Table of Contents
HPAS 2023 Maths Optional Paper 1 Question 1 (a)
Show that the roots of the equation \(P_n(x)=0\) are real and lie between \(-1\) and \(1\),
where \(P_n(x)\) is the Legendre polynomial of degree \(n\) (a positive integer).
Step 1: Rodrigues’ Formula
The Legendre polynomial is defined by \[ P_n(x) = \frac{1}{2^n n!}\,\frac{d^n}{dx^n}\!\left[(x^2 – 1)^n\right]. \]
The Legendre polynomial is defined by \[ P_n(x) = \frac{1}{2^n n!}\,\frac{d^n}{dx^n}\!\left[(x^2 – 1)^n\right]. \]
Step 2: Consider the function
Let \[ f(x) = (x^2 – 1)^n. \] This function has roots at \(x = -1\) and \(x = 1\), each with multiplicity \(n\).
Let \[ f(x) = (x^2 – 1)^n. \] This function has roots at \(x = -1\) and \(x = 1\), each with multiplicity \(n\).
Step 3: First application of Rolle’s Theorem
Since \(f(-1) = f(1) = 0\), by Rolle’s Theorem, the derivative \[ f'(x) = \frac{d}{dx}(x^2 – 1)^n \] must have at least one root in the interval \((-1,1)\).
Since \(f(-1) = f(1) = 0\), by Rolle’s Theorem, the derivative \[ f'(x) = \frac{d}{dx}(x^2 – 1)^n \] must have at least one root in the interval \((-1,1)\).
Step 4: Repeated application of Rolle’s Theorem
– The function \(f'(x)\) also has zeros at \(-1\) and \(1\) (with multiplicity \(n-1\)). – Applying Rolle’s Theorem again, \(f”(x)\) must have at least two roots in \((-1,1)\). – Continuing this process: after \(k\) differentiations, \[ \frac{d^k}{dx^k}(x^2 – 1)^n \] has at least \(k\) distinct roots in \((-1,1)\).
– The function \(f'(x)\) also has zeros at \(-1\) and \(1\) (with multiplicity \(n-1\)). – Applying Rolle’s Theorem again, \(f”(x)\) must have at least two roots in \((-1,1)\). – Continuing this process: after \(k\) differentiations, \[ \frac{d^k}{dx^k}(x^2 – 1)^n \] has at least \(k\) distinct roots in \((-1,1)\).
Step 5: When \(k = n\)
The \(n\)-th derivative \[ \frac{d^n}{dx^n}(x^2 – 1)^n \] has at least \(n\) distinct real roots in \((-1,1)\).
The \(n\)-th derivative \[ \frac{d^n}{dx^n}(x^2 – 1)^n \] has at least \(n\) distinct real roots in \((-1,1)\).
Step 6: Relation with Legendre polynomial
Since \(P_n(x)\) is proportional to this \(n\)-th derivative, it follows that \(P_n(x) = 0\) also has \(n\) distinct real roots, all lying strictly inside the interval \((-1,1)\).
Since \(P_n(x)\) is proportional to this \(n\)-th derivative, it follows that \(P_n(x) = 0\) also has \(n\) distinct real roots, all lying strictly inside the interval \((-1,1)\).
Final Answer: All roots of the Legendre polynomial \(P_n(x) = 0\) are real, distinct, and located in the open interval \((-1,1)\).
HPAS 2023 Maths Optional Paper 1 Question 1 (b)
Prove that the set \( A = \{x : f(x) = 0\} \) is closed.
Step 1: Complement of \(A\)
The complement of \(A\) is \[ A^c = \{x : f(x) \neq 0\}. \] This can be written as \[ A^c = \{x : f(x) > 0\} \;\cup\; \{x : f(x) < 0\}. \] So, to prove that \(A\) is closed, it is enough to show that \(A^c\) is open.
The complement of \(A\) is \[ A^c = \{x : f(x) \neq 0\}. \] This can be written as \[ A^c = \{x : f(x) > 0\} \;\cup\; \{x : f(x) < 0\}. \] So, to prove that \(A\) is closed, it is enough to show that \(A^c\) is open.
Step 2: Show that \(\{x : f(x) > 0\}\) is open
Let \(x_0 \in \{x : f(x) > 0\}\). Then \(f(x_0) > 0\). Since \(f\) is continuous at \(x_0\), for \(\epsilon = f(x_0) > 0\) there exists \(\delta > 0\) such that \[ |x – x_0| < \delta \;\;\Rightarrow\;\; |f(x) - f(x_0)| < \epsilon. \] This implies \[ 0 < f(x) < 2f(x_0), \] so all points in the interval \((x_0 - \delta, x_0 + \delta)\) also satisfy \(f(x) > 0\). Hence, \(\{x : f(x) > 0\}\) is an open set.
