HPAS 2024 Mathematics Optional Paper 1 – Complete Solved PYQ
This page contains complete, step-by-step solutions to HPAS 2024 Mathematics Optional Paper 1 for Himachal Pradesh Administrative Service aspirants. The solutions cover Linear Algebra, Calculus, Differential Equations, Vector Analysis, Statics and Dynamics, with equations rendered using KaTeX.
Use the question index below to jump directly to any HPAS Maths Optional Paper 1 previous-year question and its solution.
HPAS 2024 Maths Optional Paper-1 Question 1(a)
Show that similar matrices have the same minimal polynomial.
Solution
Step 1: Recall what similar matrices mean
Let A and B be similar n\times n matrices. By definition, there is an invertible matrix P such that
B=P^{-1}AP.Let m_A(x) and m_B(x) denote the minimal polynomials of A and B. The minimal polynomial is the monic polynomial of least positive degree that annihilates the matrix.
Step 2: Show that polynomial evaluation is preserved by similarity
From B=P^{-1}AP, repeated multiplication gives
B^2=P^{-1}A^2P,\qquad B^3=P^{-1}A^3P,\qquad \ldots,\qquad B^k=P^{-1}A^kP.Therefore, for any polynomial
f(x)=c_0+c_1x+\cdots+c_kx^k,we obtain
f(B)=P^{-1}f(A)P.This identity is the key point: a polynomial vanishes at A exactly when the same polynomial vanishes at the similar matrix B.
Step 3: Use the minimal polynomial of A
By definition,
m_A(A)=0.Applying the identity from Step 2,
m_A(B)=P^{-1}m_A(A)P=P^{-1}(0)P=0.Thus m_A(x) is an annihilating polynomial for B. Since m_B(x) is the monic annihilating polynomial of smallest degree, it must divide every polynomial that annihilates B. Hence
m_B(x)\mid m_A(x).Step 4: Reverse the argument
Similarity is symmetric. From B=P^{-1}AP, we also have
A=PBP^{-1}.Using exactly the same argument with A and B interchanged gives
m_A(x)\mid m_B(x).Step 5: Conclude equality
The two minimal polynomials are monic and each divides the other. Therefore they are identical.
HPAS 2024 Maths Optional Paper-1 Question 1(b)
Find the asymptotes of the curve:
x^{3}+x^{2}y+xy^{2}+y^{3}+2x^{2}+3xy-4y^{2}+7x+2y=0Solution
Step 1: Separate the homogeneous parts of the curve
The curve is
x^3+x^2y+xy^2+y^3+2x^2+3xy-4y^2+7x+2y=0.For a cubic curve, write the degree-three and degree-two homogeneous parts as
F_3(x,y)=x^3+x^2y+xy^2+y^3, F_2(x,y)=2x^2+3xy-4y^2.To find a non-vertical asymptote of the form y=mx+c, first determine its slope m from the highest-degree terms.
Step 2: Find the possible slopes
Put x=1 and y=m in F_3:
\phi_3(m)=F_3(1,m)=1+m+m^2+m^3.Factor:
1+m+m^2+m^3=(1+m)(1+m^2).Hence
(1+m)(1+m^2)=0.The only real root is
m=-1.The roots m=\pm i are complex and therefore do not give real asymptotes.
Step 3: Find the intercept corresponding to m=-1
For a simple root m of \phi_3(m)=0, the intercept is
c=-\frac{\phi_2(m)}{\phi_3'(m)},where
\phi_2(m)=F_2(1,m)=2+3m-4m^2and
\phi_3'(m)=1+2m+3m^2.At m=-1,
\phi_2(-1)=2-3-4=-5, \phi_3'(-1)=1-2+3=2.Therefore,
c=-\frac{-5}{2}=\frac52.Step 4: Write the asymptote
Substitute m=-1 and c=5/2 into y=mx+c:
y=-x+\frac52.Equivalently,
2x+2y-5=0.There is no vertical real asymptote because the coefficient of y^3 in the highest-degree homogeneous part is non-zero.
HPAS 2024 Maths Optional Paper-1 Question 1(c)
Compute the general solution of the non-linear differential equation y = xy' + (y')^2 where y' = \frac{dy}{dx}.
Solution
Step 1: Recognize the Clairaut form
Put
p=\frac{dy}{dx}.The equation becomes
y=xp+p^2.This is a Clairaut equation of the standard form y=xp+f(p), with f(p)=p^2.
