UPHESC Assistant Professor Mathematics · 100 questions

UPHESC Assistant Professor Mathematics 2025 PYQs

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100 questions.

Q61.

The residue at z = 0 of

z \cos (\frac{1}{z})

z \cos (\frac{1}{z})

का z = 0 पर अवशेष होगा

A1
B $\frac{1}{2}$
C $- \frac{1}{2}$
D0

Show answer

Concept: Residue = coefficient of $1 / z$ in the Laurent series.

$\cos \frac{1}{z} = 1 - \frac{1}{2 z^{2}} + \frac{1}{24 z^{4}} - \dots$
$z \cos \frac{1}{z} = z - \frac{1}{2 z} + \frac{1}{24 z^{3}} - \dots$
Coefficient of $1 / z$ : $- \frac{1}{2}$ .

Short trick: Essential singularity → always expand, never use pole formulas.

अवधारणा: अवशेष = लॉरेंट श्रेणी में $1 / z$ का गुणांक।

$\cos \frac{1}{z} = 1 - \frac{1}{2 z^{2}} + \dots$
$z \cos \frac{1}{z} = z - \frac{1}{2 z} + \dots$
$1 / z$ का गुणांक $- \frac{1}{2}$ ।

याद रखें: अनिवार्य विचित्रता → श्रेणी प्रसार करें।

UPHESC Assistant Professor MathematicsMathematicsResidues (Laurent series)UPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q62.

The transformation

w = \frac{5 - 4 z}{4 z - 2}

transforms the circle |z| = 1 into a circle in w – plane whose centre is

रूपान्तरण

w = \frac{5 - 4 z}{4 z - 2}

वृत्त |z| = 1 को w – समतल में एक वृत्त में रूपान्तरित करता है जिसका केन्द्र है

A $(- \frac{1}{2}, 0)$
B $(\frac{1}{2}, 0)$
C $(0, \frac{1}{2})$
D $(0, - \frac{1}{2})$

Show answer

Concept: Real coefficients ⇒ the image circle is symmetric about the real axis; its diameter joins the images of $z = \pm 1$ .

$z = 1$ : $w = \frac{1}{2}$ ; $z = - 1$ : $w = \frac{9}{- 6} = - \frac{3}{2}$ .
Centre $= \frac{1}{2} (\frac{1}{2} - \frac{3}{2}) = - \frac{1}{2}$ , radius 1.
(Directly: $z = \frac{5 + 2 w}{4 (w + 1)}$ , $| z | = 1$ ⇒ $| 5 + 2 w | = 4 | w + 1 |$ ⇒ $| w + \frac{1}{2} | = 1$ .)

Short trick: Map the two real points $z = \pm 1$ ; midpoint = centre.

अवधारणा: वास्तविक गुणांक ⇒ प्रतिबिंब वृत्त वास्तविक अक्ष के सापेक्ष सममित; $z = \pm 1$ के प्रतिबिंब व्यास के सिरे।

$z = 1 \to w = \frac{1}{2}$ ; $z = - 1 \to w = - \frac{3}{2}$ ।
केन्द्र $= - \frac{1}{2}$ , त्रिज्या 1।
सीधे: $| 5 + 2 w | = 4 | w + 1 |$ ⇒ $| w + \frac{1}{2} | = 1$ ।

याद रखें: $z = \pm 1$ के प्रतिबिंबों का मध्य बिंदु = केन्द्र।

UPHESC Assistant Professor MathematicsMathematicsBilinear (Möbius) transformationsUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q63.

If C is the circle |z| = 2, then the value of the integral

\int_{C} \frac{e^{2 z}}{(z + 1)^{4}} d z

यदि C, वृत्त |z| = 2 हो, तो समाकल

\int_{C} \frac{e^{2 z}}{(z + 1)^{4}} d z

का मान होगा

A $\frac{8 π i}{3 e}$
B $\frac{8 π i}{3 e^{2}}$
C $\frac{π i}{3 e^{2}}$
D $\frac{6 π i}{5 e^{2}}$

Show answer

Concept: $\oint_{C} \frac{f (z)}{(z - a)^{n + 1}} d z = \frac{2 π i}{n!} f^{(n)} (a)$ .

$a = - 1$ lies inside $| z | = 2$ ; $n = 3$ , $f = e^{2 z}$ , $f^{‴} = 8 e^{2 z}$ .
Value $= \frac{2 π i}{6} \cdot 8 e^{- 2} = \frac{8 π i}{3 e^{2}}$ .

Short trick: Power 4 ⇒ third derivative over 3!.

अवधारणा: $\oint \frac{f (z)}{(z - a)^{n + 1}} d z = \frac{2 π i}{n!} f^{(n)} (a)$ ।

$a = - 1$ वृत्त के अंदर; $n = 3$ , $f^{‴} = 8 e^{2 z}$ ।
मान $= \frac{2 π i}{6} \cdot 8 e^{- 2} = \frac{8 π i}{3 e^{2}}$ ।

याद रखें: घात 4 ⇒ तृतीय अवकलज / 3!।

UPHESC Assistant Professor MathematicsMathematicsCauchy integral formula for derivativesUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q64.

