UPHESC Assistant Professor Mathematics · 100 questions

UPHESC Assistant Professor Mathematics 2025 PYQs

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70 questions in Mathematics.

Q41.

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

Q42.

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

Q43.

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

Q44.

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

Q45.

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

Q46.

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

Q47.

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

Q48.

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

Q49.

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

Q50.

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

Q51.

Which of the following sets in

R^{2}

is not a convex set ?

R^{2}

में निम्नलिखित में से कौन समुच्चय उत्तल समुच्चय नहीं है ?

A ${(x_{1}, x_{2}) : x_{1}^{2} + x_{2}^{2} \leq 1}$
B ${(x_{1}, x_{2}) : x_{1}^{2} + x_{2}^{2} \geq 1, x_{1}^{2} + x_{2}^{2} \leq 4}$
C ${(x_{1}, x_{2}) : x_{1}^{2} + x_{2}^{2} \leq 4}$
D ${(x_{1}, x_{2}) : x_{1} \geq 0}$

Show answer

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

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

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

अवधारणा: समुच्चय उत्तल है यदि किन्हीं दो बिंदुओं को मिलाने वाला रेखाखंड उसी में हो।

चक्रिका (A), (C) व अर्धतल (D) उत्तल हैं।
(B) वलय है: $(1.5, 0)$ व $(- 1.5, 0)$ इसमें हैं पर मध्य बिंदु $(0, 0)$ नहीं।
अतः (B) उत्तल नहीं।

याद रखें: 'छेद' वाला समुच्चय उत्तल नहीं होता।

UPHESC Assistant Professor MathematicsMathematicsConvex setsUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q52.

Newton-Raphson formula of successive approximation to find the appropriate value of a root of the equation f(x) = 0 is

समीकरण f(x) = 0 के मूल का उपयुक्त मान प्राप्त करने के लिए क्रमिक सन्निकटन का न्यूटन-रैफ्सन का सूत्र है

A $x_{n + 1} = x_{n} - \frac{f (x_{n})}{f^{'} (x_{n})}$
B $x_{n + 1} = x_{n} - \frac{f^{'} (x_{n})}{f (x_{n})}$
C $x_{n + 1} = x_{n} + \frac{f (x_{n})}{f^{'} (x_{n})}$
D $x_{n + 1} = x_{n} + \frac{f^{'} (x_{n})}{f (x_{n})}$

Show answer

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

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

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

अवधारणा: $(x_{n}, f (x_{n}))$ पर स्पर्शरेखा x-अक्ष को जहाँ काटे वह अगला सन्निकटन।

स्पर्शरेखा: $y - f (x_{n}) = f^{'} (x_{n}) (x - x_{n})$ ।
$y = 0$ : $x_{n + 1} = x_{n} - \frac{f (x_{n})}{f^{'} (x_{n})}$ ।

याद रखें: 'घटा f बटा f-डैश'।

UPHESC Assistant Professor MathematicsMathematicsNewton–Raphson methodUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q53.

The maximum value of

z = 5 x_{1} + 7 x_{2}

Subject to constraints

x_{1} + x_{2} \leq 4,

3 x_{1} + 4 x_{2} \leq 24,

10 x_{1} + 7 x_{2} \leq 35,

and

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

, is

प्रतिबन्धों

x_{1} + x_{2} \leq 4,

3 x_{1} + 4 x_{2} \leq 24,

10 x_{1} + 7 x_{2} \leq 35,

और

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

,
के अधीन

z = 5 x_{1} + 7 x_{2}

का महत्तम मान है

A0
B17.5
C28
D34.8

Show answer

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

$3 x_{1} + 4 x_{2} \leq 24$ is redundant (never binding inside the others).
Corners: $(0, 0)$ , $(3.5, 0)$ , $(\frac{7}{3}, \frac{5}{3})$ , $(0, 4)$ .
$z$ : 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.

