UPHESC Assistant Professor Mathematics · 100 questions

UPHESC Assistant Professor Mathematics 2025 PYQs

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100 questions.

Q41.

What is the interval in which the series

\sum_{n = 1}^{\infty} \frac{1}{1 + n^{2} x}

is uniformly convergent ?

वह कौन-सा अन्तराल है जिसमें श्रेणी

\sum_{n = 1}^{\infty} \frac{1}{1 + n^{2} x}

एकसमान रूप से अभिसारी है ?

A $(0, \frac{1}{4})$
B $(0, \frac{1}{2})$
C(0, 1)
D $[1, \infty)$

Show answer

Concept: Weierstrass M-test; near $x = 0$ the terms do not go to 0 uniformly.

On $[1, \infty)$ : $\frac{1}{1 + n^{2} x} \leq \frac{1}{n^{2}} = M_{n}$ , $\sum M_{n} < \infty$ → uniformly convergent.
On $(0, a)$ : $sup_{x} \frac{1}{1 + n^{2} x} = 1$ for each $n$ , so $u_{n} ↛ 0$ uniformly → not uniformly convergent.
Hence only $[1, \infty)$ works.

Short trick: Uniform convergence needs $sup | u_{n} | \to 0$ ; intervals touching 0 fail.

अवधारणा: वायरस्ट्रास M-परीक्षण; $x = 0$ के पास पद एकसमान रूप से 0 नहीं होते।

$[1, \infty)$ पर $\frac{1}{1 + n^{2} x} \leq \frac{1}{n^{2}}$ , $\sum \frac{1}{n^{2}}$ अभिसारी → एकसमान अभिसरण।
$(0, a)$ पर हर $n$ के लिए $sup = 1$ → एकसमान नहीं।
अतः केवल $[1, \infty)$ ।

याद रखें: 0 को छूने वाले अंतराल असफल होते हैं।

UPHESC Assistant Professor MathematicsMathematicsUniform convergence of seriesUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q42.

Arrange the following in a sequence.
I.

{a_{n}}

is a Cauchy sequence
II.

{a_{1} + a_{2} + \dots + a_{n}}

is convergent
III.

\sum_{n = 1}^{\infty} a_{n}

is convergent
IV.

lim_{n \to \infty} a_{n} = 0

Choose the correct answer from the code given below :
Code :

निम्नलिखित को एक अनुक्रम में व्यवस्थित कीजिए ।
I.

{a_{n}}

एक कौशी अनुक्रम है
II.

{a_{1} + a_{2} + \dots + a_{n}}

अभिसारी है
III.

\sum_{n = 1}^{\infty} a_{n}

अभिसारी है
IV.

lim_{n \to \infty} a_{n} = 0

नीचे दिए गए कूट में से सही उत्तर चुनिए :
कूट :

AI, II, IV, III
BI, IV, III, II
CII, III, IV, I
DI, III, II, IV

Show answer

Concept: A series converges by definition when its sequence of partial sums converges.

II (partial sums converge) ⇔ III ( $\sum a_{n}$ converges).
III ⇒ IV: $a_{n} = S_{n} - S_{n - 1} \to 0$ .
IV ⇒ I: $a_{n}$ converges (to 0), and every convergent sequence in $R$ is Cauchy. Chain: II, III, IV, I.

Short trick: Partial sums → series → $n$ th term → 0 → Cauchy.

अवधारणा: श्रेणी अभिसारी है जब उसके आंशिक योगों का अनुक्रम अभिसारी हो।

II ⇔ III।
III ⇒ IV: $a_{n} = S_{n} - S_{n - 1} \to 0$ ।
IV ⇒ I: $a_{n}$ अभिसारी है, अतः कौशी। क्रम: II, III, IV, I।

याद रखें: आंशिक योग → श्रेणी → $n$ वाँ पद → 0 → कौशी।

UPHESC Assistant Professor MathematicsMathematicsConvergence of series and sequencesUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q43.

