৪৯তম বিসিএস ⎯ ফলিত গণিত [৫৬১]

প্রশ্ন:


সমাধান:

প্রশ্ন: প্রশ্নবোধক স্থানে কোন সংখ্যাটি বসবে?

সমাধান:
(২য় কলাম × ৩য় কলাম) - ১ম কলাম = ৪র্থ কলাম

(6 × 10) - 2 = 60 - 2 = 58
(7 × 11) - 3 = 77 - 3 = 74
(8 × 12) - 4 = 96 - 4 = 92

সুতরাং, প্রশ্নবোধক স্থানে 92 সংখ্যাটি বসবে।

For general coplanar forces, equilibrium requires both translational equilibrium (ΣFx​=0, ΣFy​=0) and rotational equilibrium (ΣM=0).

Convention: anticlockwise = positive, clockwise = negative (though in vector form direction decides).

Distance between forces = 2r = 0.5 m.
Torque = Fd = 200.5=10 N·m.

Stable → small displacement → restoring force brings it back (ball in bowl).
Top of hill → unstable.
Cube edge → astatic/neutral (small disturbance may topple).
Cylinder on curved side → neutral/asstatic (may roll slowly).

No force acts to move it back or further → astatic (neutral) equilibrium.

3 forces → triangle of forces
4 forces → quadrilateral of forces
n forces → n-sided polygon of forces.
It is not necessary to form a Square or Parallelogram.

The virtual displacement causes potential energy to increase, so: ΔU>0
Using the relation δW=−ΔU:
δW=−(positive)=negative. Hence, virtual work is negative.
A) Positive: Wrong, Because Virtual work is positive if the displacement reduces potential energy, not increases.
C) Zero: Wrong, Because Only in neutral equilibrium, where potential energy does not change.
D) Cannot be determined: Wrong, Because Wrong; for conservative forces, virtual work is directly related to ΔU.

-4W∂(GA) - T∂(BD) = 0
The weight located upward related to the line. So The virtual work done by weight is negative.

For a uniform lamina, CG coincides with the centroid (intersection of medians).

Centroid along base of triangle: 1/3 from vertex opposite base. For a right triangle with right angle at rectangle top-right, the centroid along x-axis from left =7a/3
total areas: rectangle 2a², triangle a²/2.

Horizontal tension T_0​ is constant. Vertical tension increases with distance from the lowest point: T_y​=ws=T_0​sinh(x/c).

Equal distances in equal times → uniform motion → acceleration = 0.

At mean position, x=0x=0x=0, velocity is maximum. At extreme displacement, acceleration is maximum; velocity is zero, acceleration points toward mean.

Period slightly decreases for underdamped oscillations.

Polar coordinates describe position using radius (r) and angle (θ).
For circular or orbital motion, r is constant and only θ changes with time → very simple representation.

Central forces (like gravity) act along the radius, so writing motion in r,θ separates the radial dynamics from the angular motion. Conservation of angular momentum gives r²θ^˙= constant, so one can reduce the two-variable problem to a single radial equation (with an effective potential). Options B–D are false: accelerations aren’t constant (B), Cartesian can exactly describe curves (C), and polar doesn’t eliminate differential equations or time dependence of motion (D).

For circular orbits (e = 0), periapsis and apoapsis distances are equal, so the satellite’s distance from Earth remains constant.

Stokes’ law assumption:

F_d​=6πηrv is valid only when Reynolds number is very small Re≪1), i.e., very small spheres or very slow motion. In that regime:

Other options:

A: Incorrect, Stokes’ law does not give exact drag.
C: Incorrect, drag depends on radius (F∝r in Stokes, ∝r² in quadratic drag).
D: Incorrect, terminal velocity depends on viscosity; higher η → lower v_t​.

Spherical velocity has three terms: radial, polar (θ), and azimuthal (φ).

This is a standard derivative relation of Bessel functions.

For large x, Bessel functions oscillate with decreasing amplitude.

