Solution:

Given:

A=\begin{bmatrix} 1&1&1&-1\\ 1&2&3&4\\ 3&4&5&2 \end{bmatrix}.

Required: Find \operatorname{rank}(A).

Method: Reduce the matrix to row-echelon form. The rank equals the number of non-zero rows in echelon form.

Apply the elementary row operations

R_2\to R_2-R_1,\qquad R_3\to R_3-3R_1.

Thus

\begin{bmatrix} 1&1&1&-1\\ 1&2&3&4\\ 3&4&5&2 \end{bmatrix} \sim \begin{bmatrix} 1&1&1&-1\\ 0&1&2&5\\ 0&1&2&5 \end{bmatrix}.

Now perform

R_3\to R_3-R_2.

We get

\begin{bmatrix} 1&1&1&-1\\ 0&1&2&5\\ 0&0&0&0 \end{bmatrix}.

There are exactly two non-zero rows, hence

\operatorname{rank}(A)\le2.

To verify that the rank is not less than 2, consider the minor formed by the first two rows and first two columns:

\begin{vmatrix} 1&1\\ 1&2 \end{vmatrix} =2-1=1\ne0.

Therefore a non-zero minor of order 2 exists, so

\operatorname{rank}(A)\ge2.

Final Answer:

\boxed{\operatorname{rank}(A)=2}

Solution:

We are given

W=\{(a,b,c)\in\mathbb R^3:a-3b+4c=0\}.

Subspace criterion: A non-empty subset W of a vector space is a subspace if it is closed under vector addition and scalar multiplication.

1. Non-empty.

The zero vector belongs to W, because

0-3(0)+4(0)=0.

2. Closure under addition.

Let

u=(a_1,b_1,c_1)\in W,\qquad v=(a_2,b_2,c_2)\in W.

Then

a_1-3b_1+4c_1=0,

and

a_2-3b_2+4c_2=0.

For u+v=(a_1+a_2,b_1+b_2,c_1+c_2),

\begin{aligned} &(a_1+a_2)-3(b_1+b_2)+4(c_1+c_2)\\ &=(a_1-3b_1+4c_1)+(a_2-3b_2+4c_2)\\ &=0+0=0. \end{aligned}

Hence u+v\in W.

3. Closure under scalar multiplication.

Let \lambda\in\mathbb R and u=(a_1,b_1,c_1)\in W. Then

\lambda u=(\lambda a_1,\lambda b_1,\lambda c_1).

Moreover,

\lambda a_1-3\lambda b_1+4\lambda c_1 = \lambda(a_1-3b_1+4c_1) =0.

Hence \lambda u\in W.

Thus W is non-empty and is closed under the vector-space operations.

Final Answer:

\boxed{W\text{ is a subspace of }V_3(\mathbb R).}

Solution:

The function is

f(x)= \begin{cases} \dfrac{x-|x|}{x},&x\ne0,\\ 1,&x=0. \end{cases}

Step 1: Simplify the function on either side of the origin.

If x>0, then |x|=x, so

f(x)=\frac{x-x}{x}=0.

If x<0[/katex], then [katex]|x|=-x[/katex], so</p> [katex display="true"] f(x)=\frac{x-(-x)}{x} =\frac{2x}{x} =2.

Hence

\lim_{x\to0^-}f(x)=2, \qquad \lim_{x\to0^+}f(x)=0.

Since the left-hand and right-hand limits are different,

\lim_{x\to0}f(x)

does not exist.

Also f(0)=1. Therefore the continuity condition

\lim_{x\to0}f(x)=f(0)

is not satisfied.

Thus f is discontinuous at x=0.

Differentiability: Every differentiable function is continuous. Since f is not continuous at the origin, it cannot be differentiable there.

Equivalently, the difference quotient itself confirms this:

\frac{f(h)-f(0)}{h} = \begin{cases} -\dfrac1h,&h>0,\\[1mm] \dfrac1h,&h<0, \end{cases} [/katex] <p>which has no finite limit as [katex]h\to0.

Final Answer:

\boxed{f\text{ is neither continuous nor differentiable at }x=0.}

Solution:

The equation is

x^{2}+4y^{2}+z^{2}-4yz+2zx-4xy-2x+4y-2z-3=0.

Group the quadratic terms:

x^2+4y^2+z^2-4xy+2xz-4yz = (x-2y+z)^2.

Also the linear terms satisfy

-2x+4y-2z = -2(x-2y+z).

Therefore the equation becomes

(x-2y+z)^2 -2(x-2y+z)-3=0.

Put

u=x-2y+z.

Then

u^2-2u-3=0.

Factorizing,

(u-3)(u+1)=0.

Hence

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

Thus the given second-degree equation represents two distinct parallel planes:

x-2y+z=3

and

x-2y+z=-1.

They are parallel because both have the same normal vector (1,-2,1).

Final Answer:

\boxed{x-2y+z=3,\qquad x-2y+z=-1.}

The surface is a pair of parallel planes.

Solution:

The differential equation is

(D^4+2D^3-3D^2)y=x^3+2\sin x.

Step 1: Complementary function.

The auxiliary equation is

m^4+2m^3-3m^2=0.

Factor:

m^2(m^2+2m-3) = m^2(m-1)(m+3) =0.

The roots are

m=0,0,1,-3.

Hence

y_c=C_1+C_2x+C_3e^x+C_4e^{-3x}.