The maximum value of

\sin x \cdot \sin y \cdot \sin (x + y)

is attained at

\sin x \cdot \sin y \cdot \sin (x + y)

का अधिकतम मान प्राप्त होता है

A $(\frac{π}{4}, \frac{π}{4})$ $(\frac{π}{4}, \frac{π}{4})$ पर
B $(\frac{π}{3}, \frac{π}{3})$ $(\frac{π}{3}, \frac{π}{3})$ पर
C $(\frac{2 π}{3}, \frac{2 π}{3})$ $(\frac{2 π}{3}, \frac{2 π}{3})$ पर
D $(\frac{3 π}{4}, \frac{3 π}{4})$ $(\frac{3 π}{4}, \frac{3 π}{4})$ पर

Show answer

Concept: Set $f_{x} = f_{y} = 0$ and use the second-derivative test.

$f_{x} = \sin y \sin (2 x + y) = 0$ , $f_{y} = \sin x \sin (x + 2 y) = 0$ → $2 x + y = π, x + 2 y = π$ → $x = y = π / 3$ .
$f (π / 3, π / 3) = {(\frac{\sqrt{3}}{2})}^{3} = \frac{3 \sqrt{3}}{8}$ (max; $r t - s^{2} > 0, r < 0$ ).
At $(2 π / 3, 2 π / 3)$ the value is $- \frac{3 \sqrt{3}}{8}$ (minimum).

Short trick: Symmetric function → try $x = y$ ; 'equilateral triangle' angles $π / 3$ give max.

अवधारणा: $f_{x} = f_{y} = 0$ रखकर द्वितीय अवकलज परीक्षण।

$2 x + y = π, x + 2 y = π$ → $x = y = π / 3$ ।
$f = {(\frac{\sqrt{3}}{2})}^{3} = \frac{3 \sqrt{3}}{8}$ (उच्चिष्ठ)।
$(2 π / 3, 2 π / 3)$ पर निम्निष्ठ $- \frac{3 \sqrt{3}}{8}$ ।

याद रखें: समबाहु त्रिभुज के कोण $π / 3$ पर अधिकतम।

UPHESC Assistant Professor MathematicsMathematicsMaxima of functions of two variablesUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q44.

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

\sum u_{n}

, where

u_{n} = \frac{3^{n}}{2^{n} + 3^{n}}

is divergent.
Reason (R) :

lim_{n \to \infty} u_{n} \neq 0

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

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

\sum u_{n}

, जहाँ

u_{n} = \frac{3^{n}}{2^{n} + 3^{n}}

, अपसारी है ।
तर्क (R) :

lim_{n \to \infty} u_{n} \neq 0

.
उपर्युक्त कथनों के आलोक में निम्नांकित विकल्पों में से सर्वाधिक उपयुक्त 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 $u_{n} ↛ 0$ then $\sum u_{n}$ diverges.

$u_{n} = \frac{1}{(2 / 3)^{n} + 1} \to 1 \neq 0$ → (R) true.
By the $n$ th-term test the series diverges → (A) true.
(R) is exactly the reason for (A).

Short trick: Divide by the dominant term $3^{n}$ .

अवधारणा: यदि $u_{n} ↛ 0$ तो $\sum u_{n}$ अपसारी।

$u_{n} = \frac{1}{(2 / 3)^{n} + 1} \to 1 \neq 0$ → (R) सत्य।
n-वाँ पद परीक्षण से श्रेणी अपसारी → (A) सत्य।
(R), (A) की सही व्याख्या है।

याद रखें: प्रभावी पद $3^{n}$ से भाग दें।

UPHESC Assistant Professor MathematicsMathematicsnth term test for divergenceUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q45.

The sequence

{f_{n}}

, where

f_{n} (x) = n x e^{- n x^{2}}

converges uniformly on

अनुक्रम

{f_{n}}

, जहाँ

f_{n} (x) = n x e^{- n x^{2}}

एक समान अभिसरण करेगा

A[0, 1][0, 1] पर
B[0, 2][0, 2] पर
C[0, 3][0, 3] पर
D[1, 2][1, 2] पर

Show answer

Concept: $f_{n} \to f$ uniformly ⇔ $M_{n} = sup | f_{n} - f | \to 0$ .

