UPHESC Assistant Professor Mathematics · 100 questions

UPHESC Assistant Professor Mathematics 2025 PYQs

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70 questions in Mathematics.

Q21.

The linear map

T : V_{3} \to V_{3}

defined by

T (e_{1}) = e_{1} - e_{2}

T (e_{2}) = 2 e_{2} + e_{3}

T (e_{3}) = e_{1} + e_{2} + e_{3}

, then the map T is

रैखिक प्रतिचित्रण

T : V_{3} \to V_{3}

T (e_{1}) = e_{1} - e_{2}

T (e_{2}) = 2 e_{2} + e_{3}

T (e_{3}) = e_{1} + e_{2} + e_{3}

द्वारा परिभाषित है, तो प्रतिचित्रण T है

AOne-one and ontoएकैकी तथा आच्छादक
BOne-one and intoएकैकी तथा अन्त:क्षेपी
CMany-one and ontoबहु-एकी तथा आच्छादक
DNeither one-one nor ontoन तो एकैकी और न ही आच्छादक

Show answer

Concept: For $T : V \to V$ (finite-dim), one-one ⇔ onto ⇔ $det [T] \neq 0$ .

Matrix (columns = images): $[\begin{matrix} 1 & 0 & 1 \\ - 1 & 2 & 1 \\ 0 & 1 & 1 \end{matrix}]$ .
$det = 1 (2 - 1) - 0 + 1 (- 1 - 0) = 0$ .
So T is singular: neither one-one nor onto (rank 2).

Short trick: Same dimension: either both one-one & onto, or neither.

अवधारणा: $T : V \to V$ के लिए एकैकी ⇔ आच्छादक ⇔ $det \neq 0$ ।

आव्यूह $[\begin{matrix} 1 & 0 & 1 \\ - 1 & 2 & 1 \\ 0 & 1 & 1 \end{matrix}]$ ।
$det = 1 - 1 = 0$ ।
अतः न एकैकी न आच्छादक।

याद रखें: समान विमा में दोनों गुण साथ-साथ।

UPHESC Assistant Professor MathematicsMathematicsLinear maps – injective/surjectiveUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q22.

α

and

β

are orthogonal unit vectors, then distance between

α

and

β

यदि

α

और

β

लम्बवत इकाई सदिश है, तो

α

और

β

के बीच की दूरी है

A2
B $\sqrt{2}$
C1
D0

Show answer

Concept: $d (α, β) = ∥ α - β ∥$ .

$∥ α - β ∥^{2} = ∥ α ∥^{2} + ∥ β ∥^{2} - 2 ⟨ α, β ⟩ = 1 + 1 - 0 = 2$ .
Distance $= \sqrt{2}$ .

Short trick: Pythagoras: diagonal of a unit square.

अवधारणा: $d (α, β) = ∥ α - β ∥$ ।

$∥ α - β ∥^{2} = 1 + 1 - 0 = 2$ ।
दूरी $= \sqrt{2}$ ।

याद रखें: इकाई वर्ग का विकर्ण।

UPHESC Assistant Professor MathematicsMathematicsInner product spacesUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q23.

The rank and nullity of linear transformation

T : R^{2} \to R^{3}

defined by

T (x_{1}, x_{2}) = (x_{1} + x_{2}, x_{1} - x_{2}, x_{2})

are respectively

T (x_{1}, x_{2}) = (x_{1} + x_{2}, x_{1} - x_{2}, x_{2})

से परिभाषित रैखिक रूपान्तरण

T : R^{2} \to R^{3}

की कोटि तथा शून्यता क्रमशः है

A3, 0
B0, 3
C0, 2
D2, 0

Show answer

Concept: rank + nullity = dim domain = 2.

$T (x) = 0$ ⇒ $x_{2} = 0$ and $x_{1} + x_{2} = 0$ ⇒ $x = 0$ , so nullity 0.
rank $= 2 - 0 = 2$ .

Short trick: Rank can never exceed dim of domain (2) – rules out (A).

