BE Civil Engineering (IOE, TU) Engineering Mathematics II (IOE, SH 451) Question Paper 2080 Nepal

Step 1 — Write in matrix form $A\mathbf{x}=\mathbf{b}$.

$$A=\begin{pmatrix}2&1&-1\\1&-1&2\\3&2&1\end{pmatrix},\quad \mathbf{x}=\begin{pmatrix}x\\y\\z\end{pmatrix},\quad \mathbf{b}=\begin{pmatrix}3\\8\\7\end{pmatrix}.$$

Step 2 — Determinant (non-singularity). Expand along row 1:

$$\det A = 2\big((-1)(1)-(2)(2)\big)-1\big((1)(1)-(2)(3)\big)+(-1)\big((1)(2)-(-1)(3)\big)$$ $$= 2(-1-4)-1(1-6)-1(2+3)=2(-5)-1(-5)-1(5)=-10+5-5=-10.$$

Since $\det A=-10\neq 0$, $A$ is non-singular and a unique solution exists.

Step 3 — Cofactors.

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

Step 4 — Adjoint (transpose of cofactor matrix).

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

Step 5 — Inverse.

$$A^{-1}=\frac{1}{\det A}\operatorname{adj}A=-\frac{1}{10}\begin{pmatrix}-5&-3&1\\5&5&-5\\5&-1&-3\end{pmatrix}.$$

Step 6 — Solve $\mathbf{x}=A^{-1}\mathbf{b}$.

$$\operatorname{adj}A\cdot\mathbf{b}=\begin{pmatrix}-5(3)-3(8)+1(7)\\5(3)+5(8)-5(7)\\5(3)-1(8)-3(7)\end{pmatrix}=\begin{pmatrix}-15-24+7\\15+40-35\\15-8-21\end{pmatrix}=\begin{pmatrix}-32\\20\\-14\end{pmatrix}.$$ $$\mathbf{x}=-\frac{1}{10}\begin{pmatrix}-32\\20\\-14\end{pmatrix}=\begin{pmatrix}3.2\\-2\\1.4\end{pmatrix}.$$

Step 7 — Verify in equation 1: $2(3.2)+(-2)-(1.4)=6.4-2-1.4=3$ ✓.

$$\boxed{x=\tfrac{16}{5}=3.2,\quad y=-2,\quad z=\tfrac{7}{5}=1.4.}$$

Step 1 — Data from the line. The line passes through $A(1,-2,0)$ with direction $\mathbf{d}=(2,-1,3)$.

Step 2 — Second in-plane vector. Since $P(2,-1,3)$ lies in the plane and $A$ is on the contained line,

$$\overrightarrow{AP}=(2-1,\,-1-(-2),\,3-0)=(1,1,3).$$

Step 3 — Normal vector $\mathbf{n}=\mathbf{d}\times\overrightarrow{AP}$:

$$\mathbf{n}=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&-1&3\\1&1&3\end{vmatrix}=\mathbf{i}((-1)(3)-(3)(1))-\mathbf{j}((2)(3)-(3)(1))+\mathbf{k}((2)(1)-(-1)(1)).$$ $$\mathbf{n}=\mathbf{i}(-3-3)-\mathbf{j}(6-3)+\mathbf{k}(2+1)=(-6,-3,3).$$

Divide by $-3$: take $\mathbf{n}=(2,1,-1)$.

Step 4 — Plane equation through $P(2,-1,3)$:

$$2(x-2)+1(y+1)-1(z-3)=0\ \Rightarrow\ 2x+y-z+( -4+1+3)=0\ \Rightarrow\ 2x+y-z=0.$$

Check: $A(1,-2,0)$: $2(1)+(-2)-0=0$ ✓; $P$: $2(2)+(-1)-3=0$ ✓.

$$\boxed{\text{(a) Plane: } 2x+y-z=0.}$$

Step 5 — Distance from origin. For plane $ax+by+cz+d=0$, $d=0$ here, so the plane passes through the origin.

$$\text{Distance}=\frac{|2(0)+0-0+0|}{\sqrt{2^2+1^2+(-1)^2}}=\frac{0}{\sqrt6}=0.$$ $$\boxed{\text{(b) Distance from origin} = 0\ \text{units (the plane passes through the origin).}}$$

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

$$m^2-3m+2=0\ \Rightarrow\ (m-1)(m-2)=0\ \Rightarrow\ m=1,\,2.$$ $$y_c=C_1e^{x}+C_2e^{2x}.$$

Step 2 — Particular integral for $4e^{3x}$. Try $y_{p1}=Ae^{3x}$. Then $y_{p1}'=3Ae^{3x}$, $y_{p1}''=9Ae^{3x}$. Substitute: $(9A-9A+2A)e^{3x}=2Ae^{3x}=4e^{3x}\Rightarrow A=2.$ So $y_{p1}=2e^{3x}$.

