5}=5v_1$ ✓; $Av_2=\binom{4-2}{2-6}=\binom{2}{-4}=2v_2$ ✓.

Step 1 — Form of the 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 $f(x)=x$.

$$b_n=\frac{2}{\pi}\int_0^{\pi} x\sin(nx)\,dx.$$

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

$$\int_0^{\pi} x\sin(nx)\,dx = \left[-\frac{x\cos(nx)}{n}\right]_0^{\pi} + \frac{1}{n}\int_0^{\pi}\cos(nx)\,dx.$$

First term: $-\dfrac{\pi\cos(n\pi)}{n} = -\dfrac{\pi(-1)^n}{n}$. Second term: $\dfrac{1}{n}\left[\dfrac{\sin(nx)}{n}\right]_0^{\pi}=0$ (since $\sin(n\pi)=0$).

So

$$\int_0^{\pi} x\sin(nx)\,dx = -\frac{\pi(-1)^n}{n}=\frac{\pi(-1)^{n+1}}{n}.$$

Step 3 — Coefficient.

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

Step 4 — Series.

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