The negative sign in front of x² distinguishes modified Bessel’s equation from standard Bessel’s equation.

J₀​(x) starts at 1 for x=0 and oscillates with decreasing amplitude as x increases.

The two linearly independent solutions are the Legendre functions of first and second kind.

∣sinx∣ is an even function (∣=∣sinx∣), so the Fourier series contains cosine terms and the constant term.

Fourier series can be differentiated term by term if the function and its derivative are piecewise continuous. This is used to solve differential equations

(a) Sup norm → valid metric.
(b) Not a metric: can vanish for non-identical functions (e.g. f(x)=x,g(x)=1−x).
(c) L1L^1L1-metric, valid.
So (b) is correct.

• If $r<1$, then only $y=x$ satisfies $d(x,y)<r$.
  So the ball contains only the center point.

Finite sets have isolated points → not perfect.
[0,1] is closed and has no isolated points → perfect.
(0,1) not closed.
{1/n} has 0 as limit point but 0 not included → not closed.

Cantor set is uncountable (same cardinality as real numbers).
It is closed (constructed as nested closed sets).
Has Lebesgue measure 0 (total length removed = 1).
It is nowhere dense.

Alternating sign does not affect convergence to 0.

Even subsequence → 1+1/(2n)→1
Odd subsequence → −1+1/(2n−1)→−1
Others → all subsequences converge to same limit

(0)=f(π)=0 → satisfies f(a)=f(b)
f(x)f(x)f(x) is continuous & differentiable
Hence Rolle’s theorem applies

Non-differentiability occurs at corners, cusps, vertical tangents, or discontinuities.

Continuity means:

Function defined.
Limit exists.
Both equal.

(A) Not continuous at (0,0) (denominator = 0).
(B) Undefined on x=0. Not continuous everywhere.
(C) Undefined at (0,0). Not continuous everywhere.

• ​ f(z)=x2+y2x+iy→ continuous for all z≠0
  

• Compute derivative: check along multiple paths (x-axis, y-axis) → limit changes → derivative does not exist → nowhere differentiable

• x²−y²:u_xx​+u_yy​=2−2=0

• excos⁡y:uxx+uyy=excos⁡y−excos⁡y=0
  

singularity at z=2

Mobius transformations preserve cross ratios, which is a key invariant property.

Factor: f(z)=(z−i)3(z+i)3

We use Cauchy’s derivative formula:

f′(a)=12πi∫Cf(z)(z−a)2dz.

Here f(z)=ez, a=0. So

f′(0)=12πi∫∣z∣=1ezz2dz.

But by the formula, this integral equals the derivative at 0 directly:

f′(0)=e0=1.

• Near z=0, sin⁡z∼z → f(z)∼1/z.

• Finite power of z → pole of order 1.

• The only singularity is at $z=2$, which is outside the contour $|z|=1$.

• By Cauchy’s theorem, integral over a closed contour enclosing no singularities = 0.

• Expand 1/(z−1)=−1−z−z2−…
  

• Multiply by 1/z2 to get principal part: 1/z2−1/z+…
  

Integral: ∮Cezz3dz∮C​z3ez​dz

Step 1: Expand ezez

ez=1+z+z22!+…

Step 2: Divide by z3z3

ezz3=1z3+1z2+12z+…

Step 3: Residue = coefficient of 1/z = 1/2

Step 4: Apply residue theorem

∮Cezz3dz=2πi⋅12=πi

Option A: True for exact interpolation, but not a limitation per se.
Option B: Correct — adding a new point means you must rebuild the Lagrange polynomial.
Option C: False — Lagrange can interpolate any set of discrete points, linear or nonlinear.
Option D: False — Lagrange does not require derivatives.

• Newton-Raphson and Modified Newton require the derivative of the function.

• Halley’s method requires both the first and second derivatives.

• Secant method approximates the derivative using two previous points, so it does not require the explicit derivative.