Let \(x_0 \in \{x : f(x) > 0\}\). Then \(f(x_0) > 0\). Since \(f\) is continuous at \(x_0\), for \(\epsilon = f(x_0) > 0\) there exists \(\delta > 0\) such that \[ |x – x_0| < \delta \;\;\Rightarrow\;\; |f(x) - f(x_0)| < \epsilon. \] This implies \[ 0 < f(x) < 2f(x_0), \] so all points in the interval \((x_0 - \delta, x_0 + \delta)\) also satisfy \(f(x) > 0\). Hence, \(\{x : f(x) > 0\}\) is an open set.
Step 3: Show that \(\{x : f(x) < 0\}\) is open
Let \(x_0 \in \{x : f(x) < 0\}\). Then \(f(x_0) < 0\). Since \(f\) is continuous at \(x_0\), choose \(\epsilon = -f(x_0) > 0\). Then there exists \(\delta > 0\) such that \[ |x – x_0| < \delta \;\;\Rightarrow\;\; |f(x) - f(x_0)| < \epsilon. \] This gives \[ 2f(x_0) < f(x) < 0, \] which means all points in some interval around \(x_0\) also satisfy \(f(x) < 0\). Hence, \(\{x : f(x) < 0\}\) is an open set.
Let \(x_0 \in \{x : f(x) < 0\}\). Then \(f(x_0) < 0\). Since \(f\) is continuous at \(x_0\), choose \(\epsilon = -f(x_0) > 0\). Then there exists \(\delta > 0\) such that \[ |x – x_0| < \delta \;\;\Rightarrow\;\; |f(x) - f(x_0)| < \epsilon. \] This gives \[ 2f(x_0) < f(x) < 0, \] which means all points in some interval around \(x_0\) also satisfy \(f(x) < 0\). Hence, \(\{x : f(x) < 0\}\) is an open set.
Step 4: Conclusion
Since both \(\{x : f(x) > 0\}\) and \(\{x : f(x) < 0\}\) are open sets, their union \[ A^c = \{x : f(x) > 0\} \cup \{x : f(x) < 0\} \] is also open. Therefore, the original set \[ A = \{x : f(x) = 0\} \] is closed, because the complement of a closed set is open.
Since both \(\{x : f(x) > 0\}\) and \(\{x : f(x) < 0\}\) are open sets, their union \[ A^c = \{x : f(x) > 0\} \cup \{x : f(x) < 0\} \] is also open. Therefore, the original set \[ A = \{x : f(x) = 0\} \] is closed, because the complement of a closed set is open.
Final Answer: The set \(A = \{x : f(x) = 0\}\) is closed, since its complement is open.
HPAS 2023 Maths Optional Paper 1 Question 1 (c)
For what values of \(a\) and \(b\), is the vector field
\[
\mathbf{F} = (x+z)\mathbf{i} + a(y+z)\mathbf{j} + b(x+y)\mathbf{k}
\]
a conservative field?
The vector field is given by
\[
\mathbf{F} = F_1 \mathbf{i} + F_2 \mathbf{j} + F_3 \mathbf{k},
\]
where
\[
F_1 = x+z, \quad F_2 = a(y+z)=ay+az, \quad F_3 = b(x+y)=bx+by.
\]
A vector field is conservative if its curl is zero:
\[
\nabla \times \mathbf{F} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
F_1 & F_2 & F_3
\end{vmatrix}.
\]
Expanding,
\[
\nabla \times \mathbf{F} =
\left(\frac{\partial F_3}{\partial y} – \frac{\partial F_2}{\partial z}\right)\mathbf{i}
– \left(\frac{\partial F_1}{\partial z} – \frac{\partial F_3}{\partial x}\right)\mathbf{j}
+ \left(\frac{\partial F_2}{\partial x} – \frac{\partial F_1}{\partial y}\right)\mathbf{k}.
\]
Step 1: Compute partial derivatives
\[ \frac{\partial F_3}{\partial y} = b, \quad \frac{\partial F_2}{\partial z} = a, \quad \frac{\partial F_1}{\partial z} = 1, \] \[ \frac{\partial F_3}{\partial x} = b, \quad \frac{\partial F_2}{\partial x} = 0, \quad \frac{\partial F_1}{\partial y} = 0. \] Step 2: Substitute into curl
\[ \nabla \times \mathbf{F} = (b-a)\mathbf{i} + (b-1)\mathbf{j} + 0\mathbf{k}. \] Step 3: Condition for conservative field
\[ b-a = 0, \quad b-1 = 0, \quad 0=0. \] Step 4: Solve for \(a\) and \(b\)
From \(b-1=0 \Rightarrow b=1\). Substituting into \(b-a=0\): \(1-a=0 \Rightarrow a=1\).
\[ \frac{\partial F_3}{\partial y} = b, \quad \frac{\partial F_2}{\partial z} = a, \quad \frac{\partial F_1}{\partial z} = 1, \] \[ \frac{\partial F_3}{\partial x} = b, \quad \frac{\partial F_2}{\partial x} = 0, \quad \frac{\partial F_1}{\partial y} = 0. \] Step 2: Substitute into curl
\[ \nabla \times \mathbf{F} = (b-a)\mathbf{i} + (b-1)\mathbf{j} + 0\mathbf{k}. \] Step 3: Condition for conservative field
\[ b-a = 0, \quad b-1 = 0, \quad 0=0. \] Step 4: Solve for \(a\) and \(b\)
From \(b-1=0 \Rightarrow b=1\). Substituting into \(b-a=0\): \(1-a=0 \Rightarrow a=1\).