The analytic function f(z) = u + iv, of which the real part is

u = e^{x} (x \cos y - y \sin y)

, is

वैश्लेषिक फलन f(z) = u + iv, जिसका वास्तविक भाग

u = e^{x} (x \cos y - y \sin y)

है, होगा

A $z e^{z} + C$
B $e^{z} + C$
C $z e^{- z} + C$
D $e^{- z} + C$

Show answer

Concept: Milne-Thomson: $f^{'} (z) = u_{x} (z, 0) - i u_{y} (z, 0)$ .

$u_{x} (z, 0) = e^{z} (z + 1)$ , $u_{y} (z, 0) = 0$ .
$f (z) = \int e^{z} (z + 1) d z = z e^{z} + C$ .
Check: $z e^{z} = (x + i y) e^{x} (\cos y + i \sin y)$ , real part $e^{x} (x \cos y - y \sin y)$ ✓.

Short trick: Put $y = 0$ : $u (x, 0) = x e^{x}$ ⇒ $f (z) = z e^{z}$ .

अवधारणा: मिल्न-थॉमसन: $f^{'} (z) = u_{x} (z, 0) - i u_{y} (z, 0)$ ।

$u_{x} (z, 0) = e^{z} (z + 1)$ , $u_{y} (z, 0) = 0$ ।
$f (z) = z e^{z} + C$ ।
जाँच: वास्तविक भाग $e^{x} (x \cos y - y \sin y)$ ✓।

याद रखें: $y = 0$ रखें: $u = x e^{x}$ ⇒ $f = z e^{z}$ ।

UPHESC Assistant Professor MathematicsMathematicsAnalytic functions (Milne-Thomson)UPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q65.

The value of the integral

\int_{0}^{2 π} \frac{d θ}{5 + 4 \sin θ}

समाकल का मान

\int_{0}^{2 π} \frac{d θ}{5 + 4 \sin θ}

है

A $\frac{π}{3}$
B $\frac{2 π}{3}$
C $\frac{π}{2}$
D $2 π$

Show answer

Concept: $\int_{0}^{2 π} \frac{d θ}{a + b \sin θ} = \frac{2 π}{\sqrt{a^{2} - b^{2}}}$ for $a > | b |$ .

$a = 5, b = 4$ : $\sqrt{25 - 16} = 3$ .
Value $= \frac{2 π}{3}$ (by residues with $z = e^{i θ}$ , pole $z = - i / 2$ inside).

Short trick: Remember $2 π / \sqrt{a^{2} - b^{2}}$ (same for cos).

अवधारणा: $\int_{0}^{2 π} \frac{d θ}{a + b \sin θ} = \frac{2 π}{\sqrt{a^{2} - b^{2}}}$ ।

$a = 5, b = 4$ : $\sqrt{9} = 3$ ।
मान $= \frac{2 π}{3}$ ।

याद रखें: सूत्र $2 π / \sqrt{a^{2} - b^{2}}$ ।

UPHESC Assistant Professor MathematicsMathematicsContour integration (trigonometric)UPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q66.

Given below are two statements, one is labeled as Assertion (A) and other is labeled as Reason (R).
Assertion (A) : The quotient

p (z) / q (z)

has a pole of order m at

z_{0}

.
Reason (R) :

p (z_{0}) \neq 0

and q has a zero of order m at

z_{0}

.
In the light of the above statements, choose the correct answer from the options given below :

नीचे दो कथन दिए गए है , एक को अभिकथन (A) और दूसरे को तर्क (R) कहा गया है ।
अभिकथन (A) : भागफल

p (z) / q (z)

में एक m क्रम का ध्रुव

z_{0}

पर है ।
तर्क (R) :

p (z_{0}) \neq 0

, तथा q में एक m क्रम का,

z_{0}

पर एक शून्य है ।
उपर्युक्त कथनों के आलोक में निम्नांकित विकल्पों में से सही उत्तर चुनें :

ABoth (A) and (R) are correct and (R) is the correct explanation of (A)(A) और (R) दोनों सही हैं और (R), (A) की सही व्याख्या है
BBoth (A) and (R) are correct but (R) is not the correct explanation of (A)(A) और (R) दोनों सही हैं परन्तु (R), (A) की सही व्याख्या नहीं है
C(A) is correct but (R) is not correct(A) सही है परन्तु (R) गलत है
D(A) is not correct but (R) is correct(A) गलत है परन्तु (R) सही है

Show answer

Concept: If $q (z) = (z - z_{0})^{m} g (z)$ with $g (z_{0}) \neq 0$ and $p (z_{0}) \neq 0$ , then $\frac{p}{q} = \frac{1}{(z - z_{0})^{m}} \cdot \frac{p}{g}$ with $\frac{p}{g}$ analytic and non-zero at $z_{0}$ .

