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৪৯তম বিসিএস ⎯ ফলিত গণিত [৫৬১]

পরীক্ষা৪৯তম বিসিএস ⎯ ফলিত গণিত [৫৬১]তারিখতারিখ অনির্ধারিতসময়45 minutes৩৯ বৈধ · অসম্পূর্ণ
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Exam - 11 Topics: Numerical Analysis (a) Solution of algebraic and transcendental equation, Interpolation. (b) Numerical solution of linear and non-linear system of equations. [Source: Class - 08 and Relevant Books]
ঘনত্ব
উত্তর
উত্তরিতবর্তমানপুনরায় দেখুনঅসম্পূর্ণ

৪৯তম বিসিএস ⎯ ফলিত গণিত [৫৬১]

৪৯তম বিসিএস ⎯ ফলিত গণিত [৫৬১] · তারিখ অনির্ধারিত · ৪০ প্রশ্ন

.
The equation ex − 3x = 0 is an example of:
  1. Linear equation
  2. Algebraic equation
  3. Transcendental equation
  4. Quadratic equation
সঠিক উত্তর:
Transcendental equation
উত্তর
সঠিক উত্তর:
Transcendental equation
ব্যাখ্যা

.
In the bisection method, the condition for applying the method on interval [a,b] is:
  1. f(a) = f(b)
  2. f(a)f(b) > 0
  3. f(a) + f(b) = 0
  4. f(a)f(b) < 0
সঠিক উত্তর:
f(a)f(b) < 0
উত্তর
সঠিক উত্তর:
f(a)f(b) < 0
ব্যাখ্যা

The Intermediate Value Theorem says: if f(x) is continuous and f(a), f(b) have opposite signs, then at least one root lies in [a,b].
If f(a)f(b)>0, both values are positive or negative, so no root guaranteed.
Thus, for bisection, we must have sign change → root inside.

.
Newton-Raphson formula is:
    সঠিক উত্তর:
    উত্তর
    সঠিক উত্তর:
    ব্যাখ্যা

    .
    If f(a)f(b) < 0, the next approximation xr is computed. If f(a)f(xr) < 0, the new interval becomes:
    1. [xr,b]
    2. [a,xr]
    3. [a,b]
    4. None
    সঠিক উত্তর:
    [a,xr]
    উত্তর
    সঠিক উত্তর:
    [a,xr]
    ব্যাখ্যা

    A root always lies where the function changes sign.
    If f(a) and f(xr) have opposite signs, then the root is in [a,xr]
    Otherwise, the root lies in [xr,b]
    So the correct updated interval depends on sign check, here it is [a,xr].

    .
    Which of the following is true for the False Position method?
    1. Requires derivative of f(x)
    2. Always halves the interval like bisection
    3. Root is approximated by linear interpolation
    4. Only works for polynomial equations
    সঠিক উত্তর:
    Root is approximated by linear interpolation
    উত্তর
    সঠিক উত্তর:
    Root is approximated by linear interpolation
    ব্যাখ্যা

    The method does not need derivatives → (a) is false.
    Unlike bisection, the interval is not always halved → (b) is false.
    It works for any continuous function with sign change, not just polynomials → (d) is false.
    Correct → (c): The root is approximated by drawing a straight line between endpoints and finding where it cuts the x-axis.

    .
    The bisection method is guaranteed to converge if:
    1. f(x) is differentiable in [a, b]
    2. f(x) is discontinuous in [a, b]
    3. f(x) is continuous in [a, b] and f(a)f(b) < 0
    4. In [a, b], f(a)f(b) < 0
    সঠিক উত্তর:
    f(x) is continuous in [a, b] and f(a)f(b) < 0
    উত্তর
    সঠিক উত্তর:
    f(x) is continuous in [a, b] and f(a)f(b) < 0
    ব্যাখ্যা

    Convergence is guaranteed by the Intermediate Value Theorem, which requires continuity and a sign change. Differentiability is not necessary.

    .
    The convergence rate of the bisection method is:
    1. Quadratic
    2. Super-linear
    3. Exponential
    4. Linear
    সঠিক উত্তর:
    Linear
    উত্তর
    সঠিক উত্তর:
    Linear
    ব্যাখ্যা

    The error reduces by half in each step, so convergence is linear (slow) compared to Newton-Raphson (quadratic).

    .
    Which method generally converges fastest?
    1. Bisection
    2. Regula-Falsi
    3. Newton-Raphson
    4. Secant
    সঠিক উত্তর:
    Newton-Raphson
    উত্তর
    সঠিক উত্তর:
    Newton-Raphson
    ব্যাখ্যা

    Bisection and Regula-Falsi have linear convergence (slow).
    Secant is faster, with super-linear convergence.
    Newton-Raphson has quadratic convergence (error reduces very fast, squared each step).
    Thus, Newton-Raphson is usually the fastest (if derivative exists & initial guess is good).