Step 3 — Particular integral for $\sin x$. Try $y_{p2}=B\cos x+C\sin x$.

$$y_{p2}'=-B\sin x+C\cos x,\quad y_{p2}''=-B\cos x-C\sin x.$$

Substitute into LHS $y''-3y'+2y$:

$$(-B\cos x-C\sin x)-3(-B\sin x+C\cos x)+2(B\cos x+C\sin x).$$

Collect $\cos x$: $(-B-3C+2B)=(B-3C)$. Collect $\sin x$: $(-C+3B+2C)=(3B+C)$. Match to RHS $0\cdot\cos x+1\cdot\sin x$:

$$B-3C=0,\qquad 3B+C=1.$$

From the first, $B=3C$. Then $3(3C)+C=10C=1\Rightarrow C=\tfrac{1}{10},\ B=\tfrac{3}{10}.$ So $y_{p2}=\tfrac{3}{10}\cos x+\tfrac{1}{10}\sin x$.

Step 4 — General solution.

$$\boxed{y=C_1e^{x}+C_2e^{2x}+2e^{3x}+\frac{3}{10}\cos x+\frac{1}{10}\sin x.}$$

Step 1 — Symmetry. $f(x)=x^2$ is even, so all sine coefficients $b_n=0$. The series is

$$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos nx.$$

Step 2 — Compute $a_0$.

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

Step 3 — Compute $a_n$.

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

Using integration 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 $\pi$: $\sin n\pi=0$, $\cos n\pi=(-1)^n$, so the value is $\dfrac{2\pi(-1)^n}{n^2}$. At $0$ everything vanishes. Thus

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

Step 4 — Fourier series.

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

Step 5 — Deduction. Put $x=\pi$ (a point of continuity of the periodic extension, with $f(\pi)=\pi^2$):

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

Therefore

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

Part (a) — Ratio test on $\sum \dfrac{n!}{n^n}$.

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

$$\frac{u_{n+1}}{u_n}=\frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\frac{(n+1)\,n^n}{(n+1)^{n+1}}=\frac{n^n}{(n+1)^n}=\left(\frac{n}{n+1}\right)^n=\frac{1}{\left(1+\frac1n\right)^n}.$$

Taking the limit,

$$\lim_{n\to\infty}\frac{u_{n+1}}{u_n}=\frac{1}{e}\approx 0.3679<1.$$

By the ratio test, since the limit $<1$, the series converges.

Part (b) — Power series $\sum \dfrac{(x-2)^n}{n\,3^n}$.

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

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

For convergence we need $\dfrac{|x-2|}{3}<1\Rightarrow |x-2|<3$.

Radius of convergence: $\boxed{R=3}$ (centred at $x=2$), giving the open interval $-1<x<5$.

Endpoint tests.

• At $x=5$: $\sum\dfrac{3^n}{n\,3^n}=\sum\dfrac{1}{n}$ — harmonic series, diverges.
• At $x=-1$: $\sum\dfrac{(-3)^n}{n\,3^n}=\sum\dfrac{(-1)^n}{n}$ — alternating harmonic, converges (conditionally).

Interval of convergence:

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

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

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

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

$$\begin{pmatrix}-1&1\\2&-2\end{pmatrix}\begin{pmatrix}v_1\\v_2\end{pmatrix}=0\ \Rightarrow\ -v_1+v_2=0\ \Rightarrow\ v_2=v_1.$$

Eigenvector: $\mathbf{v}_1=\begin{pmatrix}1\\1\end{pmatrix}.$

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

$$\begin{pmatrix}2&1\\2&1\end{pmatrix}\begin{pmatrix}v_1\\v_2\end{pmatrix}=0\ \Rightarrow\ 2v_1+v_2=0\ \Rightarrow\ v_2=-2v_1.$$

Eigenvector: $\mathbf{v}_2=\begin{pmatrix}1\\-2\end{pmatrix}.$

$$\boxed{\lambda_1=5,\ \mathbf{v}_1=(1,1)^T;\qquad \lambda_2=2,\ \mathbf{v}_2=(1,-2)^T.}$$

Step 1 — Radius. The sphere passes through the origin $O(0,0,0)$ with centre $C(2,-3,4)$, so

$$r=|OC|=\sqrt{2^2+(-3)^2+4^2}=\sqrt{4+9+16}=\sqrt{29}.$$

Step 2 — Equation of the sphere.

$$(x-2)^2+(y+3)^2+(z-4)^2=29.$$

Expanding: $x^2-4x+4+y^2+6y+9+z^2-8z+16=29$, i.e.

$$\boxed{x^2+y^2+z^2-4x+6y-8z=0.}$$

(The absence of a constant term confirms it passes through the origin.)

Step 3 — Tangent plane at the origin. The radius vector at the origin is $\overrightarrow{OC}=(2,-3,4)$, which is the normal to the tangent plane. The tangent plane through $O(0,0,0)$ with normal $(2,-3,4)$ is

$$2(x-0)-3(y-0)+4(z-0)=0.$$ $$\boxed{2x-3y+4z=0.}$$

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

$$\frac{dy}{dx}=mx^{m-1},\qquad \frac{d^2y}{dx^2}=m(m-1)x^{m-2}.$$

Step 2 — Substitute.