So $p / q$ has a pole of order exactly $m$ at $z_{0}$ – this is the standard theorem.
(R) gives precisely the hypothesis that produces (A), so (R) explains (A).

Short trick: Pole order of $p / q$ = zero order of $q$ − zero order of $p$ .

अवधारणा: $q (z) = (z - z_{0})^{m} g (z)$ , $g (z_{0}) \neq 0$ , $p (z_{0}) \neq 0$ ⇒ $\frac{p}{q} = \frac{1}{(z - z_{0})^{m}} \cdot \frac{p}{g}$ ।

अतः $p / q$ का $z_{0}$ पर ठीक m क्रम का ध्रुव है – मानक प्रमेय।
(R) ही (A) का कारण है, अतः सही व्याख्या।

याद रखें: $p / q$ के ध्रुव का क्रम = $q$ के शून्य का क्रम − $p$ के शून्य का क्रम।

UPHESC Assistant Professor MathematicsMathematicsZeros and polesUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q67.

In the case of spherical symmetry the equation of continuity is

गोलीय सममिति की दशा में सातत्य का समीकरण है

A $\frac{\partial ρ}{\partial t} + \frac{1}{r} \frac{\partial (ρ q r^{2})}{\partial r} = 0$
B $\frac{\partial ρ}{\partial t} + \frac{1}{r^{2}} \frac{\partial (ρ q r^{2})}{\partial r} = 0$
C $\frac{\partial ρ}{\partial t} + \frac{1}{r} \frac{\partial (ρ q r)}{\partial r} = 0$
DNone of the aboveउपर्युक्त में से कोई नहीं

Show answer

Concept: $\frac{\partial ρ}{\partial t} + \nabla \cdot (ρ \vec{q}) = 0$ ; in spherical polars with only radial velocity $q (r, t)$ , $\nabla \cdot (ρ q \hat{r}) = \frac{1}{r^{2}} \frac{\partial}{\partial r} (r^{2} ρ q)$ .

Mass balance in the shell between $r$ and $r + d r$ (area $4 π r^{2}$ ) gives the same result.
Hence $\frac{\partial ρ}{\partial t} + \frac{1}{r^{2}} \frac{\partial (ρ q r^{2})}{\partial r} = 0$ .
(C) with $\frac{1}{r} \frac{\partial (ρ q r)}{\partial r}$ is the cylindrical-symmetry form.

Short trick: Spherical: $r^{2}$ inside and $1 / r^{2}$ outside; cylindrical: $r$ and $1 / r$ .

अवधारणा: $\frac{\partial ρ}{\partial t} + \nabla \cdot (ρ \vec{q}) = 0$ ; गोलीय ध्रुवीय निर्देशांक में $\nabla \cdot (ρ q \hat{r}) = \frac{1}{r^{2}} \frac{\partial}{\partial r} (r^{2} ρ q)$ ।

$r$ व $r + d r$ के बीच के गोलीय कोश (क्षेत्रफल $4 π r^{2}$ ) में द्रव्यमान संतुलन से भी यही मिलता है।
अतः विकल्प (B)।
(C) बेलनाकार सममिति का रूप है।

याद रखें: गोलीय: अंदर $r^{2}$ , बाहर $1 / r^{2}$ ; बेलनाकार: $r$ व $1 / r$ ।

UPHESC Assistant Professor MathematicsMathematicsEquation of continuityUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q68.

The number of generalized coordinates required to describe the motion of a rigid body with one of its points fixed is

एक दृढ़ पिण्ड, जिसका एक बिन्दु नियत है, की गति का वर्णन करने के लिए आवश्यक सामान्यीकृत निर्देशांकों की संख्या है

Show answer

Concept: A free rigid body has 6 degrees of freedom (3 translation + 3 rotation).

Fixing one point removes the 3 translational freedoms.
Only rotation about the point remains: 3 coordinates (e.g. Euler angles $θ, ϕ, ψ$ ).

Short trick: Fixed point ⇒ Euler angles ⇒ 3.

अवधारणा: मुक्त दृढ़ पिंड की 6 स्वातंत्र्य कोटियाँ (3 स्थानांतरण + 3 घूर्णन)।

एक बिंदु नियत करने से 3 स्थानांतरण समाप्त।
केवल घूर्णन बचता है: 3 (ऑयलर कोण)।

याद रखें: नियत बिंदु ⇒ ऑयलर कोण ⇒ 3।

UPHESC Assistant Professor MathematicsMathematicsDegrees of freedom of a rigid bodyUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q69.

Let f, g, h be any three functions of (q, p). If [f, g] denotes the Poisson bracket of f and g, which of the following relations does not always hold good ?

मान लीजिए कि (q, p) के f, g, h कोई तीन फलन हैं । यदि f और g के प्वायसन कोष्ठक को [f, g] से निरूपित किया जाय तो निम्नलिखित सम्बन्धों में से कौन-सा हमेशा सही नहीं है ?