    .
    Interpolation is used to:
    1. Find values beyond given data range
    2. Estimate values between given data points
    3. Solve differential equations
    4. Approximate integrals
    সঠিক উত্তর:
    Estimate values between given data points
    উত্তর
    সঠিক উত্তর:
    Estimate values between given data points
    ব্যাখ্যা

    Interpolation = estimating unknown values within the range of given data.
    Extrapolation = predicting outside the given range.
    Example: If data is given at x=1,2,3, interpolation estimates value at x=2.5

    ১০.
    Newton’s forward interpolation is best applied when:
    1. Data points are not equally spaced
    2. x is near the end of the data
    3. x is near the beginning of the data
    4. None of these
    সঠিক উত্তর:
    x is near the beginning of the data
    উত্তর
    সঠিক উত্তর:
    x is near the beginning of the data
    ব্যাখ্যা

    Newton’s forward formula uses forward difference table starting at x0.
    It works well when x is close to the starting point.
    For data near the end, backward interpolation is better.

    ১১.
    Newton–Raphson method is applied to solve f(x) = x3 - 2x - 5 = 0. Starting with x0 = 2, the next approximation x1 is:
    1. 1.9
    2. 2.1
    3. 31/12
    4. None of these
    সঠিক উত্তর:
    2.1
    উত্তর
    সঠিক উত্তর:
    2.1
    ব্যাখ্যা

    ১২.
    A major disadvantage of Newton-Raphson method is:
    1. Always slower than bisection
    2. Needs function to be continuous
    3. May fail if f′(x) = 0 near the roo
    4. Cannot be used for polynomials
    সঠিক উত্তর:
    May fail if f′(x) = 0 near the roo
    উত্তর
    সঠিক উত্তর:
    May fail if f′(x) = 0 near the roo
    ব্যাখ্যা

    (a) is wrong: Newton–Raphson is usually faster than bisection.
    (b) is too weak: continuity alone is not enough, it also needs a derivative.
    (d) is wrong: it can solve polynomials easily.
    Correct → (c): If the derivative is zero or very small, denominator blows up → formula fails or gives poor results.

    ১৩.
    Newton–Raphson method is best described as:
    1. An open method
    2. A bracketing method
    3. A direct method
    4. A graphical method only
    সঠিক উত্তর:
    An open method
    উত্তর
    সঠিক উত্তর:
    An open method
    ব্যাখ্যা

    Bracketing methods (like bisection/false position) require an interval.
    Newton–Raphson only needs an initial guess, not an interval → it is an open method.
    It is not “direct” since it’s iterative.
    Though based on tangent geometry, it is more than a graphical method.

    ১৪.
    Given data:
    x : 1,  2,  3 f(x) : 1,  8,  27
    Estimate f(1.5) using Newton’s Forward Interpolation.
    1. 2.5
    2. 2.25
    3. 4.5
    4. 3.375
    সঠিক উত্তর:
    3.375
    উত্তর
    সঠিক উত্তর:
    3.375
    ব্যাখ্যা

    Here f(x) = x3 At x = 1.5, exact value is 1.5 = 3.3751.
    Forward difference method gives same result.

    ১৫.
    Which interpolation method requires the least number of arithmetic operations if only 3 data points are given?
    1. Lagrange interpolation
    2. Newton forward
    3. Newton backward
    4. All are same
    সঠিক উত্তর:
    All are same
    উত্তর
    সঠিক উত্তর:
    All are same
    ব্যাখ্যা

    With 3 points, polynomial is quadratic. Any method produces the same quadratic with almost equal effort.

    ১৬.
    If f(x) = sin⁡(x) is tabulated at x = 0°, 30°, 60°, then interpolation at 45° will give:
    1. Exact value of sin⁡45°
    2. Approximation
    3. Error-free only if more points are used
    4. Zero value
    সঠিক উত্তর:
    Approximation
    উত্তর
    সঠিক উত্তর:
    Approximation
    ব্যাখ্যা

    If the original function itself is a polynomial of degree ≤ (number of points – 1), interpolation will be exact.
    Since sine is not a polynomial, the interpolation will only be an approximation.

    ১৭.
    Apply one Newton–Raphson iteration to f(x) = x2 - 4 with x0 = 3. The next approximation is:
    1. 21/5
    2. 23/6
    3. 13/6
    4. 9/5
    সঠিক উত্তর:
    13/6
    উত্তর
    সঠিক উত্তর:
    13/6
    ব্যাখ্যা

    ১৮.
    If the data points are not equally spaced, which interpolation method should be used?
    1. Newton forward
    2. Newton backward
    3. Lagrange
    4. All of these
    সঠিক উত্তর:
    Lagrange
    উত্তর
    সঠিক উত্তর:
    Lagrange
    ব্যাখ্যা

    Both forward and backward formulas require equal spacing. Lagrange works for unequal spacing.