Step 2: Differentiate with respect to x
Differentiating both sides, remembering that p is a function of x,
\frac{dy}{dx}=p+x\frac{dp}{dx}+2p\frac{dp}{dx}.Since dy/dx=p,
p=p+(x+2p)\frac{dp}{dx}.Thus
(x+2p)\frac{dp}{dx}=0.Step 3: Obtain the general solution
One possibility is
\frac{dp}{dx}=0.Therefore p=C, where C is an arbitrary constant. Substituting in y=xp+p^2,
y=Cx+C^2.This one-parameter family is the required general solution.
Step 4: Find the singular solution
The second factor gives
x+2p=0\quad\Longrightarrow\quad p=-\frac{x}{2}.Substitute this into the original equation:
y=x\left(-\frac{x}{2}\right)+\left(-\frac{x}{2}\right)^2=-\frac{x^2}{2}+\frac{x^2}{4}=-\frac{x^2}{4}.This is the envelope of the family y=Cx+C^2 and is the singular solution.
General solution: \boxed{y=Cx+C^2}. Singular solution: \boxed{y=-\frac{x^2}{4}}.
HPAS 2024 Maths Optional Paper-1 Question 1(d)
Show that the vector field defined by the vector function \vec{V} = xyz(yz\mathbf{i} + xz\mathbf{j} + xy\mathbf{k}) is conservative.
Solution
Step 1: Expand the vector field into components
The given field is
\vec V=xyz\left(yz\mathbf{i}+xz\mathbf{j}+xy\mathbf{k}\right).Therefore,
\vec V=xy^2z^2\mathbf{i}+x^2yz^2\mathbf{j}+x^2y^2z\mathbf{k}.So its components are
P=xy^2z^2,\qquad Q=x^2yz^2,\qquad R=x^2y^2z.Step 2: Look for a scalar potential
A vector field is conservative if there exists a scalar function \phi(x,y,z) such that
\nabla\phi=\vec V.From the first component,
\frac{\partial\phi}{\partial x}=xy^2z^2.Integrating with respect to x, treating y and z as constants,
\phi=\frac12x^2y^2z^2+g(y,z),where g(y,z) is independent of x.
Step 3: Compare the y-component
Differentiate the potential with respect to y:
\frac{\partial\phi}{\partial y}=x^2yz^2+\frac{\partial g}{\partial y}.This must equal Q=x^2yz^2. Hence
\frac{\partial g}{\partial y}=0.Therefore g can depend only on z; write g(y,z)=h(z).
Step 4: Compare the z-component
Now
\phi=\frac12x^2y^2z^2+h(z).Differentiate with respect to z:
\frac{\partial\phi}{\partial z}=x^2y^2z+h'(z).This must equal R=x^2y^2z, so
h'(z)=0.Thus h(z)=C, a constant.
Step 5: State the potential and conclusion
We have found a scalar potential
\phi(x,y,z)=\frac12x^2y^2z^2+Csuch that \nabla\phi=\vec V. Hence the field is conservative.
HPAS 2024 Maths Optional Paper-1 Question 2(a)
Show that the inner product space is a normed vector space but the converse is not true.
Solution
Step 1: Construct a norm from the inner product
Let V be an inner product space. Define
\|x\|=\sqrt{\langle x,x\rangle}.We show that this function satisfies all the axioms of a norm.
Step 2: Verify positivity and definiteness
By the positivity property of an inner product,
\langle x,x\rangle\ge 0.Therefore \|x\|\ge0. Also,
\|x\|=0\iff \langle x,x\rangle=0\iff x=0.Step 3: Verify homogeneity
For any scalar \alpha,
\begin{aligned}\|\alpha x\|^2&=\langle \alpha x,\alpha x\rangle\\&=|\alpha|^2\langle x,x\rangle. \end{aligned}Taking non-negative square roots gives
\|\alpha x\|=|\alpha|\,\|x\|.Step 4: Verify the triangle inequality
Using the inner product expansion,
\|x+y\|^2=\|x\|^2+2\operatorname{Re}\langle x,y\rangle+\|y\|^2.By the Cauchy-Schwarz inequality,
|\langle x,y\rangle|\le \|x\|\,\|y\|.Hence
\|x+y\|^2\le \|x\|^2+2\|x\|\|y\|+\|y\|^2=(\|x\|+\|y\|)^2.Taking square roots,
\|x+y\|\le \|x\|+\|y\|.Thus every inner product space is a normed vector space.
Step 5: Show that the converse is false
A norm comes from an inner product only if it satisfies the parallelogram identity
\|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2.Consider \mathbb{R}^2 with the supremum norm
\|(x_1,x_2)\|_\infty=\max\{|x_1|,|x_2|\}.Take x=(1,0) and y=(0,1). Then
\|x\|_\infty=\|y\|_\infty=1,and
\|x+y\|_\infty=\|x-y\|_\infty=1.Therefore the left side of the parallelogram identity is 1^2+1^2=2, while the right side is 2(1^2+1^2)=4. The identity fails.