Step 2: Particular integral for x^3.

Because the operator contains D^2, take

y_{p1}=Ax^5+Bx^4+Cx^3+Dx^2.

Substitution in (D^4+2D^3-3D^2)y_{p1} gives

-60Ax^3+(120A-36B)x^2 +(120A+48B-18C)x +(24B+12C-6D).

Equating this to x^3 yields

-60A=1, 120A-36B=0, 120A+48B-18C=0, 24B+12C-6D=0.

Solving successively,

A=-\frac1{60},\qquad B=-\frac1{18},\qquad C=-\frac7{27},\qquad D=-\frac{20}{27}.

Therefore

y_{p1} = -\frac{x^5}{60} -\frac{x^4}{18} -\frac{7x^3}{27} -\frac{20x^2}{27}.

Step 3: Particular integral for 2\sin x.

Take

y_{p2}=A\sin x+B\cos x.

Then

(D^4+2D^3-3D^2)y_{p2} = (4A+2B)\sin x+(-2A+4B)\cos x.

Equating with 2\sin x,

4A+2B=2,\qquad -2A+4B=0.

Hence

A=\frac25,\qquad B=\frac15.

Thus

y_{p2}=\frac25\sin x+\frac15\cos x.

Adding the complementary function and both particular integrals gives the complete solution.

Final Answer:

\boxed{ \begin{aligned} y={}&C_1+C_2x+C_3e^x+C_4e^{-3x}\\ &-\frac{x^5}{60} -\frac{x^4}{18} -\frac{7x^3}{27} -\frac{20x^2}{27} +\frac25\sin x+\frac15\cos x. \end{aligned}}

Solution:

Let the two equal uniform rods, each of length 2a, be rigidly joined at their common end A, the angle between them being \alpha. In the symmetrical position on the smooth sphere, the bisector of the angle passes through the centre O of the sphere.

Let T be the point of contact of one rod with the sphere. Since a radius to a point of contact is perpendicular to the tangent rod, triangle AOT is right-angled at T. Also

\angle OAT=\frac{\alpha}{2}.

Therefore

\sin\frac{\alpha}{2} = \frac{OT}{OA} = \frac{r}{OA},

so

OA=r\csc\frac{\alpha}{2}. \tag{1}

Let G_1 and G_2 be the centres of gravity of the two rods. Since each rod has length 2a,

AG_1=AG_2=a.

The combined centre of gravity G is the midpoint of G_1G_2 and lies on the angle bisector. From the right triangle formed by resolving AG_1 along the angle bisector,

AG=a\cos\frac{\alpha}{2}. \tag{2}

The relative position of G and O determines the stability.

From (1) and (2), G lies below O when

AG>AO.

That is,

a\cos\frac{\alpha}{2} > r\csc\frac{\alpha}{2}.

Multiplying by \sin(\alpha/2),

a\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}>r.

Using 2\sin(\alpha/2)\cos(\alpha/2)=\sin\alpha,

\frac a2\sin\alpha>r,

or

\sin\alpha>\frac{2r}{a}. \tag{3}

Stability test by potential energy.

Let

d=AG-AO.

For a small angular displacement \theta of the rigid system about the sphere centre, the height of the combined centre of gravity relative to O changes as -d\cos\theta, apart from an additive constant. If the total weight of the rods is W, the variable part of the potential energy is

V(\theta)=-Wd\cos\theta.

At the symmetrical position \theta=0,

V'(0)=0,

and

V''(0)=Wd.

Thus:

• If d>0, then V''(0)>0 and the equilibrium is stable.
• If d<0[/katex], then [katex]V''(0)<0[/katex] and the equilibrium is unstable.</li> </ul><p>By (3), [katex]d>0 is exactly equivalent to \sin\alpha>2r/a.

  Final Answer:

  \boxed{ \begin{array}{ll} \sin\alpha>\dfrac{2r}{a} &\Rightarrow\text{stable equilibrium},\\[2mm] \sin\alpha<\dfrac{2r}{a} &\Rightarrow\text{unstable equilibrium}. \end{array}} [/katex] <p>The equality case is the limiting neutral case.</p></div> </div><div id="q2a" class="math-qna-wrapper"> <div class="question-box"> <h2>HPAS 2013 Maths Optional Paper-1 Question 2(a)</h2> <p>Find the characteristic roots and characteristic vectors of the following matrix :</p> [katex display="true"] \begin{bmatrix}1&2&3\\ 0&-4&2\\ 0&0&7\end{bmatrix}

Solution:

Let

A= \begin{bmatrix} 1&2&3\\ 0&-4&2\\ 0&0&7 \end{bmatrix}.

Because A is upper triangular, its characteristic roots are its diagonal entries. Explicitly,

\det(A-\lambda I) = (1-\lambda)(-4-\lambda)(7-\lambda).

Thus the characteristic roots are

\lambda_1=1,\qquad \lambda_2=-4,\qquad \lambda_3=7.

Eigenvector for \lambda=1.

Solve (A-I)X=0:

\begin{bmatrix} 0&2&3\\ 0&-5&2\\ 0&0&6 \end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix}=0.

From the last row z=0, then y=0, while x is arbitrary. Take x=1. Hence

v_1=(1,0,0)^T.

Eigenvector for \lambda=-4.

Solve (A+4I)X=0:

\begin{bmatrix} 5&2&3\\ 0&0&2\\ 0&0&11 \end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix}=0.