अवधारणा: सुसंगत क्षेत्र के कोनों पर $z$ का मान निकालें।

$3 x_{1} + 4 x_{2} \leq 24$ अनावश्यक है।
कोने: $(0, 0)$ , $(3.5, 0)$ , $(\frac{7}{3}, \frac{5}{3})$ , $(0, 4)$ ।
$z$ : 0, 17.5, 23.3, 28 → अधिकतम 28।

याद रखें: सबसे बड़ा गुणांक $x_{2}$ का – पहले $x_{2}$ -अक्ष वाला कोना देखें।

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

Q54.

The Linear Programming Problem (LPP)
Max

z = 2 x_{1} - x_{2}

Subject to constraints

x_{1} - x_{2} \leq 1,

2 x_{1} + x_{2} \geq 6,

and

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

, has

रैखिक प्रोग्रामन समस्या (LPP)
Max

z = 2 x_{1} - x_{2}

प्रतिबन्धों के अधीन

x_{1} - x_{2} \leq 1,

2 x_{1} + x_{2} \geq 6,

और

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

, का

ANo solutionकोई हल नहीं है
BUnbounded solutionअपरिबद्ध हल है
CSeveral alternative optimal solutionsकई वैकल्पिक इष्ट हल हैं
DOnly one optimal solutionकेवल एक इष्टतम हल हैं

Show answer

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

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

Short trick: $z = (x_{1} - x_{2}) + x_{1} \leq 1 + x_{1}$ – and $x_{1}$ is free to grow.

अवधारणा: सुसंगत क्षेत्र में ऐसी दिशा देखें जिसमें $z$ असीमित बढ़े।

क्षेत्र अरिक्त व अपरिबद्ध है।
रेखा $x_{2} = x_{1} - 1$ पर $z = x_{1} + 1 \to \infty$ ।
अतः अपरिबद्ध हल।

याद रखें: $z = (x_{1} - x_{2}) + x_{1}$ , और $x_{1}$ असीमित बढ़ सकता है।

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

Q55.

The third divided difference of the function

\frac{1}{x}

for the points a, b, c, d is

बिन्दुओं a, b, c, d के लिए फलन

\frac{1}{x}

का तृतीय विभाजित अन्तर है

A $\frac{a b c + a b d + a c d + b c d}{a^{2} b^{2} c^{2} d^{2}}$
B $- (\frac{a b c + a b d + a c d + b c d}{a^{2} b^{2} c^{2} d^{2}})$
C $\frac{1}{a b c d}$
D $- \frac{1}{a b c d}$

Show answer

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

$f [a, b] = \frac{1 / b - 1 / a}{b - a} = - \frac{1}{a b}$ .
$f [a, b, c] = \frac{- \frac{1}{b c} + \frac{1}{a b}}{c - a} = \frac{1}{a b c}$ .
$f [a, b, c, d] = \frac{\frac{1}{b c d} - \frac{1}{a b c}}{d - a} = - \frac{1}{a b c d}$ .

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

अवधारणा: $f [a, b] = \frac{f (b) - f (a)}{b - a}$ , पुनरावृत्ति से।

$f [a, b] = - \frac{1}{a b}$ ।
$f [a, b, c] = \frac{1}{a b c}$ ।
$f [a, b, c, d] = - \frac{1}{a b c d}$ ।

याद रखें: $1 / x$ का $k$ वाँ विभाजित अंतर $= (- 1)^{k} /$ (सभी बिंदुओं का गुणनफल)।

UPHESC Assistant Professor MathematicsMathematicsDivided differencesUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q56.

By applying the Lagrange’s interpolation formula in the following table

x	0	1	2	5
y	2	3	12	147

The polynomial approximating to y(x) is

निम्नलिखित सारणी में लैग्रांज अंतर्वेशन सूत्र का प्रयोग करने के पश्चात

x	0	1	2	5
y	2	3	12	147

y(x) का प्राप्त निकटतम बहुपद है

A $x^{3} + x^{2} - x + 2$
B $x^{3} + x^{2} + x + 2$
C $x^{3} - x^{2} - x + 2$
D $x^{3} - x^{2} + x - 2$

Show answer

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

$y (0) = 2$ rules out (D).
$y (1) = 3$ : (A) $1 + 1 - 1 + 2 = 3$ ✓, (B) 5 ✗, (C) 1 ✗.
(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.