    ১৯.
    For equally spaced data points, if we want to interpolate the value of f(x) near the middle of the table, the most suitable method is:
    1. Newton’s Forward Interpolation
    2. Newton’s Backward Interpolation
    3. Gauss’s Central Difference Formula
    4. Lagrange Interpolation
    সঠিক উত্তর:
    Gauss’s Central Difference Formula
    উত্তর
    সঠিক উত্তর:
    Gauss’s Central Difference Formula
    ব্যাখ্যা

    Central difference interpolation (Gauss forward/backward) is designed for approximations near the middle.

    ২০.
    Which of the following can cause Newton–Raphson to fail?
    1. Poor initial guess
    2. f’(x) 0 near the root
    3. Function not differentiable
    4. All of the above
    সঠিক উত্তর:
    All of the above
    উত্তর
    সঠিক উত্তর:
    All of the above
    ব্যাখ্যা

    Poor guess may cause divergence.
    If derivative is very small, denominator becomes unstable.
    If f(x) is not differentiable, method cannot be applied.
    So all reasons can cause failure.

    ২১.
    For equally spaced data, which interpolation method is best when the required value of xxx is near the beginning of the table?
    1. Newton’s Forward Interpolation
    2. Newton’s Backward Interpolation
    3. Lagrange’s Interpolation
    4. Gauss Backward
    সঠিক উত্তর:
    Newton’s Forward Interpolation
    উত্তর
    সঠিক উত্তর:
    Newton’s Forward Interpolation
    ব্যাখ্যা

    Forward interpolation uses forward differences, suitable when xxx is near the first tabulated point.

    ২২.
    For equally spaced data, Newton’s forward and backward interpolation formulas give:
    1. Different interpolating polynomials
    2. Same interpolating polynomial
    3. Approximate but not exact values
    4. None of these
    সঠিক উত্তর:
    Same interpolating polynomial
    উত্তর
    সঠিক উত্তর:
    Same interpolating polynomial
    ব্যাখ্যা

    Both are just different forms of writing the same unique polynomial through the data points
    Though the formulas look different, they are simply two different ways of constructing the same polynomial.
    Whether you start expanding from the beginning (forward) or from the end (backward), you eventually arrive at the same interpolating polynomial that uniquely fits all points.

    ২৩.
    In Newton’s forward formula, if h = 1,  x0 = 2,  f(x0) = 4,  Δf(x0) = 3,  Δ2f(x0) = 2, find f(2.5).
    1. 5.75
    2. 6
    3. 6.25
    4. 6.5
    অনির্ধারিত
    ব্যাখ্যা

    সঠিক উত্তর: 5.25
    অপশনে সঠিক উত্তর না থাকায় প্রশ্নটি বাতিল করা হলো। 
    ---------- 

    ২৪.
    x : 0, 1, 2 f(x) : 1, 3, 7
    Find f(1.5) using Lagrange interpolation.
    1. 4.25
    2. 4.5
    3. 4.75
    4. 5
    সঠিক উত্তর:
    4.75
    উত্তর
    সঠিক উত্তর:
    4.75
    ব্যাখ্যা

    Polynomial fitting points is f(x)=x2+x+1.
    At x=1.5:  f(1.5)=2.25+1.5+1=4.75

    ২৫.
    Which method gives the exact solution of a linear system in a finite number of steps (ignoring round-off error)?
    1. Jacobi Iteration
    2. Gauss–Seidel Iteration
    3. Gaussian Elimination
    4. Successive Substitution
    সঠিক উত্তর:
    Gaussian Elimination
    উত্তর
    সঠিক উত্তর:
    Gaussian Elimination
    ব্যাখ্যা

    Iterative methods converge gradually, but Gaussian elimination (a direct method) solves the system exactly in finite steps.

    ২৬.
    In Jacobi iteration, the value of each variable at the new iteration is computed using:
    1. Updated values from the same iteration
    2. Values from the previous iteration only
    3. Random approximations
    4. Determinant-based formula
    সঠিক উত্তর:
    Values from the previous iteration only
    উত্তর
    সঠিক উত্তর:
    Values from the previous iteration only
    ব্যাখ্যা

    Jacobi updates all variables simultaneously using only the previous iteration’s values, unlike Gauss–Seidel which reuses updated values.