A[f, g] = – [g, f]
B[ [ f, g], h] + [ [ g, h], f] + [ [ h, f], g] = 0
C $[q_{r}, q_{s}] = 0$
D $[q_{r}, p_{s}] = 0$

Show answer

Concept: Fundamental Poisson brackets: $[q_{r}, q_{s}] = 0$ , $[p_{r}, p_{s}] = 0$ , $[q_{r}, p_{s}] = δ_{r s}$ .

(A) antisymmetry and (B) Jacobi identity always hold.
(C) always holds.
(D) fails when $r = s$ : $[q_{r}, p_{r}] = 1$ .

Short trick: $[q, p] = 1$ – the canonical pair never commutes.

अवधारणा: मूल प्वायसन कोष्ठक: $[q_{r}, q_{s}] = 0$ , $[p_{r}, p_{s}] = 0$ , $[q_{r}, p_{s}] = δ_{r s}$ ।

(A) प्रतिसममिति, (B) जैकोबी सर्वसमिका सदैव सत्य।
(C) सदैव सत्य।
(D) $r = s$ पर असत्य: $[q_{r}, p_{r}] = 1$ ।

याद रखें: $[q, p] = 1$ ।

UPHESC Assistant Professor MathematicsMathematicsPoisson bracketsUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q70.

Given below are two statements, one is labeled as Assertion (A) and other is labeled as Reason (R).
Assertion (A) : Equation of continuity for a fluid is

\frac{\partial ρ}{\partial t} + \nabla \cdot (ρ \vec{v}) = 0

.
Reason (R) : The equation of continuity for incompressible fluid is

\nabla \times \vec{v} = \vec{0}

.
In the light of the above statements, choose the correct answer from the options given below :

नीचे दो कथन दिए गए है , एक को अभिकथन (A) और दूसरे को तर्क (R) कहा गया है ।
अभिकथन (A) : किसी तरल पदार्थ के लिए सातत्य का समीकरण

\frac{\partial ρ}{\partial t} + \nabla \cdot (ρ \vec{v}) = 0

है ।
तर्क (R) : असंपीड्य तरल पदार्थ के लिए सातत्य का समीकरण

\nabla \times \vec{v} = \vec{0}

होता है ।
उपर्युक्त कथनों के आलोक में निम्नलिखित विकल्पों में से सही उत्तर चुनें :

ABoth (A) and (R) are correct and (R) is the correct explanation of (A)(A) और (R) दोनों सही हैं और (R), (A) की सही व्याख्या है
BBoth (A) and (R) are correct but (R) is not the correct explanation of (A)(A) और (R) दोनों सही हैं परन्तु (R), (A) की सही व्याख्या नहीं है
C(A) is correct but (R) is not correct(A) सही है परन्तु (R) गलत है
D(A) is not correct but (R) is correct(A) गलत है परन्तु (R) सही है

Show answer

Concept: General continuity equation: $ρ_{t} + \nabla \cdot (ρ \vec{v}) = 0$ – (A) true.

For incompressible flow $ρ$ = const, so it reduces to $\nabla \cdot \vec{v} = 0$ (divergence, not curl).
$\nabla \times \vec{v} = \vec{0}$ is the condition for irrotational flow – so (R) is false.

Short trick: Incompressible ↔ div = 0; Irrotational ↔ curl = 0.

अवधारणा: सामान्य सातत्य समीकरण $ρ_{t} + \nabla \cdot (ρ \vec{v}) = 0$ – (A) सत्य।

असंपीड्य तरल में $ρ$ अचर, अतः $\nabla \cdot \vec{v} = 0$ (डाइवर्जेंस, कर्ल नहीं)।
$\nabla \times \vec{v} = \vec{0}$ अघूर्णी प्रवाह की शर्त है – (R) असत्य।

याद रखें: असंपीड्य ↔ div = 0; अघूर्णी ↔ curl = 0।

UPHESC Assistant Professor MathematicsMathematicsEquation of continuityUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q71.

For the system described by the Lagrangian

L = {\dot{q}}_{1}^{2} + {\dot{q}}_{2}^{2} + {\dot{q}}_{1} {\dot{q}}_{2} - q_{1}^{2} - q_{2}^{2}

, which one of the following statements is not true ?

उस तंत्र के लिए जो लैग्रान्जियन

L = {\dot{q}}_{1}^{2} + {\dot{q}}_{2}^{2} + {\dot{q}}_{1} {\dot{q}}_{2} - q_{1}^{2} - q_{2}^{2}

से प्रदर्शित है, निम्नलिखित कथनों में से कौन-सा एक सत्य नहीं है ?