HPAS 2024 Maths Optional Paper-1 Question 2(b)
If the plane x+y+z=1 cuts the cylinder x^2+y^2=1 in an ellipse, then determine the points on the ellipse that lie closest to and farthest from the origin.
Solution
Step 1: Write the distance to be optimized
A point (x,y,z) on the ellipse satisfies
x^2+y^2=1,\qquad x+y+z=1.The square of its distance from the origin is
D^2=x^2+y^2+z^2.Because x^2+y^2=1 on the cylinder,
D^2=1+z^2.Thus it is enough to find the smallest and largest possible values of |z|.
Step 2: Express z in terms of x+y
From the plane,
z=1-x-y.On the unit circle x^2+y^2=1, the quantity x+y satisfies
-\sqrt2\le x+y\le \sqrt2.This follows from Cauchy-Schwarz:
|x+y|\le \sqrt{1^2+1^2}\sqrt{x^2+y^2}=\sqrt2.Step 3: Find the closest points
The minimum possible distance occurs when z^2 is minimum. Since x+y=1 is possible on the unit circle, we can have z=0.
Solve
x+y=1,\qquad x^2+y^2=1.Using y=1-x,
x^2+(1-x)^2=1\Longrightarrow 2x(x-1)=0.Hence x=0 or x=1, giving
(0,1,0),\qquad(1,0,0).At either point, D^2=1, so the minimum distance is 1.
Step 4: Find the farthest point
Since z=1-(x+y), the largest possible value of |z| occurs at the endpoint x+y=-\sqrt2, for which
z=1+\sqrt2.Equality x+y=-\sqrt2 on the unit circle occurs when
x=y=-\frac{1}{\sqrt2}.Thus the farthest point is
\left(-\frac1{\sqrt2},-\frac1{\sqrt2},1+\sqrt2\right).Its squared distance is
D_{\max}^2=1+(1+\sqrt2)^2=4+2\sqrt2.Step 5: State the extremal distances
Therefore
D_{\min}=1,\qquad D_{\max}=\sqrt{4+2\sqrt2}.Closest points: \boxed{(1,0,0)\text{ and }(0,1,0)}.
Farthest point: \boxed{\left(-\frac1{\sqrt2},-\frac1{\sqrt2},1+\sqrt2\right)}.
HPAS 2024 Maths Optional Paper-1 Question 3(a)
Find the volume of the solid enclosed between the surfaces x^2+y^2=a^2 and x^2+z^2=a^2.
Solution
Step 1: Understand the common solid
The surfaces
x^2+y^2=a^2,\qquad x^2+z^2=a^2are two circular cylinders of radius a whose axes are perpendicular. Their common region is a symmetric Steinmetz solid.
Because the solid is symmetric with respect to all three coordinate planes, it is enough to calculate the volume in the first octant and multiply by 8.
Step 2: Determine the limits in the first octant
In the first octant, x,y,z are non-negative. For a fixed x, both cylinder inequalities give
0\le y\le \sqrt{a^2-x^2},\qquad 0\le z\le \sqrt{a^2-x^2}.Also, 0\le x\le a.
Step 3: Set up the triple integral
The first-octant volume is
V_{1}=\int_0^a\int_0^{\sqrt{a^2-x^2}}\int_0^{\sqrt{a^2-x^2}}dz\,dy\,dx.Hence the total volume is
V=8V_1.Step 4: Integrate with respect to z and y
First,
\int_0^{\sqrt{a^2-x^2}}dz=\sqrt{a^2-x^2}.Therefore,
\begin{aligned}V&=8\int_0^a\int_0^{\sqrt{a^2-x^2}}\sqrt{a^2-x^2}\,dy\,dx\\&=8\int_0^a(a^2-x^2)\,dx. \end{aligned}Step 5: Evaluate the final integral
\begin{aligned}V&=8\left[a^2x-\frac{x^3}{3}\right]_0^a\\&=8\left(a^3-\frac{a^3}{3}\right)\\&=\frac{16a^3}{3}. \end{aligned}HPAS 2024 Maths Optional Paper-1 Question 3(b)
Find the equation of the plane passing through the line of intersection of the planes a_1x+b_1y+c_1z+d_1=0 and a_2x+b_2y+c_2z+d_2=0 and perpendicular to the xy-plane.
Solution
Step 1: Write the two given planes compactly
Let
P_1=a_1x+b_1y+c_1z+d_1=0, P_2=a_2x+b_2y+c_2z+d_2=0.Every plane that passes through the line of intersection of P_1=0 and P_2=0 is obtained from a non-trivial linear combination of these two plane equations.