  Modified Newton Method

  xn+1=xn−mf(xn)f′(xn)​

  • m is the multiplicity of the root (if known).

  • Still requires the first derivative, sometimes adjusted for multiple roots.

  Halley’s Method

  xn+1=xn−2f(xn)f′(xn)2(f′(xn))2−f(xn)f′′(xn)​

  • Requires first and second derivatives.

  • Cubic convergence (faster than Newton-Raphson).

First equation: x1=4−y0=4−0=4, second equation: y1=x1−2=4−2=2, then x1 updated again → average iteration, simple approximation gives x1=2

A) Incorrect — function values alone are not enough; need derivatives.
B) Correct — Jacobian contains all first-order partial derivatives for the system.
C) Hessian involves second derivatives; used in optimization, not standard Newton-Raphson for systems.
D) Determinant alone is insufficient; it’s part of the process when inverting the Jacobian

LU factorization saves work by reusing $L$ and $U$ when solving Ax=b1,Ax=b2,…

• Trapezoidal is exact only for linear functions.

• Simpson’s 1/3 Rule integrates polynomials of degree ≤2 exactly.

• Simpson’s 3/8 Rule integrates polynomials of degree ≤3 exactly.

• Therefore, both B and C are valid.

Global truncation error ∝ h, meaning the error decreases linearly with smaller step size h.

Formula:

yn+1=yn+hf(xn,yn)At x0=0, y0=1:

f(x0,y0)=02+2(1)=2So,

y1=y0+0.1×f(x0,y0)
y1=1+0.1(2)=1.2

y′(x)=x²+2y, y′′(x)=2x+2y′=2x+2(x²+2y)

At x=0, y=1:

y′(0)=2,  y′′(0)=4.Then

y(0.1)=1+0.1(2)+(0.1)22(4)=1+0.2+0.02=1.22

It guesses missing initial slopes and integrates the ODE as an Initial Value Problem.

Streamlines represent instantaneous velocity direction, so they change as the flow changes with time.

Explanation:

ωz=∂v∂x−∂u∂y=0−0=0 Hence flow is irrotational.
Velocity components are independent of time. So it is steady flow.

Stream function: ψ=x2+y2

Velocity components:

u=∂ψ∂y=2y,v=−∂ψ∂x=−2x,w=0Check incompressibility:
∂u/∂x+∂v/∂y+∂w/∂z=0+0+0=0

Check rotation (vorticity):
ωz=∂v/∂x−∂u/∂y=−2−2=−4≠0 → rotational

Flow type: Circular motion in x–y plane → solid-body rotation

Vortex line → tangent to local vorticity vector at each point.

• Conservative force → work done depends only on position, not path.

• Gravity is classic example: F⃗=−∇(ρgz)
  

• Kinetic energy: 12ρq2=0.5⋅1000⋅100=50 kPa
  

• Potential energy: ρgz=1000⋅10⋅5=50 kPa
  

• Total Bernoulli constant:

p+KE+PE=100+50+50=200 kPa

ψ=r2sin⁡θcos⁡θ, tangential velocity:

qθ=−∂ψ∂r=−2rsin⁡θcos⁡θ

Bernoulli’s principle: zero velocity → maximum pressure.

The complex velocity for a cylinder in uniform flow U∞​ and doublet μ=U∞a2 is:

$$SetV(z)=0\to U\infty -\mu /z2=0\to z2=a2\to z=\pm a$$

A doublet is a source-sink pair, net mass flux is zero → circulation around a closed curve enclosing it is zero (irrotational)

By Stokes’ theorem, the circulation around a closed curve equals the surface integral of vorticity over the enclosed area.
So, circulation measures the total rotation (vorticity) within that area.

Circulation = total rotation, Vorticity = local rotation of fluid particles.

A doublet (or dipole) is the limit of a source–sink pair of equal and opposite strengths +m and −m as the distance l→0, with ml=μ (the doublet strength) remaining finite and constant.