A $p_{1} + p_{2} = 3 ({\dot{q}}_{1} + {\dot{q}}_{2})$
B ${\dot{p}}_{1} + {\dot{p}}_{2} = - 2 (q_{1} + q_{2})$
C $(q_{1} + q_{2})$ executes simple harmonic motion given by ${\ddot{q}}_{1} + {\ddot{q}}_{2} = - \frac{2}{3} (q_{1} + q_{2})$ $(q_{1} + q_{2})$ सरल आवर्त गति करता है और जो ${\ddot{q}}_{1} + {\ddot{q}}_{2} = - \frac{2}{3} (q_{1} + q_{2})$ से दिया जाता है
D ${\dot{p}}_{1} / {\dot{p}}_{2} = q_{2} / q_{1}$

Show answer

Concept: $p_{i} = \partial L / \partial {\dot{q}}_{i}$ and Lagrange's equations ${\dot{p}}_{i} = \partial L / \partial q_{i}$ .

$p_{1} = 2 {\dot{q}}_{1} + {\dot{q}}_{2}$ , $p_{2} = 2 {\dot{q}}_{2} + {\dot{q}}_{1}$ ⇒ $p_{1} + p_{2} = 3 ({\dot{q}}_{1} + {\dot{q}}_{2})$ – (A) true.
${\dot{p}}_{1} = - 2 q_{1}$ , ${\dot{p}}_{2} = - 2 q_{2}$ ⇒ (B) true.
From (A),(B): $3 ({\ddot{q}}_{1} + {\ddot{q}}_{2}) = - 2 (q_{1} + q_{2})$ – (C) true.
${\dot{p}}_{1} / {\dot{p}}_{2} = q_{1} / q_{2}$ , not $q_{2} / q_{1}$ – (D) false.

Short trick: ${\dot{p}}_{i} = \partial L / \partial q_{i} = - 2 q_{i}$ – ratio keeps the same order.

अवधारणा: $p_{i} = \partial L / \partial {\dot{q}}_{i}$ , ${\dot{p}}_{i} = \partial L / \partial q_{i}$ ।

$p_{1} + p_{2} = 3 ({\dot{q}}_{1} + {\dot{q}}_{2})$ – (A) सत्य।
${\dot{p}}_{1} = - 2 q_{1}, {\dot{p}}_{2} = - 2 q_{2}$ – (B) सत्य।
${\ddot{q}}_{1} + {\ddot{q}}_{2} = - \frac{2}{3} (q_{1} + q_{2})$ – (C) सत्य।
${\dot{p}}_{1} / {\dot{p}}_{2} = q_{1} / q_{2}$ – (D) असत्य।

याद रखें: अनुपात में क्रम वही रहता है: $q_{1} / q_{2}$ ।

UPHESC Assistant Professor MathematicsMathematicsLagrangian mechanicsUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q72.

The relation between velocity potential

ϕ

and stream function

ψ

are
a.

\frac{\partial ϕ}{\partial x} = \frac{\partial ψ}{\partial y}

\frac{\partial ϕ}{\partial y} = \frac{\partial ψ}{\partial x}

\frac{\partial ϕ}{\partial x} = - \frac{\partial ψ}{\partial y}

\frac{\partial ϕ}{\partial y} = - \frac{\partial ψ}{\partial x}

Choose the correct answer from the options given below :

वेग विभव

ϕ

और धारा फलन

ψ

के बीच सम्बन्ध है
a.

\frac{\partial ϕ}{\partial x} = \frac{\partial ψ}{\partial y}

\frac{\partial ϕ}{\partial y} = \frac{\partial ψ}{\partial x}

\frac{\partial ϕ}{\partial x} = - \frac{\partial ψ}{\partial y}

\frac{\partial ϕ}{\partial y} = - \frac{\partial ψ}{\partial x}

नीचे दिए गए विकल्पों में से सही उत्तर चुनिए :

Aa, b, c onlyकेवल a, b, c
Bb, c, d onlyकेवल b, c, d
Ca, c onlyकेवल a, c
Da, d onlyकेवल a, d

Show answer

Concept: $w = ϕ + i ψ$ is the complex potential, an analytic function, so $ϕ, ψ$ satisfy the Cauchy–Riemann equations.

$u = - ϕ_{x} = - ψ_{y}$ and $v = - ϕ_{y} = ψ_{x}$ .
Hence $ϕ_{x} = ψ_{y}$ (a) and $ϕ_{y} = - ψ_{x}$ (d).
(b) and (c) contradict these.

Short trick: φ and ψ are like u and v of an analytic function: $u_{x} = v_{y}, u_{y} = - v_{x}$ .

अवधारणा: सम्मिश्र विभव $w = ϕ + i ψ$ वैश्लेषिक है, अतः कॉशी–रीमान समीकरण लागू।

$u = - ϕ_{x} = - ψ_{y}$ , $v = - ϕ_{y} = ψ_{x}$ ।
अतः $ϕ_{x} = ψ_{y}$ (a), $ϕ_{y} = - ψ_{x}$ (d)।
(b), (c) गलत।

याद रखें: $u_{x} = v_{y}, u_{y} = - v_{x}$ जैसा।

UPHESC Assistant Professor MathematicsMathematicsVelocity potential and stream functionUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q73.