Thus z=0 and

5x+2y=0.

Taking x=-2 gives y=5. Therefore

v_2=(-2,5,0)^T.

Eigenvector for \lambda=7.

Solve (A-7I)X=0:

\begin{bmatrix} -6&2&3\\ 0&-11&2\\ 0&0&0 \end{bmatrix} \begin{bmatrix}x\\y\\z\end{bmatrix}=0.

From the second row,

-11y+2z=0 \quad\Rightarrow\quad z=\frac{11}{2}y.

Choose y=12, so z=66. The first equation gives

-6x+24+198=0 \quad\Rightarrow\quad x=37.

Hence

v_3=(37,12,66)^T.

Final Answer:

\boxed{\lambda=1,-4,7}

with corresponding characteristic vectors, for example,

\boxed{ (1,0,0)^T,\quad (-2,5,0)^T,\quad (37,12,66)^T.}

Solution:

The system is

\begin{cases} 2x_1+3x_2+x_3=9,\\ x_1+2x_2+3x_3=6,\\ 3x_1+x_2+2x_3=8. \end{cases}

Write it as

AX=B,

where

A= \begin{pmatrix} 2&3&1\\ 1&2&3\\ 3&1&2 \end{pmatrix}, \quad X= \begin{pmatrix} x_1\\x_2\\x_3 \end{pmatrix}, \quad B= \begin{pmatrix} 9\\6\\8 \end{pmatrix}.

Compute the determinant:

\begin{aligned} |A| &= 2\begin{vmatrix}2&3\\1&2\end{vmatrix} -3\begin{vmatrix}1&3\\3&2\end{vmatrix} +\begin{vmatrix}1&2\\3&1\end{vmatrix}\\ &=2(4-3)-3(2-9)+(1-6)\\ &=2+21-5\\ &=18\ne0. \end{aligned}

Hence A^{-1} exists. Its inverse is

A^{-1} = \frac1{18} \begin{pmatrix} 1&-5&7\\ 7&1&-5\\ -5&7&1 \end{pmatrix}.

Therefore

X=A^{-1}B.

Multiplying,

\begin{aligned} x_1 &= \frac1{18}(9-30+56) = \frac{35}{18},\\ x_2 &= \frac1{18}(63+6-40) = \frac{29}{18},\\ x_3 &= \frac1{18}(-45+42+8) = \frac5{18}. \end{aligned}

Final Answer:

\boxed{ x_1=\frac{35}{18},\qquad x_2=\frac{29}{18},\qquad x_3=\frac5{18}.}

Solution:

The mapping is

t:V_2(\mathbb R)\to V_3(\mathbb R), \qquad t(x,y)=(x,y,0).

Definition: A map t is linear if

t(\alpha u+\beta v) = \alpha t(u)+\beta t(v)

for all vectors u,v and all scalars \alpha,\beta.

Let

u=(x_1,y_1),\qquad v=(x_2,y_2).

Then

\alpha u+\beta v = (\alpha x_1+\beta x_2,\, \alpha y_1+\beta y_2).

Apply t:

\begin{aligned} t(\alpha u+\beta v) &= (\alpha x_1+\beta x_2,\, \alpha y_1+\beta y_2,\, 0)\\ &= \alpha(x_1,y_1,0) + \beta(x_2,y_2,0)\\ &= \alpha t(u)+\beta t(v). \end{aligned}

Hence the linearity condition is satisfied.

Final Answer:

\boxed{t(x,y)=(x,y,0)\text{ is a linear transformation from }V_2(\mathbb R)\text{ to }V_3(\mathbb R).}

Solution:

Let the side of the square be x and the radius of the circle be r. Let the fixed sum of their perimeters be P. Then

4x+2\pi r=P. \tag{1}

The sum of their areas is

A=x^2+\pi r^2.

From (1),

x=\frac{P-2\pi r}{4}.

Therefore

A(r) = \frac{(P-2\pi r)^2}{16} +\pi r^2.

Differentiate:

\frac{dA}{dr} = -\frac{\pi}{4}(P-2\pi r)+2\pi r.

At an extremum,

-\frac{\pi}{4}(P-2\pi r)+2\pi r=0.

Dividing by \pi and multiplying by 4,

-P+2\pi r+8r=0.

Using P=4x+2\pi r,

4x+2\pi r=2\pi r+8r,

so

x=2r.

Thus the side of the square equals the diameter of the circle.

To verify that this gives a minimum,

\frac{d^2A}{dr^2} = \frac{\pi^2}{2}+2\pi>0.

Hence the extremum is a minimum.

Final Answer:

\boxed{\text{The sum of the areas is least when }x=2r,}

i.e. when the side of the square equals the diameter of the circle.

Solution:

The curve is

x^2(x^2-4a^2)=y^2(x^2-a^2).

Solving for y^2,

y^2 = \frac{x^2(x^2-4a^2)}{x^2-a^2}. \tag{1}

1. Symmetry.

Only even powers of x and y occur. Therefore the curve is symmetric about both coordinate axes.

2. Nature of the origin.

Bring all terms to one side:

x^4-x^2y^2-4a^2x^2+a^2y^2=0.

The lowest-degree terms near the origin are

a^2(y^2-4x^2)=0.

Thus the tangents at the origin are

y=2x,\qquad y=-2x.

