UPHESC Assistant Professor Mathematics 2025 PYQs

Concept: Arrange the works chronologically.

1. Aryabhatiya – Aryabhata, 499 CE.
2. Panchasiddhantika – Varahamihira, 6th century (c. 505–575 CE).
3. Brahmasphuta Siddhanta – Brahmagupta, 628 CE.
4. Trisatika – Sridharacharya, c. 8th–9th century.

Short trick: A-V-B-S: Aryabhata → Varahamihira → Brahmagupta → Sridhara.

Concept: Famous numbers and dates.

1. Taxi-cab (Hardy–Ramanujan) number $1729=1^3+{12}^3=9^3+{10}^3$ → III.
2. Ramanujan born 22 Dec 1887 → I.
3. Kaprekar's constant 6174 → II.
4. Bhaskaracharya II born 1114 CE → IV.

Short trick: 1729 taxi, 1887 Ramanujan (National Mathematics Day 22 Dec), 6174 Kaprekar, 1114 Bhaskara II.

Concept: Round bracket = end point excluded (<), square bracket = included (≤).

1. (a, b) → $a<x<b$ (III).
2. (a, b] → $a<x\le b$ (IV).
3. [a, b) → $a\le x<b$ (I).
4. [a, b] → $a\le x\le b$ (II).

Short trick: Square bracket 'holds' the end point.

Concept: Every metric space is Hausdorff and normal (in fact perfectly normal), hence $T_1,T_2,T_3,T_4$.

1. $T_2$: for $x\ne y$ the balls of radius $d(x,y)/2$ are disjoint; $T_2\Rightarrow T_1$.
2. $T_4$: for disjoint closed $A,B$, $U=\{x:d(x,A)<d(x,B)\}$ and $V=\{x:d(x,B)<d(x,A)\}$ are disjoint open sets containing them.
3. $T_4+T_1\Rightarrow T_3$. So all four statements are true – none is incorrect, so no option can be chosen as printed (question treated as bonus).

Short trick: Metric ⇒ $T_6$ (perfectly normal) ⇒ every $T_i$, $i\le 4$.

Concept: A topology must contain ϕ, X and be closed under unions and finite intersections.

1. (B): $\{a,b\}\cap \{b,c\}=\{b\}$ is missing → not a topology.
2. (A): union = X, intersection = ϕ ✓. (C): adds {b}, closed ✓. (D): nested chain ✓.

Short trick: Check pairwise intersections first – the usual failure.

Concept: Totally bounded: for every ε>0 there is a finite ε-net.

1. Union of the finite $\frac{1}{n}$-nets ($n=1,2,\dots$) is a countable dense set → separable.
2. Not necessarily compact: $(0,1)$ or $\mathbb{Q}\cap [0,1]$ is totally bounded but not compact (compact = complete + totally bounded).
3. $\mathbb{Q}\cap [0,1]$ is not even locally compact.

Short trick: Totally bounded ⇒ separable; + complete ⇒ compact.

Concept: Gaussian curvature $K={\kappa }_1{\kappa }_2$ (product of principal curvatures).

1. On a sphere every normal section is a great circle: ${\kappa }_1={\kappa }_2=\frac{1}{r}$.
2. $K=\frac{1}{r^2}$ (constant positive).

Short trick: Sphere: $K=1/r^2$; plane & cylinder: $K=0$.

Concept: $\tau =\frac{[{\overrightarrow{r}}^{\prime }{\overrightarrow{r}}^{″}{\overrightarrow{r}}^{‴}]}{|{\overrightarrow{r}}^{\prime }\times {\overrightarrow{r}}^{″}{|}^2}$.

1. ${\overrightarrow{r}}^{\prime }=(-a\sin u,a\cos u,b)$, ${\overrightarrow{r}}^{″}=(-a\cos u,-a\sin u,0)$, ${\overrightarrow{r}}^{‴}=(a\sin u,-a\cos u,0)$.
2. $[{\overrightarrow{r}}^{\prime }{\overrightarrow{r}}^{″}{\overrightarrow{r}}^{‴}]=a^{2}b$, $|{\overrightarrow{r}}^{\prime }\times {\overrightarrow{r}}^{″}{|}^2=a^2(a^2+b^2)$.
3. $\tau =\frac{b}{a^2+b^2}$ (and $\kappa =\frac{a}{a^2+b^2}$).

Short trick: Helix: κ has $a$, τ has $b$ on top; both over $a^2+b^2$.

Concept: $K={\kappa }_1{\kappa }_2$; for a sphere both principal curvatures are $1/a$.

1. $K=\frac{1}{a}\cdot \frac{1}{a}=\frac{1}{a^2}$.
2. Also $K=\frac{LN-M^2}{EG-F^2}$ gives the same constant value.

Short trick: Units check: curvature ~ 1/length, Gaussian curvature ~ 1/length².

Concept: By the fundamental theorem of curves, κ and τ determine the curve up to rigid motion.

1. The circular helix $(a\cos u,a\sin u,bu)$ has constant $\kappa =\frac{a}{a^2+b^2}$, $\tau =\frac{b}{a^2+b^2}$.
2. Hence any curve with constant κ ≠ 0 and τ ≠ 0 is a circular helix.
3. (Lancret: only the ratio κ/τ constant ⇒ general/cylindrical helix.)

Short trick: κ, τ both constant ⇒ circular helix; only κ/τ constant ⇒ general helix.