The Hamiltonian corresponding to the Lagrangian

L = a {\dot{x}}^{2} + b {\dot{y}}^{2} - k x y

लैग्रान्जियन

L = a {\dot{x}}^{2} + b {\dot{y}}^{2} - k x y

के संगत हैमिल्टोनियन है

A $\frac{p_{x}^{2}}{2 a} + \frac{p_{y}^{2}}{2 b} + k x y$
B $\frac{p_{x}^{2}}{4 a} + \frac{p_{y}^{2}}{4 b} - k x y$
C $\frac{p_{x}^{2}}{4 a} + \frac{p_{y}^{2}}{4 b} + k x y$
D $\frac{p_{x}^{2}}{4 a} - \frac{p_{y}^{2}}{4 b} + k x y$

Show answer

Concept: $H = \sum p_{i} {\dot{q}}_{i} - L$ ; for quadratic kinetic energy $H = T + V$ .

$p_{x} = 2 a \dot{x}$ , $p_{y} = 2 b \dot{y}$ ⇒ $\dot{x} = \frac{p_{x}}{2 a}$ , $\dot{y} = \frac{p_{y}}{2 b}$ .
$T = a {\dot{x}}^{2} + b {\dot{y}}^{2} = \frac{p_{x}^{2}}{4 a} + \frac{p_{y}^{2}}{4 b}$ , $V = k x y$ .
$H = \frac{p_{x}^{2}}{4 a} + \frac{p_{y}^{2}}{4 b} + k x y$ .

Short trick: $a {\dot{x}}^{2}$ ↔ $p^{2} / 4 a$ (not $2 a$ : there is no ½).

अवधारणा: $H = \sum p_{i} {\dot{q}}_{i} - L = T + V$ ।

$p_{x} = 2 a \dot{x}$ , $p_{y} = 2 b \dot{y}$ ।
$T = \frac{p_{x}^{2}}{4 a} + \frac{p_{y}^{2}}{4 b}$ , $V = k x y$ ।
$H = \frac{p_{x}^{2}}{4 a} + \frac{p_{y}^{2}}{4 b} + k x y$ ।

याद रखें: $a {\dot{x}}^{2}$ ↔ $p^{2} / 4 a$ ।

UPHESC Assistant Professor MathematicsMathematicsHamiltonian from LagrangianUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q74.

The extremal of the functional

\int_{1}^{2} \frac{\sqrt{1 + (y^{'})^{2}}}{x} d x

subject to the conditions y(1) = 0, y(2) = 1 is

फलनक

\int_{1}^{2} \frac{\sqrt{1 + (y^{'})^{2}}}{x} d x

का शर्तों y(1) = 0, y(2) = 1 के अधीन चरम वक्र है

A $x^{2} + y^{2} = 5$
B $(x - 2)^{2} + y^{2} = 5$
C $x^{2} + (y - 2)^{2} = 5$
D $(x - 1)^{2} + (y - 2)^{2} = 5$

Show answer

Concept: $F$ has no $y$ ⇒ $\partial F / \partial y^{'} =$ const.

$\frac{y^{'}}{x \sqrt{1 + y^{' 2}}} = c$ ⇒ $y^{'} = \frac{c x}{\sqrt{1 - c^{2} x^{2}}}$ ⇒ $y - d = - \frac{\sqrt{1 - c^{2} x^{2}}}{c}$ .
So $x^{2} + (y - d)^{2} = R^{2}$ (circles centred on the y-axis).
$(1, 0)$ : $1 + d^{2} = R^{2}$ ; $(2, 1)$ : $4 + (1 - d)^{2} = R^{2}$ ⇒ $d = 2$ , $R^{2} = 5$ .

Short trick: Check the end points in each option: only $x^{2} + (y - 2)^{2} = 5$ passes through (1,0) and (2,1).

अवधारणा: $F$ में $y$ नहीं ⇒ $\partial F / \partial y^{'} =$ अचर।

$\frac{y^{'}}{x \sqrt{1 + y^{' 2}}} = c$ ⇒ $x^{2} + (y - d)^{2} = R^{2}$ ।
$(1, 0)$ : $1 + d^{2} = R^{2}$ ; $(2, 1)$ : $4 + (1 - d)^{2} = R^{2}$ ।
$d = 2$ , $R^{2} = 5$ ।

याद रखें: विकल्पों में सिरे के बिंदु (1,0), (2,1) रखकर जाँचें।

UPHESC Assistant Professor MathematicsMathematicsCalculus of variationsUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q75.

If L{F(t)} = f(s), then the value of

L {t^{n} F (t)}

यदि L{F(t)} = f(s), तो

L {t^{n} F (t)}

का मान है

A $(- 1)^{n} \frac{d^{n}}{d s^{n}} f (s)$
B $\frac{d^{n}}{d s^{n}} f (s)$
C $(- 1)^{n - 1} \frac{d^{n}}{d s^{n}} f (s)$
D $(- 1)^{n + 1} \frac{d^{n}}{d s^{n}} f (s)$

Show answer

Concept: Differentiating $f (s) = \int_{0}^{\infty} e^{- s t} F (t) d t$ under the integral brings down $- t$ each time.