They are real and distinct; hence the origin is an ordinary double point (node).

3. Regions in which real points occur.

From (1), real points require

\frac{x^2(x^2-4a^2)}{x^2-a^2}\ge0.

A sign analysis gives:

• |x|<a[/katex]: real points occur.</li> <li>[katex]a<|x|<2a[/katex]: no real points occur.</li> <li>[katex]|x|\ge2a: real points occur.

4. Intercepts.

Putting y=0,

x^2(x^2-4a^2)=0,

so the curve meets the x-axis at

x=0,\qquad x=\pm2a.

Putting x=0 gives y=0, so the only y-axis intercept is the origin.

5. Vertical asymptotes.

As x\to a^-, the denominator x^2-a^2\to0^- while the numerator tends to -3a^4. Hence y^2\to+\infty. Therefore

x=a

is a vertical asymptote. By symmetry, so is

x=-a.

6. Oblique asymptotes.

From (1),

y^2-x^2 = -\frac{3a^2x^2}{x^2-a^2}.

As |x|\to\infty,

y^2-x^2\to-3a^2.

Hence y/x\to\pm1, and the oblique asymptotes are

y=x,\qquad y=-x.

7. Tangents at (\pm2a,0).

Let

F=x^2(x^2-4a^2)-y^2(x^2-a^2).

At (2a,0), F_y=0 but F_x\ne0. Hence the tangent is vertical:

x=2a.

Similarly, at (-2a,0) the tangent is x=-2a.

Trace: There are four inner branches issuing from the node at the origin along y=\pm2x, tending to the vertical asymptotes x=\pm a. No real curve exists for a<|x|<2a[/katex]. Four outer branches begin at [katex](\pm2a,0)[/katex] with vertical tangents and tend to the oblique asymptotes [katex]y=\pm x[/katex].</p><p><strong>Final Answer:</strong></p> <p>The origin is a <strong>node</strong> with tangents</p> [katex display="true"]\boxed{y=\pm2x}

and the principal asymptotes are

\boxed{x=\pm a,\qquad y=\pm x.}

Solution:

Let the cone have semi-vertical angle \alpha and base radius r.

Let \rho denote the distance of a point on the surface from the axis. Then the cone may be written as

z=\rho\cot\alpha, \qquad 0\le\rho\le r.

Using cylindrical coordinates, the surface element of a surface z=z(\rho) is

dS= \sqrt{1+\left(\frac{dz}{d\rho}\right)^2} \,\rho\,d\rho\,d\theta.

Here

\frac{dz}{d\rho}=\cot\alpha.

Hence

\sqrt{1+\cot^2\alpha} = \csc\alpha.

Therefore the curved surface area is

\begin{aligned} S &= \int_0^{2\pi}\int_0^r \csc\alpha\;\rho\,d\rho\,d\theta\\ &= 2\pi\csc\alpha \left[\frac{\rho^2}{2}\right]_0^r\\ &= \pi r^2\csc\alpha. \end{aligned}

This also agrees with the elementary formula \pi rl, because the slant height is l=r\csc\alpha.

Final Answer:

\boxed{S_{\text{curved}}=\pi r^2\csc\alpha.}

If the base is also included, the total surface area is \pi r^2(1+\csc\alpha).

Solution:

The region lies between

y=x^2

and

y=x.

The curves meet where

x^2=x \quad\Rightarrow\quad x=0,\ 1.

For 0\le x\le1, we have x^2\le x. Thus

I= \int_0^1\int_{x^2}^{x} xy(x+y)\,dy\,dx.

Expand the integrand:

xy(x+y)=x^2y+xy^2.

Integrate first with respect to y:

\int(x^2y+xy^2)\,dy = \frac{x^2y^2}{2} +\frac{xy^3}{3}.

Hence

\begin{aligned} I &= \int_0^1 \left[ \frac{x^2y^2}{2} +\frac{xy^3}{3} \right]_{y=x^2}^{y=x} dx\\ &= \int_0^1 \left( \frac{x^4}{2} +\frac{x^4}{3} -\frac{x^6}{2} -\frac{x^7}{3} \right)dx\\ &= \int_0^1 \left( \frac56x^4-\frac12x^6-\frac13x^7 \right)dx. \end{aligned}

Therefore

\begin{aligned} I &= \left[ \frac{x^5}{6} -\frac{x^7}{14} -\frac{x^8}{24} \right]_0^1\\ &= \frac16-\frac1{14}-\frac1{24}. \end{aligned}

Taking the LCM 168,

I=\frac{28-12-7}{168} =\frac9{168} =\frac3{56}.

Final Answer:

\boxed{\displaystyle \frac3{56}}

Solution:

Given

u=\tan^{-1}\left(\frac{x^3+y^3}{x-y}\right).

Put

v=\tan u = \frac{x^3+y^3}{x-y}.

The numerator is homogeneous of degree 3 and the denominator is homogeneous of degree 1. Therefore v is homogeneous of degree

3-1=2.

By Euler's theorem for homogeneous functions,

x\frac{\partial v}{\partial x} + y\frac{\partial v}{\partial y} = 2v. \tag{1}

Since v=\tan u,

v_x=\sec^2u\,u_x, \qquad v_y=\sec^2u\,u_y.

Substitute into (1):

\sec^2u \left( xu_x+yu_y \right) = 2\tan u.