अवधारणा: कोटि + शून्यता = प्रांत की विमा = 2।

$T (x) = 0$ ⇒ $x_{2} = 0, x_{1} = 0$ → शून्यता 0।
कोटि $= 2$ ।

याद रखें: कोटि प्रांत की विमा से अधिक नहीं हो सकती।

UPHESC Assistant Professor MathematicsMathematicsRank–nullityUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q24.

The particular integral of the differential equation

\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 5 y = e^{2 x} \sin x

अवकल समीकरण

\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 5 y = e^{2 x} \sin x

का विशिष्ट समाकल है

A $- \frac{1}{10} e^{2 x} \cos x$
B $- \frac{1}{10} e^{2 x} \sin x$
C $- \frac{1}{10} e^{2 x} (\cos x - 2 \sin x)$
D $- \frac{1}{10} e^{2 x} (\sin x - 2 \cos x)$

Show answer

Concept: $\frac{1}{f (D)} e^{a x} V = e^{a x} \frac{1}{f (D + a)} V$ .

$f (D + 2) = (D + 2)^{2} - 2 (D + 2) + 5 = D^{2} + 2 D + 5$ .
With $D^{2} = - 1$ : $\frac{1}{2 D + 4} \sin x = \frac{2 D - 4}{4 D^{2} - 16} \sin x = \frac{2 \cos x - 4 \sin x}{- 20}$ .
PI $= - \frac{1}{10} e^{2 x} (\cos x - 2 \sin x)$ .

Short trick: Shift $D \to D + a$ , then $D^{2} \to - b^{2}$ and rationalise.

अवधारणा: $\frac{1}{f (D)} e^{a x} V = e^{a x} \frac{1}{f (D + a)} V$ ।

$f (D + 2) = D^{2} + 2 D + 5$ ।
$D^{2} = - 1$ : $\frac{1}{2 D + 4} \sin x = \frac{2 \cos x - 4 \sin x}{- 20}$ ।
PI $= - \frac{1}{10} e^{2 x} (\cos x - 2 \sin x)$ ।

याद रखें: $D \to D + a$ , फिर $D^{2} \to - b^{2}$ ।

UPHESC Assistant Professor MathematicsMathematicsParticular integral (exponential shift)UPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q25.

A particular integral of the partial differential equation

\frac{\partial^{2} z}{\partial x^{2}} - \frac{\partial z}{\partial y} = 2 y - x^{2}

आंशिक अवकल समीकरण

\frac{\partial^{2} z}{\partial x^{2}} - \frac{\partial z}{\partial y} = 2 y - x^{2}

का एक विशिष्ट समाकल है

A $x y^{2}$
B $x^{2} y$
C $2 x^{2} y$
D $- x^{2} y$

Show answer

Concept: Simply substitute each option into the PDE.

$z = x^{2} y$ : $z_{x x} = 2 y$ , $z_{y} = x^{2}$ .
$z_{x x} - z_{y} = 2 y - x^{2}$ ✓.
$- x^{2} y$ gives $- 2 y + x^{2}$ ; $2 x^{2} y$ gives $4 y - 2 x^{2}$ ; $x y^{2}$ gives $- 2 x y$ .

Short trick: Verification is faster than the operator method.

अवधारणा: हर विकल्प को समीकरण में रखकर जाँचें।

$z = x^{2} y$ : $z_{x x} = 2 y$ , $z_{y} = x^{2}$ ।
$z_{x x} - z_{y} = 2 y - x^{2}$ ✓।
अन्य विकल्प मेल नहीं खाते।

याद रखें: प्रतिस्थापन विधि सबसे तेज़।

UPHESC Assistant Professor MathematicsMathematicsPDE – particular integralUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q26.

The initial value problem

x \frac{d y}{d x} = 4 y, y (0) = 2

has

प्रारम्भिक मान समस्या

x \frac{d y}{d x} = 4 y, y (0) = 2

का/के

Aa unique solutionएक अद्वितीय हल है
Bno solutionकोई हल नहीं है
Ctwo solutionsदो हल हैं
Dinfinite solutionsअनन्त हल हैं

Show answer

Concept: Solve the ODE and test the initial condition.