$f^{'} (s) = \int (- t) e^{- s t} F d t = - L {t F}$ .
Repeating $n$ times: $L {t^{n} F (t)} = (- 1)^{n} f^{(n)} (s)$ .

Short trick: Each $t$ = one ' $- \frac{d}{d s}$ '.

अवधारणा: $f (s) = \int e^{- s t} F d t$ का $s$ से अवकलन हर बार $- t$ देता है।

$f^{'} (s) = - L {t F}$ ।
$n$ बार: $L {t^{n} F} = (- 1)^{n} f^{(n)} (s)$ ।

याद रखें: हर $t$ = एक ' $- \frac{d}{d s}$ '।

UPHESC Assistant Professor MathematicsMathematicsLaplace transform – multiplication by t^nUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q76.

The solution of the Fredholm integral equation

ϕ (x) = 1 + λ \int_{0}^{1} x t ϕ (t) d t

फ्रेडहोल्म समाकल समीकरण

ϕ (x) = 1 + λ \int_{0}^{1} x t ϕ (t) d t

का हल है

A $ϕ (x) = 1 + \frac{3 λ x}{2 (3 - λ)}$
B $ϕ (x) = 1 + \frac{λ x}{(3 - λ)}$
C $ϕ (x) = 1 + \frac{2 λ x}{2 - λ}$
D $ϕ (x) = 1 + \frac{3 λ x}{2}$

Show answer

Concept: Separable kernel: put $C = \int_{0}^{1} t ϕ (t) d t$ , so $ϕ = 1 + λ C x$ .

$C = \int_{0}^{1} t (1 + λ C t) d t = \frac{1}{2} + \frac{λ C}{3}$ .
$C (1 - \frac{λ}{3}) = \frac{1}{2}$ ⇒ $C = \frac{3}{2 (3 - λ)}$ .
$ϕ (x) = 1 + \frac{3 λ x}{2 (3 - λ)}$ ( $λ \neq 3$ ).

Short trick: Eigenvalue $λ = 3$ must appear in the denominator – (A) or (B); constant check picks (A).

अवधारणा: पृथक्करणीय कर्नेल: $C = \int_{0}^{1} t ϕ (t) d t$ , तब $ϕ = 1 + λ C x$ ।

$C = \frac{1}{2} + \frac{λ C}{3}$ ।
$C = \frac{3}{2 (3 - λ)}$ ।
$ϕ (x) = 1 + \frac{3 λ x}{2 (3 - λ)}$ ।

याद रखें: हर में अभिलक्षणिक मान $λ = 3$ आता है।

UPHESC Assistant Professor MathematicsMathematicsFredholm equation (separable kernel)UPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q77.

Laplace transform of the function f(t) = t cost is

फलन f(t) = t cost का लाप्लास रूपान्तर है

A $\frac{s^{2} - 1}{s^{2} + 1}$
B $\frac{s^{2} - 1}{(s^{2} + 1)^{2}}$
C $\frac{s^{2}}{(s^{2} + 1)^{2}}$
D $\frac{1}{(s^{2} + 1)^{2}}$

Show answer

Concept: $L {t F (t)} = - \frac{d}{d s} L {F}$ .

$L {\cos t} = \frac{s}{s^{2} + 1}$ .
$- \frac{d}{d s} \frac{s}{s^{2} + 1} = - \frac{(s^{2} + 1) - 2 s^{2}}{(s^{2} + 1)^{2}} = \frac{s^{2} - 1}{(s^{2} + 1)^{2}}$ .

Short trick: $t \cos a t \to \frac{s^{2} - a^{2}}{(s^{2} + a^{2})^{2}}$ , $t \sin a t \to \frac{2 a s}{(s^{2} + a^{2})^{2}}$ .

अवधारणा: $L {t F} = - \frac{d}{d s} L {F}$ ।

$L {\cos t} = \frac{s}{s^{2} + 1}$ ।
$- \frac{d}{d s} \frac{s}{s^{2} + 1} = \frac{s^{2} - 1}{(s^{2} + 1)^{2}}$ ।

याद रखें: $t \cos a t \to \frac{s^{2} - a^{2}}{(s^{2} + a^{2})^{2}}$ ।

UPHESC Assistant Professor MathematicsMathematicsLaplace transformUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q78.

Euler-Lagrange equation

\frac{d}{d x} (\frac{\partial F}{\partial y^{'}}) - \frac{\partial F}{\partial y} = 0

is a

ऑयलर-लैग्रान्ज समीकरण

\frac{d}{d x} (\frac{\partial F}{\partial y^{'}}) - \frac{\partial F}{\partial y} = 0

है

AFirst order linear differential equationप्रथम कोटि का रेखीय अवकल समीकरण
BSecond order linear differential equationद्वितीय कोटि का रेखीय अवकल समीकरण
CSecond order non-linear differential equationद्वितीय कोटि का अरेखीय अवकल समीकरण
DThird order linear differential equationतृतीय कोटि का रेखीय अवकल समीकरण

Show answer

Concept: Expanding: $F_{y^{'} y^{'}} y^{″} + F_{y^{'} y} y^{'} + F_{y^{'} x} - F_{y} = 0$ .