Therefore

xu_x+yu_y = \frac{2\tan u}{\sec^2u}.

Now

\frac{2\tan u}{\sec^2u} = 2\frac{\sin u}{\cos u}\cos^2u = 2\sin u\cos u = \sin2u.

Final Answer:

\boxed{ x\frac{\partial u}{\partial x} + y\frac{\partial u}{\partial y} = \sin2u.}

Solution:

In Lagrange's mean value theorem, for sufficiently small h, there exists \theta=\theta(h) with 0<\theta<1[/katex] such that</p> [katex display="true"] f(x+h)-f(x) = h\,f'(x+\theta h). \tag{1}

We have to show that, provided f'' is continuous at x and f''(x)\ne0,

\theta\to\frac12 \qquad(h\to0).

Subtract hf'(x) from both sides of (1):

f(x+h)-f(x)-hf'(x) = h\{f'(x+\theta h)-f'(x)\}. \tag{2}

By Taylor's theorem in Lagrange form, for some \xi\in(0,1),

f(x+h) = f(x)+hf'(x) +\frac{h^2}{2}f''(x+\xi h).

Thus the left-hand side of (2) is

\frac{h^2}{2}f''(x+\xi h). \tag{3}

Apply the ordinary mean value theorem to f' between x and x+\theta h. For some \eta\in(0,1),

f'(x+\theta h)-f'(x) = \theta h\, f''(x+\eta\theta h). \tag{4}

Using (3) and (4) in (2),

\frac{h^2}{2}f''(x+\xi h) = \theta h^2 f''(x+\eta\theta h).

For h\ne0, cancel h^2:

\theta = \frac12 \frac{f''(x+\xi h)} {f''(x+\eta\theta h)}.

Since 0<\xi,\eta,\theta<1[/katex], both [katex]\xi h\to0[/katex] and [katex]\eta\theta h\to0[/katex]. By continuity of [katex]f''[/katex],</p> [katex display="true"] f''(x+\xi h)\to f''(x), \qquad f''(x+\eta\theta h)\to f''(x).

Because f''(x)\ne0, the quotient tends to 1.

Final Answer:

\boxed{\displaystyle \lim_{h\to0}\theta=\frac12.}

Solution:

Take the general sphere

x^2+y^2+z^2+2ux+2vy+2wz+d=0.

It passes through (1,0,0), so

1+2u+d=0. \tag{1}

Passing through (0,1,0) gives

1+2v+d=0. \tag{2}

Passing through (0,0,1) gives

1+2w+d=0. \tag{3}

Hence

u=v=w=s

and

d=-1-2s.

The centre is

(-s,-s,-s)

and the square of the radius is

R^2=u^2+v^2+w^2-d.

Therefore

\begin{aligned} R^2 &=3s^2+1+2s\\ &=3\left(s+\frac13\right)^2+\frac23. \end{aligned}

This is least when

s=-\frac13.

Then

R_{\min}^2=\frac23, \qquad R_{\min}=\sqrt{\frac23},

and the centre is

\left(\frac13,\frac13,\frac13\right).

The required sphere is

\left(x-\frac13\right)^2 + \left(y-\frac13\right)^2 + \left(z-\frac13\right)^2 = \frac23.

Expanding,

3x^2+3y^2+3z^2-2x-2y-2z-1=0.

Final Answer:

\boxed{ \left(x-\frac13\right)^2+ \left(y-\frac13\right)^2+ \left(z-\frac13\right)^2 = \frac23}

with minimum radius \boxed{\sqrt{2/3}}.

Solution:

The conic is

13x^2-18xy+37y^2+2x+14y-2=0.

In the standard notation

Ax^2+2Hxy+By^2+2Gx+2Fy+C=0,

we have

A=13,\quad H=-9,\quad B=37.

Since

AB-H^2 = 13(37)-81 = 400>0

and A>0, the quadratic part is positive definite. Hence the conic is of ellipse type, provided it is real.

Centre.

The centre is obtained by setting the first derivatives of the quadratic polynomial equal to zero:

26x-18y+2=0, -18x+74y+14=0.

Divide by 2:

13x-9y+1=0, -9x+37y+7=0.

Solving,

x=-\frac14,\qquad y=-\frac14.

Thus the centre is

\left(-\frac14,-\frac14\right).

Translate the origin to the centre.

Put

x=X-\frac14,\qquad y=Y-\frac14.

Substitution gives

13X^2-18XY+37Y^2-4=0.

Hence, referred to the centre as origin,

13X^2-18XY+37Y^2=4.

The left-hand quadratic form is positive definite, so this is a real ellipse.

Final Answer:

The conic is an ellipse with centre

\boxed{\left(-\frac14,-\frac14\right)}

and, referred to its centre, its equation is

\boxed{13X^2-18XY+37Y^2=4.}

Solution:

Write the first line as

\mathbf r = (3,8,3) +\lambda(3,-1,1),

and the second line as

\mathbf r = (-3,-7,6) +\mu(-3,2,4).

Let

A=(3,8,3),\qquad B=(-3,-7,6),

and direction vectors

\mathbf d_1=(3,-1,1), \qquad \mathbf d_2=(-3,2,4).

A vector perpendicular to both lines is

\mathbf n = \mathbf d_1\times\mathbf d_2.