$\frac{d y}{y} = \frac{4 d x}{x}$ → $y = c x^{4}$ .
Every solution has $y (0) = 0$ , so $y (0) = 2$ is impossible.
(Picard's theorem fails as $f = 4 y / x$ is not defined at $x = 0$ .)

Short trick: Condition at a singular point: $y (0) = 0$ → infinitely many, $y (0) \neq 0$ → none.

अवधारणा: समीकरण हल करके प्रारंभिक शर्त जाँचें।

$y = c x^{4}$ ।
हर हल के लिए $y (0) = 0$ , अतः $y (0) = 2$ असंभव।
$x = 0$ पर $f = 4 y / x$ परिभाषित नहीं।

याद रखें: $y (0) = 0$ → अनंत हल, $y (0) \neq 0$ → कोई हल नहीं।

UPHESC Assistant Professor MathematicsMathematicsExistence of solutions of IVPUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q27.

A singular solution of the differential equation

{(\frac{d y}{d x})}^{3} - 4 x y \frac{d y}{d x} + 8 y^{2} = 0

अवकल समीकरण

{(\frac{d y}{d x})}^{3} - 4 x y \frac{d y}{d x} + 8 y^{2} = 0

का एक विशिष्ट हल है

A $27 y = 4 x^{2}$
B $9 y^{2} = 4 x^{3}$
C $3 y = 4 x^{2}$
D $27 y = 4 x^{3}$

Show answer

Concept: A singular solution must satisfy the ODE but contain no arbitrary constant.

Try $y = \frac{4 x^{3}}{27}$ : $p = \frac{4 x^{2}}{9}$ .
$p^{3} = \frac{64 x^{6}}{729}$ , $4 x y p = \frac{192 x^{6}}{729}$ , $8 y^{2} = \frac{128 x^{6}}{729}$ ; $64 - 192 + 128 = 0$ ✓.
(It comes from eliminating $p$ between the equation and its $p$ -discriminant; general solution $y = c (x - c)^{2}$ .)

Short trick: Quick check: substitute options – only $27 y = 4 x^{3}$ satisfies the ODE.

अवधारणा: विचित्र हल समीकरण को संतुष्ट करता है पर उसमें स्वेच्छ अचर नहीं होता।

$y = \frac{4 x^{3}}{27}$ , $p = \frac{4 x^{2}}{9}$ रखें।
$64 - 192 + 128 = 0$ ✓।
व्यापक हल $y = c (x - c)^{2}$ का आवरण।

याद रखें: विकल्प प्रतिस्थापित करें।

UPHESC Assistant Professor MathematicsMathematicsSingular solutionsUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q28.

The general solution of

\frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}} = 0

is of the form

\frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}} = 0

के व्यापक हल का रूप है

A $u = f (x + y) + g (x - y)$
B $u = f (x + i y) + g (x - i y)$
C $u = c f (x + i y)$
D $u = g (x + y)$

Show answer

Concept: For $(D^{2} + D^{' 2}) u = 0$ , auxiliary $m^{2} + 1 = 0$ ⇒ $m = \pm i$ .

Solution $u = f (y + i x) + g (y - i x)$ , equivalently $f (x + i y) + g (x - i y)$ .
(A) is the solution of the wave equation $u_{x x} - u_{y y} = 0$ .

Short trick: Laplace ↔ $x \pm i y$ ; wave ↔ $x \pm y$ .

अवधारणा: $(D^{2} + D^{' 2}) u = 0$ , सहायक समीकरण $m^{2} + 1 = 0$ ⇒ $m = \pm i$ ।

हल $u = f (x + i y) + g (x - i y)$ ।
(A) तरंग समीकरण का हल है।

याद रखें: लाप्लास ↔ $x \pm i y$ ।

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

Q29.

Match List – I with List – II.