Highest derivative is $y^{″}$ → second order.
The coefficients $F_{y^{'} y^{'}}$ etc. depend on $y, y^{'}$ in general → non-linear (e.g. $F = \sqrt{1 + y^{' 2}}$ ).
It is linear only for special (quadratic) $F$ .

Short trick: E–L: second order, in general non-linear.

अवधारणा: प्रसार: $F_{y^{'} y^{'}} y^{″} + F_{y^{'} y} y^{'} + F_{y^{'} x} - F_{y} = 0$ ।

उच्चतम अवकलज $y^{″}$ → द्वितीय कोटि।
गुणांक सामान्यतः $y, y^{'}$ पर निर्भर → अरेखीय।
केवल विशेष (द्विघाती) $F$ के लिए रेखीय।

याद रखें: ऑयलर–लैग्रांज: द्वितीय कोटि, सामान्यतः अरेखीय।

UPHESC Assistant Professor MathematicsMathematicsEuler–Lagrange equationUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q79.

The maximum distance between the curve y = x and

y = x^{2}

on the interval [0, 1] is

अंतराल [0, 1] पर वक्र y = x तथा

y = x^{2}

के बीच की अधिकतम दूरी है

A1
B1/2
C1/4
D1/6

Show answer

Concept: Distance between the curves measured vertically: $d (x) = x - x^{2}$ on [0,1].

$d^{'} (x) = 1 - 2 x = 0$ ⇒ $x = \frac{1}{2}$ ; $d^{″} = - 2 < 0$ → maximum.
$d (\frac{1}{2}) = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$ .
(1/6 is the area between the curves, a common distractor. The shortest perpendicular gap, $\frac{x - x^{2}}{\sqrt{2}}$ , would give $\frac{1}{4 \sqrt{2}}$ , which is not offered – the question means the vertical gap.)

Short trick: Max of $x - x^{2}$ is at the midpoint: $\frac{1}{4}$ .

अवधारणा: ऊर्ध्वाधर दूरी $d (x) = x - x^{2}$ , [0,1] पर।

$d^{'} (x) = 1 - 2 x = 0$ ⇒ $x = \frac{1}{2}$ , उच्चिष्ठ।
$d = \frac{1}{4}$ ।
(1/6 वक्रों के बीच का क्षेत्रफल है। लम्बवत दूरी $\frac{x - x^{2}}{\sqrt{2}}$ लें तो $\frac{1}{4 \sqrt{2}}$ आता, जो विकल्पों में नहीं है – यहाँ ऊर्ध्वाधर दूरी अभिप्रेत है।)

याद रखें: $x - x^{2}$ का अधिकतम $\frac{1}{4}$ ।

UPHESC Assistant Professor MathematicsMathematicsMaxima – distance between curvesUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q80.

The solution of the Linear Programming Problem (LPP)
Max

z = x_{1} + x_{2}

Subject to the constraints

x_{1} - x_{2} \geq 0

- 3 x_{1} + x_{2} \geq 3

x_{1}, x_{2} \geq 0

is/are

रैखिक प्रोग्रामन समस्या (LPP)
Max

z = x_{1} + x_{2}

प्रतिबन्धों के अधीन

x_{1} - x_{2} \geq 0

- 3 x_{1} + x_{2} \geq 3

x_{1}, x_{2} \geq 0

का समाधान है

AFeasibleव्यवहार्य
BBoundedपरिबद्ध
CInfinitely manyअनंत
DNone of the aboveउपर्युक्त में से कोई नहीं

Show answer

Concept: Check whether the constraints have any common point.

$x_{1} - x_{2} \geq 0$ ⇒ $x_{2} \leq x_{1}$ ; $- 3 x_{1} + x_{2} \geq 3$ ⇒ $x_{2} \geq 3 + 3 x_{1}$ .
Together $3 + 3 x_{1} \leq x_{1}$ ⇒ $x_{1} \leq - \frac{3}{2}$ , contradicting $x_{1} \geq 0$ .
Feasible region is empty → no feasible solution; none of (A)–(C) holds.

Short trick: Empty region ⇒ 'infeasible' ⇒ none of the above.

अवधारणा: देखें कि प्रतिबंधों का कोई उभयनिष्ठ बिंदु है या नहीं।

$x_{2} \leq x_{1}$ तथा $x_{2} \geq 3 + 3 x_{1}$ ।
अतः $x_{1} \leq - \frac{3}{2}$ , जो $x_{1} \geq 0$ के विरुद्ध है।
सुसंगत क्षेत्र रिक्त → कोई व्यवहार्य हल नहीं।

याद रखें: रिक्त क्षेत्र ⇒ अव्यवहार्य।

UPHESC Assistant Professor MathematicsMathematicsLPP – infeasibilityUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2