Calculate:

\begin{aligned} \mathbf n &= \begin{vmatrix} \mathbf i&\mathbf j&\mathbf k\\ 3&-1&1\\ -3&2&4 \end{vmatrix}\\ &=(-6,-15,3)\\ &=3(-2,-5,1). \end{aligned}

Now

\overrightarrow{AB} = B-A = (-6,-15,3) = \mathbf n.

Therefore the segment joining the given base points A and B is already perpendicular to both lines. Hence it is their common perpendicular and gives the shortest distance.

Its length is

\begin{aligned} AB &= \sqrt{(-6)^2+(-15)^2+3^2}\\ &= \sqrt{36+225+9}\\ &= \sqrt{270}\\ &= 3\sqrt{30}. \end{aligned}

The common perpendicular through A=(3,8,3) has direction (-2,-5,1). Therefore its equation is

\frac{x-3}{-2} = \frac{y-8}{-5} = \frac{z-3}{1}.

Final Answer:

\boxed{\text{Shortest distance}=3\sqrt{30}}

and the common perpendicular is

\boxed{ \frac{x-3}{-2} = \frac{y-8}{-5} = z-3.}

Solution:

The equation is

x^3p^2+x^2yp+a^3=0, \qquad p=\frac{dy}{dx}.

We seek a one-parameter complete family. Write the equation as

x^2p(xp+y)+a^3=0.

A convenient trial suggested by the homogeneity is

y=\frac{C}{x}+\frac{a^3}{C}, \qquad C\ne0,

where C is an arbitrary constant.

Then

p=\frac{dy}{dx} = -\frac{C}{x^2}.

Substitute in the differential equation:

x^3p^2 = x^3\frac{C^2}{x^4} = \frac{C^2}{x},

and

\begin{aligned} x^2yp &= x^2 \left( \frac{C}{x}+\frac{a^3}{C} \right) \left(-\frac{C}{x^2}\right)\\ &= -\frac{C^2}{x}-a^3. \end{aligned}

Therefore

x^3p^2+x^2yp+a^3 = \frac{C^2}{x} -\frac{C^2}{x} -a^3+a^3 =0.

Hence the general one-parameter family is

y=\frac{C}{x}+\frac{a^3}{C}. \tag{1}

Singular solution.

The envelope of (1) is obtained by differentiating with respect to C:

\frac{\partial y}{\partial C} = \frac1x-\frac{a^3}{C^2} =0.

Thus

C^2=a^3x.

Substitute in (1). Since C=\pm a^{3/2}\sqrt x,

y = \pm\frac{a^{3/2}}{\sqrt x} \pm\frac{a^{3/2}}{\sqrt x} = \pm\frac{2a^{3/2}}{\sqrt x}.

Equivalently, the singular solution is

xy^2=4a^3.

Direct substitution verifies that this envelope satisfies the original differential equation.

Final Answer:

General solution:

\boxed{y=\frac{C}{x}+\frac{a^3}{C}}

and singular solution:

\boxed{xy^2=4a^3.}

Solution:

The equation is

x^2y''+xy'-y=x^2e^x.

For x\ne0, divide by x^2:

y''+\frac1x y'-\frac1{x^2}y=e^x. \tag{1}

Step 1: Complementary solution.

The homogeneous equation is an Euler equation:

x^2y''+xy'-y=0.

Try y=x^m. Then

m(m-1)+m-1=0,

so

m^2-1=0.

Hence

m=1,-1.

Thus

y_1=x,\qquad y_2=\frac1x,

and

y_c=C_1x+\frac{C_2}{x}.

Step 2: Variation of parameters.

For (1), the Wronskian is

\begin{aligned} W &= \begin{vmatrix} x&1/x\\ 1&-1/x^2 \end{vmatrix}\\ &= -\frac1x-\frac1x\\ &= -\frac2x. \end{aligned}

The standard formula is

y_p = -y_1\int\frac{y_2R}{W}\,dx + y_2\int\frac{y_1R}{W}\,dx,

where R=e^x.

First,

\frac{y_2R}{W} = \frac{(1/x)e^x}{-2/x} = -\frac12e^x.

Therefore

-y_1\int\frac{y_2R}{W}\,dx = -x\int\left(-\frac12e^x\right)dx = \frac{x e^x}{2}.

Next,

\frac{y_1R}{W} = \frac{x e^x}{-2/x} = -\frac12x^2e^x.

Use

\int x^2e^x\,dx = e^x(x^2-2x+2).

Thus

y_2\int\frac{y_1R}{W}\,dx = -\frac{e^x}{2x}(x^2-2x+2).

Hence

\begin{aligned} y_p &= \frac{x e^x}{2} -\frac{e^x}{2x}(x^2-2x+2)\\ &= e^x\left(1-\frac1x\right). \end{aligned}

Final Answer:

\boxed{ y=C_1x+\frac{C_2}{x} +e^x\left(1-\frac1x\right).}

Solution:

We use the standard series for the Bessel function of the first kind:

J_n(x) = \sum_{r=0}^{\infty} \frac{(-1)^r}{r!\,\Gamma(n+r+1)} \left(\frac{x}{2}\right)^{n+2r}.

Consider x^{-n}J_n(x):

x^{-n}J_n(x) = \sum_{r=0}^{\infty} \frac{(-1)^r x^{2r}} {2^{n+2r}r!\,\Gamma(n+r+1)}.