Pointwise limit is 0 for every $x \geq 0$ .
$f_{n}$ has its maximum at $x = 1 / \sqrt{2 n}$ with value $\sqrt{n / 2} e^{- 1 / 2} \to \infty$ ; this point lies in any interval starting at 0 → not uniform on [0,1], [0,2], [0,3].
On [1,2], $f_{n}$ is decreasing, so $M_{n} = n e^{- n} \to 0$ → uniform.

Short trick: A 'moving hump' near 0 kills uniform convergence; stay away from 0.

अवधारणा: एकसमान अभिसरण ⇔ $M_{n} = sup | f_{n} - f | \to 0$ ।

बिंदुवार सीमा 0 है।
$x = 1 / \sqrt{2 n}$ पर उच्चिष्ठ $\sqrt{n / 2} e^{- 1 / 2} \to \infty$ ; यह बिंदु 0 से शुरू हर अंतराल में है।
[1,2] पर $M_{n} = n e^{- n} \to 0$ → एकसमान।

याद रखें: 0 के पास 'चलता कूबड़' एकसमानता नष्ट करता है।

UPHESC Assistant Professor MathematicsMathematicsUniform convergence of sequencesUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q46.

The nullity of the linear transformation

T : V_{3} (F) \to V_{2} (F)

given by

T (a_{1}, a_{2}, a_{3}) = (a_{2}, a_{3})

T (a_{1}, a_{2}, a_{3}) = (a_{2}, a_{3})

द्वारा दिये गये रैखिक रूपान्तरण

T : V_{3} (F) \to V_{2} (F)

की शून्यता है

Show answer

Concept: Nullity = dim ker T; rank + nullity = dim domain.

$\ker T = {(a_{1}, 0, 0)}$ , spanned by $(1, 0, 0)$ .
Nullity $= 1$ (rank $= 2$ , $2 + 1 = 3$ ✓).

Short trick: Count the 'lost' coordinates: only $a_{1}$ is lost.

अवधारणा: शून्यता = $\ker T$ की विमा; कोटि + शून्यता = प्रांत की विमा।

$\ker T = {(a_{1}, 0, 0)}$ ।
शून्यता $= 1$ (कोटि 2)।

याद रखें: केवल $a_{1}$ 'खोता' है।

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

Q47.

If A =

[\begin{matrix} 1 & 2 & - 1 & 0 \\ - 1 & 3 & 0 & - 4 \\ 2 & 1 & 3 & - 2 \\ 1 & 1 & 1 & - 1 \end{matrix}]

, then the rank of the matrix A is

यदि A =

[\begin{matrix} 1 & 2 & - 1 & 0 \\ - 1 & 3 & 0 & - 4 \\ 2 & 1 & 3 & - 2 \\ 1 & 1 & 1 & - 1 \end{matrix}]

हो, तो आव्यूह A की कोटि है

Show answer

Concept: Row-reduce and count non-zero rows.

$R_{2} + R_{1}$ , $R_{3} - 2 R_{1}$ , $R_{4} - R_{1}$ give rows $(0, 5, - 1, - 4)$ , $(0, - 3, 5, - 2)$ , $(0, - 1, 2, - 1)$ .
Using $(0, - 1, 2, - 1)$ as pivot: $R_{2} + 5 R_{4} \to (0, 0, 9, - 9)$ and $R_{3} - 3 R_{4} \to (0, 0, - 1, 1)$ – proportional, so one row becomes zero.
Rank $= 3$ .

Short trick: First check $| A |$ : zero ⇒ rank < 4; then find one non-zero 3×3 minor.

अवधारणा: पंक्ति-संक्रिया से सोपान रूप बनाकर अशून्य पंक्तियाँ गिनें।

$R_{2} + R_{1}$ , $R_{3} - 2 R_{1}$ , $R_{4} - R_{1}$ के बाद तीन पंक्तियाँ बचती हैं।
$(0, - 1, 2, - 1)$ से: $(0, 0, 9, - 9)$ व $(0, 0, - 1, 1)$ समानुपाती – एक पंक्ति शून्य हो जाती है।
कोटि $= 3$ ।

याद रखें: पहले $| A |$ देखें: शून्य ⇒ कोटि < 4।

UPHESC Assistant Professor MathematicsMathematicsRank of a matrixUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q48.