Step 2: Use the condition of being perpendicular to the xy-plane
The xy-plane has normal vector (0,0,1). A plane perpendicular to it must have a normal vector orthogonal to (0,0,1). Therefore its normal must have zero z-component.
So we need a linear combination of P_1 and P_2 in which the coefficient of z vanishes.
Step 3: Eliminate the z-term
Multiply the first plane by c_2 and the second by c_1, then subtract:
c_2P_1-c_1P_2=0.The coefficient of z is
c_2c_1-c_1c_2=0,as required.
Step 4: Expand the required plane
Expanding c_2P_1-c_1P_2=0 gives
(a_1c_2-a_2c_1)x+(b_1c_2-b_2c_1)y+(d_1c_2-d_2c_1)=0.This plane contains the original line of intersection because every point on that line satisfies both P_1=0 and P_2=0. Its equation has no z-term, so it is perpendicular to the xy-plane.
HPAS 2024 Maths Optional Paper-1 Question 4(a)
Show that every non-zero finite-dimensional inner product space has an orthonormal basis.
Solution
Step 1: Start with any basis
Let V be a non-zero finite-dimensional inner product space with
\dim V=n\ge1.Choose any basis
\{v_1,v_2,\ldots,v_n\}.We will convert this arbitrary basis into an orthonormal basis by the Gram-Schmidt process.
Step 2: Construct the first orthogonal vector
Set
u_1=v_1.Since a basis contains no zero vector, u_1\ne0.
Step 3: Remove from v_2 its component along u_1
Define
u_2=v_2-\frac{\langle v_2,u_1\rangle}{\langle u_1,u_1\rangle}u_1.Then
\langle u_2,u_1\rangle=0,so u_2 is orthogonal to u_1. Also u_2\ne0; otherwise v_2 would be a scalar multiple of v_1, contradicting the linear independence of the original basis.
Step 4: Continue the Gram-Schmidt construction
For k=2,3,\ldots,n, define
u_k=v_k-\sum_{j=1}^{k-1}\frac{\langle v_k,u_j\rangle}{\|u_j\|^2}u_j.The subtracted sum is the projection of v_k onto the span of the previous vectors. Therefore u_k is orthogonal to each of u_1,\ldots,u_{k-1}.
Moreover, each u_k is non-zero; otherwise v_k would belong to the span of v_1,\ldots,v_{k-1}, again contradicting that the v_i form a basis.
Step 5: Obtain an orthogonal basis
The vectors
\{u_1,u_2,\ldots,u_n\}are non-zero and mutually orthogonal. Any finite set of non-zero mutually orthogonal vectors is linearly independent. Since there are n of them in an n-dimensional space, they form a basis of V.
Step 6: Normalize the orthogonal basis
Define
e_i=\frac{u_i}{\|u_i\|},\qquad i=1,2,\ldots,n.Then every e_i has norm 1, and for i\ne j,
\langle e_i,e_j\rangle=0.Hence
\langle e_i,e_j\rangle=\delta_{ij}.HPAS 2024 Maths Optional Paper-1 Question 4(b)
Using the concept of diagonalizability, determine A^5 where
A=\begin{pmatrix}0&0&-2\\1&2&1\\1&0&3\end{pmatrix}.Solution
Step 1: Find the eigenvalues of A
Given
A=\begin{pmatrix}0&0&-2\\1&2&1\\1&0&3\end{pmatrix}.The characteristic equation is \det(A-\lambda I)=0:
\begin{aligned}\det(A-\lambda I)&=\begin{vmatrix}-\lambda&0&-2\\1&2-\lambda&1\\1&0&3-\lambda\end{vmatrix}\\&=(2-\lambda)\left[-\lambda(3-\lambda)+2\right]\\&=(2-\lambda)(\lambda^2-3\lambda+2)\\&=(2-\lambda)(\lambda-1)(\lambda-2). \end{aligned}Therefore the eigenvalues are
\lambda_1=1,\qquad \lambda_2=2,\qquad \lambda_3=2.Step 2: Find an eigenvector for \lambda=1
Solve (A-I)X=0:
\begin{pmatrix}-1&0&-2\\1&1&1\\1&0&2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}.The equations give x=-2z and y=z. Choosing z=1, one eigenvector is
X_1=\begin{pmatrix}-2\\1\\1\end{pmatrix}.Step 3: Find two independent eigenvectors for \lambda=2
Solve (A-2I)X=0:
\begin{pmatrix}-2&0&-2\\1&0&1\\1&0&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=0.The only restriction is
x+z=0,while y is free. Thus the eigenspace has dimension 2. We may choose
X_2=\begin{pmatrix}0\\1\\0\end{pmatrix},\qquad X_3=\begin{pmatrix}-1\\0\\1\end{pmatrix}.We now have three linearly independent eigenvectors, so A is diagonalizable.