Differentiate term by term:

\frac{d}{dx}\{x^{-n}J_n(x)\} = \sum_{r=1}^{\infty} \frac{(-1)^r(2r)x^{2r-1}} {2^{n+2r}r!\,\Gamma(n+r+1)}.

Since r!=r(r-1)!,

\frac{2r}{2^{n+2r}r!} = \frac{1}{2^{n+2r-1}(r-1)!}.

Put s=r-1. Then

\frac{d}{dx}\{x^{-n}J_n(x)\} = - \sum_{s=0}^{\infty} \frac{(-1)^s x^{2s+1}} {2^{n+2s+1}s!\,\Gamma(n+s+2)}.

But the right-hand side is exactly

-x^{-n}J_{n+1}(x).

Therefore

\frac{d}{dx}\{x^{-n}J_n(x)\} = -x^{-n}J_{n+1}(x).

Differentiate the product on the left:

-nx^{-n-1}J_n(x) +x^{-n}J_n'(x) = -x^{-n}J_{n+1}(x).

Multiply by x^{n+1}:

-nJ_n(x)+xJ_n'(x) = -xJ_{n+1}(x).

Hence

Final Answer:

\boxed{xJ_n'(x)=nJ_n(x)-xJ_{n+1}(x).}

Solution:

For m>0 and n>0, the Beta function is

B(m,n) = \int_0^1 x^{m-1}(1-x)^{n-1}\,dx.

Now

B(m+1,n) = \int_0^1 x^m(1-x)^{n-1}\,dx,

and

B(m,n+1) = \int_0^1 x^{m-1}(1-x)^n\,dx.

Add the two integrals:

\begin{aligned} B(m+1,n)+B(m,n+1) &= \int_0^1 \left[ x^m(1-x)^{n-1} +x^{m-1}(1-x)^n \right]dx\\ &= \int_0^1 x^{m-1}(1-x)^{n-1} \{x+(1-x)\}\,dx\\ &= \int_0^1 x^{m-1}(1-x)^{n-1}\,dx\\ &= B(m,n). \end{aligned}

Final Answer:

\boxed{B(m,n)=B(m+1,n)+B(m,n+1).}

Solution:

Given

\mathbf a = \sin\theta\,\mathbf i +\cos\theta\,\mathbf j +\theta\,\mathbf k, \mathbf b = \cos\theta\,\mathbf i -\sin\theta\,\mathbf j -3\mathbf k,

and

\mathbf c=2\mathbf i+3\mathbf j-\mathbf k.

We need

\frac{d}{d\theta} [\mathbf a\times(\mathbf b\times\mathbf c)]

at \theta=0.

Use the product rule for cross products:

\frac{d}{d\theta} [\mathbf a\times(\mathbf b\times\mathbf c)] = \mathbf a'\times(\mathbf b\times\mathbf c) + \mathbf a\times(\mathbf b'\times\mathbf c),

because \mathbf c is constant.

At \theta=0,

\mathbf a_0=(0,1,0), \qquad \mathbf a'_0=(1,0,1),

and

\mathbf b_0=(1,0,-3), \qquad \mathbf b'_0=(0,-1,0).

First compute

\mathbf b_0\times\mathbf c = \begin{vmatrix} \mathbf i&\mathbf j&\mathbf k\\ 1&0&-3\\ 2&3&-1 \end{vmatrix} = (9,-5,3).

Therefore

\mathbf a'_0\times(\mathbf b_0\times\mathbf c) = (1,0,1)\times(9,-5,3) = (5,6,-5).

Next,

\mathbf b'_0\times\mathbf c = (0,-1,0)\times(2,3,-1) = (1,0,2).

Hence

\mathbf a_0\times(\mathbf b'_0\times\mathbf c) = (0,1,0)\times(1,0,2) = (2,0,-1).

Add the two vectors:

(5,6,-5)+(2,0,-1) = (7,6,-6).

Final Answer:

\boxed{ \left. \frac{d}{d\theta} [\mathbf a\times(\mathbf b\times\mathbf c)] \right|_{\theta=0} = 7\mathbf i+6\mathbf j-6\mathbf k.}

Solution:

The surface integral is of the form

\iint_S \left( P\,dy\,dz + Q\,dz\,dx + R\,dx\,dy \right),

where

P=x^3-yz,\qquad Q=-2x^2y,\qquad R=z.

Gauss divergence theorem:

\iint_S (P\,dy\,dz+Q\,dz\,dx+R\,dx\,dy) = \iiint_V \left( \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \right)dV.

Compute the divergence:

\frac{\partial P}{\partial x}=3x^2, \frac{\partial Q}{\partial y}=-2x^2, \frac{\partial R}{\partial z}=1.

Therefore

\nabla\cdot\mathbf F = x^2+1.

The cube is

0\le x\le a,\qquad 0\le y\le a,\qquad 0\le z\le a.

Hence

\begin{aligned} I &= \int_0^a\int_0^a\int_0^a (x^2+1)\,dx\,dy\,dz\\ &= a^2\int_0^a(x^2+1)\,dx\\ &= a^2 \left[ \frac{x^3}{3}+x \right]_0^a\\ &= a^2\left(\frac{a^3}{3}+a\right). \end{aligned}

Thus

I=\frac{a^5}{3}+a^3.

Final Answer:

\boxed{\displaystyle a^3+\frac{a^5}{3}}.

Solution:

Let the resultant force have components

\mathbf R=(X,Y).