Final Answer: The vector field is conservative when \[ a = 1, \quad b = 1. \]
HPAS 2023 Maths Optional Paper 1 Question 1 (d)
Give an example of a diagonalizable matrix that does not have distinct eigenvalues,
and justify why it is diagonalizable.
Let us consider the \(3 \times 3\) matrix
\[
A=\begin{bmatrix}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 2
\end{bmatrix}.
\]
Notice that the eigenvalue \(\lambda = 1\) is repeated (it occurs twice).
Therefore, the eigenvalues are not all distinct.
We now check whether \(A\) is still diagonalizable.
Step 1: Characteristic polynomial and eigenvalues
The characteristic polynomial is
\[
\chi_A(\lambda) = \det(A – \lambda I)
= (1-\lambda)^2(2-\lambda).
\]
Hence, the eigenvalues are:
\(\lambda = 1\), with algebraic multiplicity \(2\),
\(\lambda = 2\), with algebraic multiplicity \(1\).
Step 2: Eigenspace for \(\lambda = 1\)
To find eigenvectors, solve \((A – I)\mathbf{x} = 0\):
\[
A – I = \begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 1
\end{bmatrix}.
\]
This gives the condition \(z=0\), while \(x\) and \(y\) are free.
So the eigenspace is
\[
E_1 = \left\{\begin{bmatrix}x\\y\\0\end{bmatrix} : x,y \in \mathbb{R}\right\}
= \operatorname{span}\!\left\{\begin{bmatrix}1\\0\\0\end{bmatrix}, \begin{bmatrix}0\\1\\0\end{bmatrix}\right\}.
\]
Thus, the geometric multiplicity of \(\lambda=1\) is \(2\).
Step 3: Eigenspace for \(\lambda = 2\)
Now solve \((A – 2I)\mathbf{x} = 0\):
\[
A – 2I = \begin{bmatrix}
-1 & 0 & 0\\
0 & -1 & 0\\
0 & 0 & 0
\end{bmatrix}.
\]
This gives \(x=0\) and \(y=0\), while \(z\) is free.
Hence the eigenspace is
\[
E_2 = \operatorname{span}\!\left\{\begin{bmatrix}0\\0\\1\end{bmatrix}\right\}.
\]
The geometric multiplicity of \(\lambda = 2\) is \(1\).
Step 4: Checking diagonalizability
We now have three linearly independent eigenvectors:
\[
\mathbf{v}_1=\begin{bmatrix}1\\0\\0\end{bmatrix}, \quad
\mathbf{v}_2=\begin{bmatrix}0\\1\\0\end{bmatrix}, \quad
\mathbf{v}_3=\begin{bmatrix}0\\0\\1\end{bmatrix}.
\]
Since these vectors form a basis of \(\mathbb{R}^3\), the matrix \(A\) is diagonalizable.
In fact, with
\[
P = \begin{bmatrix}\mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3\end{bmatrix} = I_3,
\qquad
D = \operatorname{diag}(1,1,2),
\]
we get \(P^{-1}AP = D\).
Step 5: Contrast example
Not every matrix with repeated eigenvalues is diagonalizable. For example,
\[
B=\begin{bmatrix}2 & 1\\ 0 & 2\end{bmatrix}
\]
has eigenvalue \(\lambda=2\) with algebraic multiplicity \(2\).
But
\[
B-2I = \begin{bmatrix}0 & 1\\ 0 & 0\end{bmatrix}
\]
gives only one independent eigenvector, so \(B\) is not diagonalizable.
Additional Note
We could have also taken the identity matrix \(I\) as an example,
since it is diagonalizable and all its eigenvalues are equal.
HPAS 2023 Physics Optional Paper 1 Question 1 (a)
A particle executes Simple Harmonic Motion (SHM) of period T = 10 s and amplitude A = 5 cm. Calculate the maximum amplitude of velocity.
Step 1: Calculate the Angular Frequency
The angular frequency ω is related to the period T by:
ω = 2π/T
Given T = 10 s:
ω = 2π/10 = π/5 rad/s
The angular frequency ω is related to the period T by:
ω = 2π/T
Given T = 10 s:
ω = 2π/10 = π/5 rad/s
Step 2: Calculate the Maximum Velocity Amplitude
The maximum velocity amplitude Vmax in SHM is given by:
Vmax = A ω
Substituting A = 5 cm and ω = π/5 rad/s:
Vmax = 5 × (π/5) = π cm/s
The maximum velocity amplitude Vmax in SHM is given by:
Vmax = A ω
Substituting A = 5 cm and ω = π/5 rad/s:
Vmax = 5 × (π/5) = π cm/s
✅ Final Answer: The maximum amplitude of velocity is π cm/s.