Step 4: Form the modal matrix and diagonal matrix
Take the eigenvectors as the columns of P:
P=\begin{pmatrix}-2&0&-1\\1&1&0\\1&0&1\end{pmatrix},\qquad D=\begin{pmatrix}1&0&0\\0&2&0\\0&0&2\end{pmatrix}.The inverse of P is
P^{-1}=\begin{pmatrix}-1&0&-1\\1&1&1\\1&0&2\end{pmatrix}.Hence
A=PDP^{-1}.Step 5: Raise the diagonal matrix to the fifth power
For a diagonal matrix, powers are obtained by raising each diagonal entry to that power:
D^5=\begin{pmatrix}1&0&0\\0&32&0\\0&0&32\end{pmatrix}.Therefore,
A^5=PD^5P^{-1}.Step 6: Multiply the matrices
First,
PD^5=\begin{pmatrix}-2&0&-32\\1&32&0\\1&0&32\end{pmatrix}.Now multiply by P^{-1}:
\begin{aligned}A^5&=\begin{pmatrix}-2&0&-32\\1&32&0\\1&0&32\end{pmatrix}\begin{pmatrix}-1&0&-1\\1&1&1\\1&0&2\end{pmatrix}\\&=\begin{pmatrix}-30&0&-62\\31&32&31\\31&0&63\end{pmatrix}. \end{aligned}HPAS 2024 Maths Optional Paper-1 Question 5(a)
Solve the differential equation:
x^2\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} + 2y = x + x^2\log x + x^3Solution
Step 1: Convert the Cauchy-Euler equation to constant coefficients
The equation is
x^2y''-2xy'+2y=x+x^2\log x+x^3.Because the right side contains \log x, we work on x\gt0. Put
x=e^z,\qquad z=\log x,\qquad D=\frac{d}{dz}.For the Cauchy-Euler substitution,
xy'=Dy,\qquad x^2y''=D(D-1)y.Thus the differential equation becomes
\left[D(D-1)-2D+2\right]y=e^z+ze^{2z}+e^{3z},or
(D^2-3D+2)y=e^z+ze^{2z}+e^{3z}.Step 2: Find the complementary function
The auxiliary equation is
m^2-3m+2=0.Factor:
(m-1)(m-2)=0.Hence
m=1,2,and
y_c=C_1e^z+C_2e^{2z}=C_1x+C_2x^2.Step 3: Particular integral for the term e^z
Since e^z corresponds to the root m=1 of the auxiliary equation, resonance occurs. Using the standard repeated-factor rule,
P_1=-ze^z.Returning to x,
P_1=-x\log x.Step 4: Particular integral for ze^{2z}
Use the shift rule F(D)e^{az}v=e^{az}F(D+a)v:
P_2=e^{2z}\frac{1}{D(D+1)}z.Since
\frac{1}{D+1}z=z-1,we obtain
\frac1D(z-1)=\frac{z^2}{2}-z.Therefore,
P_2=e^{2z}\left(\frac{z^2}{2}-z\right)=x^2\left(\frac{(\log x)^2}{2}-\log x\right).Step 5: Particular integral for e^{3z}
There is no resonance at m=3, so substitute D=3 in the operator:
P_3=\frac{e^{3z}}{(3-1)(3-2)}=\frac12e^{3z}=\frac{x^3}{2}.Step 6: Combine the complementary and particular parts
The complete solution is
y=y_c+P_1+P_2+P_3.HPAS 2024 Maths Optional Paper-1 Question 5(b)
Determine the power series solution about the origin of the differential equation:
(1-x^2)\frac{d^2y}{dx^2} - 4x\frac{dy}{dx} + 2y = 0Solution
Step 1: Assume a power series about the origin
The equation is
(1-x^2)y''-4xy'+2y=0.Since the coefficient of y'' is 1-x^2, which is non-zero at x=0, the origin is an ordinary point. Hence we seek a solution
y=\sum_{n=0}^{\infty}a_nx^n.Step 2: Differentiate the series
y'=\sum_{n=1}^{\infty}na_nx^{n-1}, y''=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}.Step 3: Substitute into the differential equation
Substituting gives
\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}-\sum_{n=2}^{\infty}n(n-1)a_nx^n-4\sum_{n=1}^{\infty}na_nx^n+2\sum_{n=0}^{\infty}a_nx^n=0.Step 4: Put every sum in powers of x^n
In the first sum replace n by n+2. Then
\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n-\sum_{n=0}^{\infty}n(n-1)a_nx^n-4\sum_{n=0}^{\infty}na_nx^n+2\sum_{n=0}^{\infty}a_nx^n=0.Step 5: Derive the recurrence relation
Equating the coefficient of x^n to zero,
(n+2)(n+1)a_{n+2}-\left[n(n-1)+4n-2\right]a_n=0.Therefore,
\boxed{a_{n+2}=\frac{n^2+3n-2}{(n+2)(n+1)}a_n}.The recurrence connects even coefficients only with even coefficients and odd coefficients only with odd coefficients. Thus a_0 and a_1 remain the two arbitrary constants expected for a second-order equation.