List – I	List – II
a. $\frac{\partial u}{\partial t} = c^{2} \frac{\partial^{2} u}{\partial x^{2}}$	I. Elliptic
b. $\frac{\partial^{2} u}{\partial t^{2}} = c^{2} \frac{\partial^{2} u}{\partial x^{2}}$	II. Parabolic
c. $\frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}} = 0$	III. Hyperbolic
d. $Y \frac{\partial u}{\partial x} - X \frac{\partial u}{\partial y} = 0$	IV. $u = f (x^{2} + y^{2})$

Choose the correct answer from the code given below :

सूची – I को सूची – II के साथ सुमेलित करें ।

सूची – I	सूची – II
a. $\frac{\partial u}{\partial t} = c^{2} \frac{\partial^{2} u}{\partial x^{2}}$	I. दीर्घवृत्तीय
b. $\frac{\partial^{2} u}{\partial t^{2}} = c^{2} \frac{\partial^{2} u}{\partial x^{2}}$	II. परवलयिक
c. $\frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}} = 0$	III. अतिपरवलयिक
d. $Y \frac{\partial u}{\partial x} - X \frac{\partial u}{\partial y} = 0$	IV. $u = f (x^{2} + y^{2})$

नीचे दिए गए कूटों में से सही उत्तर चुने :

Aa – I, b – II, c – III, d – IV
Ba – I, b – III, c – IV, d – II
Ca – II, b – I, c – IV, d – III
Da – II, b – III, c – I, d – IV

Show answer

Concept: For $A u_{x x} + B u_{x y} + C u_{y y} + \dots$ : $B^{2} - 4 A C < 0$ elliptic, $= 0$ parabolic, $> 0$ hyperbolic.

Heat equation (a): $B^{2} - 4 A C = 0$ → parabolic (II).
Wave equation (b): $> 0$ → hyperbolic (III).
Laplace (c): $< 0$ → elliptic (I).
(d) Lagrange: $\frac{d x}{Y} = \frac{d y}{- X}$ gives (for $X = x, Y = y$ ) $x d x + y d y = 0$ → $u = f (x^{2} + y^{2})$ (IV).

Short trick: Heat–Parabolic, Wave–Hyperbolic, Laplace–Elliptic (H-P, W-H, L-E).

अवधारणा: $B^{2} - 4 A C < 0$ दीर्घवृत्तीय, $= 0$ परवलयिक, $> 0$ अतिपरवलयिक।

ऊष्मा समीकरण (a) → परवलयिक (II)।
तरंग समीकरण (b) → अतिपरवलयिक (III)।
लाप्लास (c) → दीर्घवृत्तीय (I)।
(d) लग्रांज विधि: $x d x + y d y = 0$ → $u = f (x^{2} + y^{2})$ (IV)।

याद रखें: ऊष्मा–परवलयिक, तरंग–अतिपरवलयिक, लाप्लास–दीर्घवृत्तीय।

UPHESC Assistant Professor MathematicsMathematicsClassification of PDEsUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q30.

The complete solution of

z = x p + y q + p^{2} + q^{2}

is, where

p = \frac{\partial z}{\partial x}, q = \frac{\partial z}{\partial y}

z = x p + y q + p^{2} + q^{2}

का पूर्ण हल, जहाँ

p = \frac{\partial z}{\partial x}, q = \frac{\partial z}{\partial y}

, है

A $z = a x + b y + a + b$
B $z = a x + b y + a^{2} + b^{2}$
C $z = x + y + a^{2} + b^{2}$
D $z = x + y + a + b$

Show answer

Concept: Clairaut's equation $z = p x + q y + f (p, q)$ has complete solution $z = a x + b y + f (a, b)$ .

Here $f (p, q) = p^{2} + q^{2}$ .
Replace $p \to a$ , $q \to b$ : $z = a x + b y + a^{2} + b^{2}$ .

Short trick: Clairaut: just put $p = a, q = b$ .

अवधारणा: क्लेरो समीकरण $z = p x + q y + f (p, q)$ का पूर्ण हल $z = a x + b y + f (a, b)$ ।

$f (p, q) = p^{2} + q^{2}$ ।
$p \to a, q \to b$ : $z = a x + b y + a^{2} + b^{2}$ ।

याद रखें: क्लेरो: $p = a, q = b$ रखें।

UPHESC Assistant Professor MathematicsMathematicsClairaut's form (PDE)UPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q31.

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

Q32.

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

Q33.

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

Q34.

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

Q35.

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

Q36.

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

Q37.

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

Q38.

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

Q39.

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

Q40.

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