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

Step 3 — Indicial (auxiliary) equation.

$$m(m-1)-2m-4=m^2-m-2m-4=m^2-3m-4=0.$$ $$(m-4)(m+1)=0\ \Rightarrow\ m=4,\ m=-1.$$

Step 4 — General solution (distinct real roots):

$$\boxed{y=C_1x^{4}+C_2x^{-1}=C_1x^4+\frac{C_2}{x}.}$$

Step 1 — Identify dominant behaviour. For large $n$,

$$u_n=\frac{2n+1}{n^3+n}\sim\frac{2n}{n^3}=\frac{2}{n^2}.$$

Choose the comparison series $v_n=\dfrac{1}{n^2}$ (a $p$-series with $p=2>1$, which converges).

Step 2 — Limit comparison test.

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

Divide numerator and denominator by $n^3$:

$$L=\lim_{n\to\infty}\frac{2+\frac1n}{1+\frac{1}{n^2}}=\frac{2}{1}=2.$$

Step 3 — Conclusion. Since $0<L=2<\infty$ (finite and non-zero), $\sum u_n$ and $\sum v_n$ behave alike. As $\sum \frac{1}{n^2}$ converges,

$$\boxed{\sum_{n=1}^{\infty}\frac{2n+1}{n^3+n}\ \text{converges.}}$$

Step 1 — Form of half-range sine series. On $0<x<L$ 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 $b_n=\displaystyle\int_0^{2}x\sin\frac{n\pi x}{2}\,dx.$

Step 2 — Integrate by parts. Let $a=\dfrac{n\pi}{2}$. Then

$$\int_0^2 x\sin(ax)\,dx=\left[-\frac{x\cos ax}{a}+\frac{\sin ax}{a^2}\right]_0^2=-\frac{2\cos 2a}{a}+\frac{\sin 2a}{a^2}.$$

With $2a=n\pi$: $\cos n\pi=(-1)^n$ and $\sin n\pi=0$. Thus

$$\int_0^2 x\sin(ax)\,dx=-\frac{2(-1)^n}{a}=-\frac{2(-1)^n}{n\pi/2}=-\frac{4(-1)^n}{n\pi}=\frac{4(-1)^{n+1}}{n\pi}.$$

So $b_n=\dfrac{4(-1)^{n+1}}{n\pi}.$

Step 3 — Series.

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

Explicitly, $x=\dfrac{4}{\pi}\left(\sin\dfrac{\pi x}{2}-\dfrac12\sin\pi x+\dfrac13\sin\dfrac{3\pi x}{2}-\cdots\right).$

Step 1 — Identify points and directions.

• $L_1$: point $A=(1,2,3)$, direction $\mathbf{d}_1=(2,3,4)$.
• $L_2$: point $B=(2,4,5)$, direction $\mathbf{d}_2=(3,4,5)$.

Step 2 — Connecting vector.

$$\overrightarrow{AB}=B-A=(2-1,\,4-2,\,5-3)=(1,2,2).$$

Step 3 — Cross product $\mathbf{d}_1\times\mathbf{d}_2$.

$$\mathbf{d}_1\times\mathbf{d}_2=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\2&3&4\\3&4&5\end{vmatrix}=\mathbf{i}(3\cdot5-4\cdot4)-\mathbf{j}(2\cdot5-4\cdot3)+\mathbf{k}(2\cdot4-3\cdot3).$$ $$=\mathbf{i}(15-16)-\mathbf{j}(10-12)+\mathbf{k}(8-9)=(-1,\,2,\,-1).$$

Magnitude: $|\mathbf{d}_1\times\mathbf{d}_2|=\sqrt{(-1)^2+2^2+(-1)^2}=\sqrt{6}.$

Step 4 — Scalar triple product $\overrightarrow{AB}\cdot(\mathbf{d}_1\times\mathbf{d}_2)$:

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

Step 5 — Shortest distance.

$$d=\frac{|\overrightarrow{AB}\cdot(\mathbf{d}_1\times\mathbf{d}_2)|}{|\mathbf{d}_1\times\mathbf{d}_2|}=\frac{|1|}{\sqrt6}=\frac{1}{\sqrt6}=\frac{\sqrt6}{6}\approx 0.408\ \text{units}.$$ $$\boxed{\text{(a) Shortest distance}=\dfrac{1}{\sqrt6}\approx 0.408\ \text{units}.}$$

Part (b) — Classification.

• The directions $\mathbf{d}_1=(2,3,4)$ and $\mathbf{d}_2=(3,4,5)$ are not proportional ($2/3\neq3/4$), so the lines are not parallel.
• The scalar triple product $\overrightarrow{AB}\cdot(\mathbf{d}_1\times\mathbf{d}_2)=1\neq 0$, so the three vectors are non-coplanar; hence the lines do not intersect.

Since they are neither parallel nor intersecting,

$$\boxed{\text{(b) The lines are SKEW.}}$$