Step 6: Calculate the first few coefficients
For the even series:
a_2=-a_0,\qquad a_4=-\frac23a_0,\qquad a_6=-\frac{26}{45}a_0,\ldotsFor the odd series:
a_3=\frac13a_1,\qquad a_5=\frac4{15}a_1,\qquad a_7=\frac{76}{315}a_1,\ldotsStep 7: Write the power series solution
Hence
\begin{aligned}y=&\;a_0\left(1-x^2-\frac23x^4-\frac{28}{45}x^6-\cdots\right)\\&+a_1\left(x+\frac13x^3+\frac4{15}x^5+\frac{22}{105}x^7+\cdots\right). \end{aligned}The nearest singular points of the differential equation are x=\pm1, so the power series about the origin has radius of convergence 1.
HPAS 2024 Maths Optional Paper-1 Question 6(a)
Find the work done by the force \vec{F}=(x^2-y^2)\mathbf{i}+(x+y)\mathbf{j} in moving a particle along the closed path C containing the curves x+y=0, x^2+y^2=16, and y=x in the first and the fourth quadrants.
Solution
Step 1: Express the work integral in Green’s theorem form
The force is
\vec F=(x^2-y^2)\mathbf{i}+(x+y)\mathbf{j}.Hence the work done around the closed curve C is
W=\oint_C\left[(x^2-y^2)\,dx+(x+y)\,dy\right].Let
M=x^2-y^2,\qquad N=x+y.Step 2: Apply Green’s theorem
For positive, counterclockwise orientation, Green’s theorem gives
\oint_C(M\,dx+N\,dy)=\iint_R\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dA.Now
\frac{\partial N}{\partial x}=1,\qquad \frac{\partial M}{\partial y}=-2y.Therefore,
\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}=1+2y.Step 3: Describe the region in polar coordinates
The boundary consists of the circle x^2+y^2=16 and the two lines y=x and y=-x in the first and fourth quadrants. Thus R is the right-hand circular sector
0\le r\le4,\qquad -\frac\pi4\le\theta\le\frac\pi4.Using y=r\sin\theta and dA=r\,dr\,d\theta,
W=\int_{-\pi/4}^{\pi/4}\int_0^4(1+2r\sin\theta)r\,dr\,d\theta.Step 4: Integrate with respect to r
\begin{aligned}\int_0^4(r+2r^2\sin\theta)\,dr&=\left[\frac{r^2}{2}+\frac{2r^3}{3}\sin\theta\right]_0^4\\&=8+\frac{128}{3}\sin\theta. \end{aligned}Step 5: Integrate with respect to \theta
Therefore,
W=\int_{-\pi/4}^{\pi/4}\left(8+\frac{128}{3}\sin\theta\right)d\theta.The sine term is odd and the limits are symmetric, so its integral is zero. Thus
W=8\left(\frac\pi4-\left(-\frac\pi4\right)\right)=4\pi.HPAS 2024 Maths Optional Paper-1 Question 6(b)
Evaluate the surface integral \iint_{S} \vec{F} \cdot \mathbf{n} \,dA where \vec{F} = z^2\mathbf{i} + xy\mathbf{j} - y^2\mathbf{k} and S is the portion of the surface of the cylinder x^2+y^2=36 for 0 \le z \le 4 included in the first quadrant.