If A =

[\begin{matrix} 1 & - 1 & 2 \\ 0 & 1 & 0 \\ 1 & 2 & 1 \end{matrix}]

, then the eigenvalues of A are

यदि A =

[\begin{matrix} 1 & - 1 & 2 \\ 0 & 1 & 0 \\ 1 & 2 & 1 \end{matrix}]

हो, तो A के अभिलक्षणिक मान है

A $1, 2, 2 \sqrt{2}$
B $1, 1 + \sqrt{2}, 1 - \sqrt{2}$
C1, 1, 1
D $1, 2 + \sqrt{2}, 2 - \sqrt{2}$

Show answer

Concept: Solve $| A - λ I | = 0$ ; expand along the row with zeros.

Expanding along row 2: $(1 - λ) [(1 - λ)^{2} - 2] = 0$ .
$λ = 1$ or $(1 - λ)^{2} = 2$ → $λ = 1 \pm \sqrt{2}$ .
Check: sum $= 3 = trace$ ✓.

Short trick: Trace test: eigenvalue sum must be 3 – only (B) and (C) pass; det $= - 1$ picks (B).

अवधारणा: $| A - λ I | = 0$ हल करें; शून्य वाली पंक्ति से प्रसार करें।

पंक्ति 2 से: $(1 - λ) [(1 - λ)^{2} - 2] = 0$ ।
$λ = 1, 1 \pm \sqrt{2}$ ।
जाँच: योग $= 3 =$ अनुरेख ✓।

याद रखें: अनुरेख व सारणिक से विकल्प जाँचें।

UPHESC Assistant Professor MathematicsMathematicsEigenvaluesUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q49.

The dimension of the subspace

W_{1} = {(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}) : x_{1} - x_{3} - x_{4} = 0}

R^{5}

R^{5}

की उपसमष्टि

W_{1} = {(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}) : x_{1} - x_{3} - x_{4} = 0}

की विमा है

Show answer

Concept: One independent linear equation in $R^{n}$ cuts the dimension by 1.

$x_{1} = x_{3} + x_{4}$ ; free variables $x_{2}, x_{3}, x_{4}, x_{5}$ .
dim $= 5 - 1 = 4$ .

Short trick: dim = n − (number of independent equations).

अवधारणा: $R^{n}$ में एक स्वतंत्र रैखिक समीकरण विमा 1 घटाता है।

$x_{1} = x_{3} + x_{4}$ ; स्वतंत्र चर $x_{2}, x_{3}, x_{4}, x_{5}$ ।
विमा $= 5 - 1 = 4$ ।

याद रखें: विमा = n − स्वतंत्र समीकरणों की संख्या।

UPHESC Assistant Professor MathematicsMathematicsDimension of a subspaceUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q50.

The number of non-singular linear transformations from

R^{5}

R^{3}

over

R

will be

R

पर

R^{5}

से

R^{3}

पर व्युत्क्रमणीय रैखिक रूपान्तरणों की संख्या होगी

Show answer

Concept: Non-singular (one-one) needs nullity 0, i.e. rank = dim domain.

Rank $\leq \dim R^{3} = 3$ .
Nullity $= 5 - rank \geq 2 > 0$ .
So no such map exists: 0.

Short trick: Can't inject a bigger space into a smaller one.

अवधारणा: व्युत्क्रमणीय (एकैकी) के लिए शून्यता 0 चाहिए।

कोटि $\leq 3$ ।
शून्यता $\geq 5 - 3 = 2 > 0$ ।
अतः संख्या 0।

याद रखें: बड़ी समष्टि को छोटी में एकैकी नहीं भेज सकते।

UPHESC Assistant Professor MathematicsMathematicsNon-singular linear mapsUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q51.

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

Q52.

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

Q53.

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

Q54.

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

Q55.

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

Q56.

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

Q57.

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

Q58.

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

Q59.

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

Q60.

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