UPHESC Assistant Professor Mathematics 2025 PYQs

Concept: A set is convex if the segment joining any two of its points lies in it.

1. Discs (A), (C) and the half-plane (D) are convex.
2. (B) is an annulus $1\le r\le 2$: $(1.5,0)$ and $(-1.5,0)$ are in it but the midpoint $(0,0)$ is not.
3. So (B) is not convex.

Short trick: A set with a 'hole' is never convex.

Concept: Take the tangent at $(x_n,f(x_n))$ and find where it meets the x-axis.

1. Tangent: $y-f(x_n)=f^{\prime }(x_n)(x-x_n)$.
2. Put $y=0$: $x_{n+1}=x_n-\frac{f(x_n)}{f^{\prime }(x_n)}$.

Short trick: 'Minus f over f-dash'.

Concept: Evaluate $z$ at the corner points of the feasible region.

1. $3x_1+4x_2\le 24$ is redundant (never binding inside the others).
2. Corners: $(0,0)$, $(3.5,0)$, $(\frac{7}{3},\frac{5}{3})$, $(0,4)$.
3. $z$: 0, 17.5, $\frac{70}{3}\approx 23.3$, 28 → maximum 28 at $(0,4)$.

Short trick: Largest coefficient is on $x_2$ – check the vertex on the $x_2$-axis first.

Concept: Look for a direction in the feasible region along which $z$ increases without limit.

1. Region is non-empty (e.g. $(3,2)$) and unbounded.
2. Along the line $x_2=x_1-1$ (feasible for $x_1\ge \frac{7}{3}$): $z=2x_1-(x_1-1)=x_1+1\to \mathrm{\infty }$.
3. So the LPP has an unbounded solution.

Short trick: $z=(x_1-x_2)+x_1\le 1+x_1$ – and $x_1$ is free to grow.

Concept: $f[a,b]=\frac{f(b)-f(a)}{b-a}$, built up recursively.

1. $f[a,b]=\frac{1/b-1/a}{b-a}=-\frac{1}{ab}$.
2. $f[a,b,c]=\frac{-\frac{1}{bc}+\frac{1}{ab}}{c-a}=\frac{1}{abc}$.
3. $f[a,b,c,d]=\frac{\frac{1}{bcd}-\frac{1}{abc}}{d-a}=-\frac{1}{abcd}$.

Short trick: For $1/x$: $k$th divided difference $=\frac{(-1{)}^k}{\text{product of all points}}$.

Concept: The interpolating cubic through 4 points is unique, so just check which option fits all points.

1. $y(0)=2$ rules out (D).
2. $y(1)=3$: (A) $1+1-1+2=3$ ✓, (B) 5 ✗, (C) 1 ✗.
3. (A) at $x=2$: $8+4-2+2=12$ ✓; at $x=5$: $125+25-5+2=147$ ✓.

Short trick: Uniqueness of the interpolating polynomial ⇒ verify options instead of expanding Lagrange's formula.

Concept: Newton–Raphson has quadratic convergence (simple root).

1. ${\epsilon }_{n+1}\approx \frac{f^{″}(\alpha )}{2f^{\prime }(\alpha )}{\epsilon }_n^2$.
2. Order $=2$ (bisection 1, secant ≈1.618, regula falsi 1).

Short trick: NR doubles the correct digits each step → order 2.

Concept: Twin primes differ by 2.

1. Primes in 51–100: 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
2. Pairs differing by 2: (59, 61) and (71, 73).
3. So 2 pairs.

Short trick: Twin primes > 3 are of the form $(6k-1,6k+1)$: check 60, 72 (and 54, 66, 78, 84, 90, 96 fail).

Concept: $\sigma$ is multiplicative: $\sigma (p^k)=\frac{p^{k+1}-1}{p-1}$.

1. $180=2^2\cdot 3^2\cdot 5$.
2. $\sigma =(1+2+4)(1+3+9)(1+5)=7\cdot 13\cdot 6=546$.

Short trick: Factorise, then multiply the geometric sums.

Concept: Find a small power of 2 that is ≡ 1 (mod 341).

1. $2^{10}=1024=3\cdot 341+1$ ⇒ $2^{10}\equiv 1(\mathrm{mod}341)$.
2. $2^{340}=(2^{10}{)}^{34}\equiv 1$.
3. (341 = 11·31 is the smallest Fermat pseudoprime to base 2.)

Short trick: 341 is a pseudoprime: $2^{340}\equiv 1$ even though 341 is composite.

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.