Solution
Step 1: Parametrize the curved cylindrical surface
The surface is the first-quadrant part of the cylinder
x^2+y^2=36,\qquad 0\le z\le4.A convenient parametrization is
\vec r(\theta,z)=\langle6\cos\theta,6\sin\theta,z\rangle,with
0\le\theta\le\frac\pi2,\qquad0\le z\le4.Step 2: Find the outward vector area element
Differentiate:
\vec r_\theta=\langle-6\sin\theta,6\cos\theta,0\rangle,\qquad \vec r_z=\langle0,0,1\rangle.The outward cross product is
\vec r_\theta\times\vec r_z=\langle6\cos\theta,6\sin\theta,0\rangle.Hence
\mathbf{n}\,dA=\langle6\cos\theta,6\sin\theta,0\rangle\,d\theta\,dz.Step 3: Substitute the parametrization into the vector field
The field is
\vec F=z^2\mathbf{i}+xy\mathbf{j}-y^2\mathbf{k}.On the cylinder,
x=6\cos\theta,\qquad y=6\sin\theta,\qquad xy=36\cos\theta\sin\theta.Therefore,
\vec F=\langle z^2,36\cos\theta\sin\theta,-36\sin^2\theta\rangle.Step 4: Compute the flux integrand
\begin{aligned}\vec F\cdot(\vec r_\theta\times\vec r_z)&=6z^2\cos\theta+216\cos\theta\sin^2\theta. \end{aligned}Thus
\iint_S\vec F\cdot\mathbf n\,dA=\int_0^{\pi/2}\int_0^4\left(6z^2\cos\theta+216\cos\theta\sin^2\theta\right)dz\,d\theta.Step 5: Integrate with respect to z
\int_0^4 6z^2\cos\theta\,dz=128\cos\theta,and
\int_0^4 216\cos\theta\sin^2\theta\,dz=864\cos\theta\sin^2\theta.So the flux becomes
\int_0^{\pi/2}\left(128\cos\theta+864\cos\theta\sin^2\theta\right)d\theta.Step 6: Evaluate the final integral
\int_0^{\pi/2}128\cos\theta\,d\theta=128,and, with u=\sin\theta,
\int_0^{\pi/2}864\cos\theta\sin^2\theta\,d\theta=864\int_0^1u^2\,du=288.Hence
128+288=416.HPAS 2024 Maths Optional Paper-1 Question 7(a)
A particle is projected in a plane with velocity \sqrt{\frac{\mu}{3a^6}} at a distance a from the center of force, attracting according to the law \frac{\mu}{r^7}, in a direction inclined at 30^\circ to the radius vector. Show that the orbit is r^2 = 2a^2 \cos(2\theta).
Solution
Step 1: Calculate the angular momentum constant
Let u=1/r and let h=r^2\dot\theta be the angular momentum per unit mass. The initial speed is
v_0=\sqrt{\frac{\mu}{3a^6}}.The velocity is inclined at 30^\circ to the radius vector, so its transverse component is v_0\sin30^\circ. At r=a,
h=av_0\sin30^\circ=\frac{a}{2}\sqrt{\frac{\mu}{3a^6}}.Therefore,
h^2=\frac{\mu}{12a^4}.Step 2: Write Binet’s equation for the central force
The attractive acceleration has magnitude
\frac{\mu}{r^7}=\mu u^7.Binet’s equation for an attractive central force is
u''+u=\frac{F}{h^2u^2}.Hence
u''+u=\frac{\mu u^7}{h^2u^2}=\frac{\mu}{h^2}u^5=12a^4u^5.Step 3: Integrate the orbit equation once
Multiply by 2u':
2u'u''+2uu'=24a^4u^5u'.Each side is now an exact derivative with respect to \theta. Integrating,
(u')^2+u^2=4a^4u^6+C.Step 4: Use the initial conditions to determine C
At projection, r=a, so
u=\frac1a.Also, since
\dot r=v_0\cos30^\circ,\qquad h=av_0\sin30^\circ,and u'=-\dot r/h,
u'=-\frac{v_0\cos30^\circ}{av_0\sin30^\circ}=-\frac{\sqrt3}{a}.Substitute these values into the first integral:
\frac3{a^2}+\frac1{a^2}=4a^4\frac1{a^6}+C.Thus C=0. Therefore,
(u')^2=u^2(4a^4u^4-1).Step 5: Introduce s=r^2
Set
s=\frac1{u^2}=r^2.Then
s'=-2u^{-3}u'.Squaring and using the equation for (u')^2,
\begin{aligned}(s')^2&=4u^{-6}(u')^2\\&=4u^{-4}(4a^4u^4-1)\\&=16a^4-4s^2. \end{aligned}Hence
\frac{ds}{\sqrt{4a^4-s^2}}=\pm2\,d\theta.Step 6: Integrate for the polar equation
Integrating,
\sin^{-1}\left(\frac{s}{2a^2}\right)=\pm2\theta+C_1.Equivalently, the orbit can be written as
s=2a^2\cos\left(2(\theta-\theta_0)\right).The polar axis is arbitrary in a central-force problem, so choose it so that \theta_0=0. Since s=r^2,
r^2=2a^2\cos2\theta.Step 7: Check the initial inclination
At r=a, the equation gives \cos2\theta=1/2. Taking the corresponding point \theta=\pi/6, differentiation of the orbit gives a tangent making an angle 30^\circ with the radius vector, consistent with the given projection condition.