Let its moment about the origin be M, positive counter-clockwise.

The moment of the resultant about a point (x_0,y_0) is

M_{(x_0,y_0)} = M-x_0Y+y_0X. \tag{1}

Using the three given moments:

About (2,0),

M-2Y=3. \tag{2}

About (0,2),

M+2X=4. \tag{3}

About (2,2),

M-2Y+2X=10. \tag{4}

Subtract (2) from (4):

2X=7 \quad\Rightarrow\quad X=\frac72.

Subtract (3) from (4):

-2Y=6 \quad\Rightarrow\quad Y=-3.

Use (2):

M-2(-3)=3,

so

M=-3.

The magnitude of the resultant is

\begin{aligned} R &= \sqrt{X^2+Y^2}\\ &= \sqrt{\frac{49}{4}+9}\\ &= \sqrt{\frac{85}{4}}\\ &= \frac{\sqrt{85}}{2}. \end{aligned}

Line of action.

At any point (x,y) on the resultant's line of action, its moment about that point is zero. Using (1),

M-xY+yX=0.

Substitute M=-3, X=7/2, Y=-3:

-3+3x+\frac72y=0.

Multiply by 2:

6x+7y-6=0.

Final Answer:

\boxed{R=\frac{\sqrt{85}}{2}}

and the line of action is

\boxed{6x+7y-6=0.}

Solution:

Let the amplitude be A and the angular frequency be \omega. Since the particle starts from rest, it starts from an extreme position. Taking that extreme displacement as A,

x=A\cos\omega t.

While the particle continues to move in the same direction, the distance travelled from its starting point by time t is

s(t)=A-x=A(1-\cos\omega t).

In the first second it travels distance a. Therefore

a=A(1-\cos\omega). \tag{1}

In the next second it travels an additional distance b. Thus

b=s(2)-s(1).

Hence

b = A(\cos\omega-\cos2\omega).

Put c=\cos\omega. Since

\cos2\omega=2c^2-1,

we get

\begin{aligned} b &= A(c-2c^2+1)\\ &= A(1-c)(2c+1). \end{aligned}

Using (1), A(1-c)=a, so

b=a(2c+1).

Therefore

2c+1=\frac ba,

and

\cos\omega = \frac{b-a}{2a}.

Hence

\omega = \cos^{-1}\left(\frac{b-a}{2a}\right).

The period is

T=\frac{2\pi}{\omega}.

Now determine the amplitude from (1):

A = \frac{a}{1-\cos\omega}.

Substituting \cos\omega=(b-a)/(2a),

\begin{aligned} A &= \frac{a} {1-\frac{b-a}{2a}}\\ &= \frac{a} {\frac{3a-b}{2a}}\\ &= \frac{2a^2}{3a-b}. \end{aligned}

Final Answer:

\boxed{ A=\frac{2a^2}{3a-b}}

and

\boxed{ T= \frac{2\pi}{ \cos^{-1}\left(\frac{b-a}{2a}\right)} }.

The inverse cosine is in radians; the given statement that the motion remains in the same direction through both seconds selects the physically appropriate branch.

Solution:

Important source check: The printed question asks for a term e^{+\mu t/m}. For a genuine resisting force proportional to velocity and opposite to the downward motion, the governing equation necessarily gives a decaying exponential e^{-\mu t/m}. The printed positive exponent is therefore a sign error. The correct derivation is given below without changing the wording of the question.

Let downward distance be s and downward velocity be

v=\frac{ds}{dt}.

The forces are:

• weight mg downward,
• resistance \mu v upward.

Hence Newton's second law gives

m\frac{dv}{dt}=mg-\mu v.

Rearrange:

\frac{dv}{dt} +\frac{\mu}{m}v =g.

The integrating factor is

e^{\mu t/m}.

Thus

\frac{d}{dt} \left( ve^{\mu t/m} \right) = ge^{\mu t/m}.

Integrating,

ve^{\mu t/m} = \frac{gm}{\mu}e^{\mu t/m}+C.

The particle is released from rest, so v(0)=0. Hence

C=-\frac{gm}{\mu}.

Therefore

v = \frac{gm}{\mu} \left( 1-e^{-\mu t/m} \right).

Now integrate to obtain the distance fallen:

s = \int_0^t \frac{gm}{\mu} \left( 1-e^{-\mu \tau/m} \right)d\tau.

Hence

\begin{aligned} s &= \frac{gm}{\mu} \left[ \tau+\frac{m}{\mu}e^{-\mu\tau/m} \right]_0^t\\ &= \frac{gm}{\mu} \left[ t+\frac{m}{\mu}e^{-\mu t/m} -\frac{m}{\mu} \right]. \end{aligned}

Factor gm^2/\mu^2:

s = \frac{gm^2}{\mu^2} \left( e^{-\mu t/m} -1 +\frac{\mu t}{m} \right).

Physical verification: Expanding the exponential for small t,

e^{-\mu t/m} = 1-\frac{\mu t}{m} +\frac12\frac{\mu^2t^2}{m^2} +\cdots,

so

s=\frac12gt^2+O(t^3),

which is exactly consistent with release from rest. The printed positive exponential would produce an unphysical linear term in t.

Final Answer:

\boxed{ s= \frac{gm^2}{\mu^2} \left( e^{-\mu t/m} -1 +\frac{\mu t}{m} \right).}

Thus the exponent in the source statement should be negative.