अवधारणा: 4 बिंदुओं से जाने वाला त्रिघाती बहुपद अद्वितीय है – विकल्पों की जाँच करें।

$y (0) = 2$ से (D) बाहर।
$y (1) = 3$ केवल (A) देता है।
(A): $y (2) = 12$ , $y (5) = 147$ ✓।

याद रखें: अद्वितीयता ⇒ विकल्प जाँचें।

UPHESC Assistant Professor MathematicsMathematicsLagrange interpolationUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q57.

The rate of convergence of Newton-Raphson method is

न्यूटन-रेफ्सन विधि के अभिसरण की दर है

Show answer

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

$ε_{n + 1} \approx \frac{f^{″} (α)}{2 f^{'} (α)} ε_{n}^{2}$ .
Order $= 2$ (bisection 1, secant ≈1.618, regula falsi 1).

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

अवधारणा: सरल मूल के लिए न्यूटन–रैफ्सन का अभिसरण द्विघाती है।

$ε_{n + 1} \approx \frac{f^{″}}{2 f^{'}} ε_{n}^{2}$ ।
कोटि 2 (द्विभाजन 1, छेदिका ≈1.618)।

याद रखें: हर चरण में सही अंक लगभग दोगुने।

UPHESC Assistant Professor MathematicsMathematicsOrder of convergenceUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q58.

The number of twin prime pairs between 51 and 100 is

51 तथा 100 के बीच जुड़वा अभाज्य युग्म की संख्या है

Show answer

Concept: Twin primes differ by 2.

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

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

अवधारणा: जुड़वाँ अभाज्य में अंतर 2 होता है।

51–100 के अभाज्य: 53, 59, 61, 67, 71, 73, 79, 83, 89, 97।
युग्म: (59, 61), (71, 73)।
अतः 2।

याद रखें: 3 से बड़े जुड़वाँ अभाज्य $(6 k - 1, 6 k + 1)$ रूप के।

UPHESC Assistant Professor MathematicsMathematicsTwin primesUPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q59.

The sum of all positive divisors of 180 is

180 के सभी धनात्मक विभाजकों का योग है

A58
B546
C646
D746

Show answer

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

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

Short trick: Factorise, then multiply the geometric sums.

अवधारणा: $σ$ गुणात्मक है: $σ (p^{k}) = 1 + p + \dots + p^{k}$ ।

$180 = 2^{2} \cdot 3^{2} \cdot 5$ ।
$σ = 7 \cdot 13 \cdot 6 = 546$ ।

याद रखें: गुणनखंड करें, फिर गुणोत्तर योगों का गुणन।

UPHESC Assistant Professor MathematicsMathematicsSum of divisors σ(n)UPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2

Q60.

The remainder when

2^{340}

is divided by 341, is

2^{340}

को 341 से भाग देने पर शेषफल है

Show answer

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

$2^{10} = 1024 = 3 \cdot 341 + 1$ ⇒ $2^{10} \equiv 1 (\mod 341)$ .
$2^{340} = (2^{10})^{34} \equiv 1$ .
(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.

अवधारणा: 2 की ऐसी छोटी घात खोजें जो 341 मॉड्यूलो 1 हो।

$2^{10} = 1024 = 3 \cdot 341 + 1$ ।
$2^{340} = (2^{10})^{34} \equiv 1$ ।
(341 = 11·31 आधार 2 का सबसे छोटा छद्म-अभाज्य।)

याद रखें: $2^{10} \equiv 1 (\mod 341)$ ।

UPHESC Assistant Professor MathematicsMathematicsModular arithmetic (pseudoprime 341)UPHESC Assistant Professor Mathematics 2025 · 16 Apr Shift 2