HPAS 2024 Maths Optional Paper-1 Question 7(b)
R is the resultant of forces P and Q acting on a particle. If P is reversed, with Q remaining the same, the resultant becomes R'. If R and R' are perpendicular to each other, show that P=Q.
Solution
Step 1: Write the original resultant as a vector
Let the two forces be \vec P and \vec Q. Their original resultant is
\vec R=\vec P+\vec Q.Step 2: Reverse the force P
When P is reversed, \vec P becomes -\vec P. Therefore the new resultant is
\vec R'=\vec Q-\vec P.Step 3: Use the perpendicularity condition
We are given that \vec R and \vec R' are perpendicular. Therefore,
\vec R\cdot\vec R'=0.Substitute the two expressions:
(\vec P+\vec Q)\cdot(\vec Q-\vec P)=0.Step 4: Expand the dot product
\vec P\cdot\vec Q-\vec P\cdot\vec P+\vec Q\cdot\vec Q-\vec Q\cdot\vec P=0.Because the dot product is commutative, \vec P\cdot\vec Q=\vec Q\cdot\vec P, so the cross terms cancel. Hence
Q^2-P^2=0.Thus
P^2=Q^2.Step 5: Compare the magnitudes
Force magnitudes are non-negative, so taking the positive square root gives
P=Q.HPAS 2024 Maths Optional Paper-1 Question 8(a)
Show that every real n-dimensional vector space is isomorphic to \mathbb{R}^n.
Solution
Step 1: Choose a basis of the vector space
Let V be a real vector space with
\dim V=n.Choose an ordered basis
B=\{v_1,v_2,\ldots,v_n\}.Because B is a basis, every vector v\in V has a unique representation
v=c_1v_1+c_2v_2+\cdots+c_nv_n,where c_i\in\mathbb R.
Step 2: Define the coordinate map
Define
T:V\longrightarrow\mathbb R^nby
T(v)=(c_1,c_2,\ldots,c_n).This simply sends each vector to its coordinate vector relative to the chosen basis.
Step 3: Prove linearity
Let
u=\sum_{i=1}^na_iv_i,\qquad v=\sum_{i=1}^nb_iv_i.For real scalars \alpha and \beta,
\alpha u+\beta v=\sum_{i=1}^n(\alpha a_i+\beta b_i)v_i.Therefore,
T(\alpha u+\beta v)=\alpha T(u)+\beta T(v).Hence T is linear.
Step 4: Prove injectivity
Suppose T(v)=0. Then every coordinate of v is zero:
c_1=c_2=\cdots=c_n=0.Thus v=0. Hence
\ker T=\{0\},so T is one-to-one.
Step 5: Prove surjectivity
Take any vector
(k_1,k_2,\ldots,k_n)\in\mathbb R^n.Define
v=k_1v_1+k_2v_2+\cdots+k_nv_n\in V.Then
T(v)=(k_1,k_2,\ldots,k_n).Hence every element of \mathbb R^n is in the image of T, so T is onto.
Step 6: Conclude isomorphism
Since T is linear, injective and surjective, it is a vector-space isomorphism.
HPAS 2024 Maths Optional Paper-1 Question 8(b)
The amplitude of a simple harmonic oscillator is doubled. How does this affect the time period, total energy, and maximum velocity of the oscillator?
Solution
Step 1: Represent the change in amplitude
Let the original amplitude be A. After doubling,
A'=2A.For an ideal simple harmonic oscillator, the angular frequency \omega is fixed by the system parameters and does not depend on amplitude.
Step 2: Effect on the time period
The time period is
T=\frac{2\pi}{\omega}.For a mass-spring oscillator, equivalently,
T=2\pi\sqrt{\frac{m}{k}}.No amplitude appears in this expression. Therefore doubling the amplitude does not change the period:
T'=T.Step 3: Effect on the total energy
The total energy of SHM is
E=\frac12kA^2=\frac12m\omega^2A^2.After doubling the amplitude,
E'=\frac12k(2A)^2=4\left(\frac12kA^2\right)=4E.So the total energy becomes four times its original value.
Step 4: Effect on the maximum velocity
For SHM,
v_{\max}=\omega A.Therefore,
v_{\max}'=\omega(2A)=2v_{\max}.The maximum velocity is doubled.
Time period: unchanged, T'=T.
Total energy: quadrupled, E'=4E.
Maximum velocity: doubled, v_{\max}'=2v_{\max}.