    ২৭.
    Which condition guarantees convergence of both Jacobi and Gauss–Seidel methods?
    1. Positive determinant of matrix
    2. Diagonal dominance
    3. Upper triangular form
    4. Lower triangular form
    সঠিক উত্তর:
    Diagonal dominance
    উত্তর
    সঠিক উত্তর:
    Diagonal dominance
    ব্যাখ্যা

     If the coefficient matrix is diagonally dominant (∣aii​∣>∑∣aij​∣), iterative methods converge reliably.

    ২৮.
    In multivariable Newton–Raphson, the update formula is
    1. X(k + 1) = X(k) F(X(k)
    2. X(k + 1) = J(X(k)).X(k)
    3. X(k + 1) = X(k)- J-1(X(k))F(X(k))
    4. X(k + 1) = X(k) + F(X(k))
    সঠিক উত্তর:
    X(k + 1) = X(k)- J-1(X(k))F(X(k))
    উত্তর
    সঠিক উত্তর:
    X(k + 1) = X(k)- J-1(X(k))F(X(k))
    ব্যাখ্যা

    ২৯.
    Which of the following is a disadvantage of Newton–Raphson for nonlinear systems?
    1. Very slow convergence
    2. Requires derivatives 
    3. Cannot be used for linear systems
    4. Always divergent
    সঠিক উত্তর:
    Requires derivatives 
    উত্তর
    সঠিক উত্তর:
    Requires derivatives 
    ব্যাখ্যা

    ৩০.
    Which of the following is not a direct method for solving linear systems?
    1. Gaussian elimination
    2. LU decomposition
    3. Gauss–Jordan elimination
    4. Jacobi iteration
    সঠিক উত্তর:
    Jacobi iteration
    উত্তর
    সঠিক উত্তর:
    Jacobi iteration
    ব্যাখ্যা

    Direct methods (Gaussian, LU, Gauss–Jordan) give the exact solution in finite steps. Jacobi is iterative.

    ৩১.
    For the system: 2x + y  = 5,  x − y = 1
    Solve using Cramer’s Rule.
    1. x=2, y=1
    2. x=2.05, y=0.95
    3. x=1, y=2
    4. x=2, y=3
    সঠিক উত্তর:
    x=2, y=1
    উত্তর
    সঠিক উত্তর:
    x=2, y=1
    ব্যাখ্যা

    All Exact Method give same result of a linear system. So, no need to solve it to the given method. Just solve it in easy way.  

    ৩২.
    Which method is generally fastest for large sparse linear systems?
    1. Cramer’s rule
    2. Gauss–Jordan elimination
    3. Iterative methods
    4. Direct substitution
    সঠিক উত্তর:
    Iterative methods
    উত্তর
    সঠিক উত্তর:
    Iterative methods
    ব্যাখ্যা

    Sparse systems benefit from iterative approaches because they avoid heavy matrix factorizations.

    ৩৩.
    For the system: f1(x, y) = x2 + y − 3 = 0, f2(x, y) = x + y2 − 3 = 0 The Jacobian matrix is:
      সঠিক উত্তর:
      উত্তর
      সঠিক উত্তর:
      ব্যাখ্যা

      ৩৪.
      If Newton–Raphson method is applied to nonlinear systems, convergence is usually (near root):
      1. Linear
      2. Quadratic
      3. Cubic
      4. None
      সঠিক উত্তর:
      Quadratic
      উত্তর
      সঠিক উত্তর:
      Quadratic
      ব্যাখ্যা

      Near the root, Newton–Raphson converges quadratically (error roughly squares each step).

      ৩৫.
      Apply one Jacobi iteration to the system: 3x + y = 9,  x + 2y = 8 starting from x0 = 0,  y0 = 0.
      1. x1=3, y1=4
      2. x1=2.5, y1=3.5
      3. x1=3.5, y1=3.5
      4. x1=2.5, y1=4.5
      সঠিক উত্তর:
      x1=3, y1=4
      উত্তর
      সঠিক উত্তর:
      x1=3, y1=4
      ব্যাখ্যা

      ৩৬.
      In Gauss–Seidel method, when updating variables:
      1. Only old values are used
      2. Only new values are used
      3. No iteration is performed
      4. New values are used as soon as they are available
      সঠিক উত্তর:
      New values are used as soon as they are available
      উত্তর
      সঠিক উত্তর:
      New values are used as soon as they are available
      ব্যাখ্যা

      Gauss–Seidel uses freshly updated values inside the same iteration for faster convergence.

      ৩৭.
      Consider: f1(x, y) = xy, f2(x, y) = x2 - y
      The Jacobian matrix J is:
      1. None
      সঠিক উত্তর:
      উত্তর
      সঠিক উত্তর:
      ব্যাখ্যা

      ৩৮.
      For the system: f1(x, y) = x2 + y2 - 5 = 0, f2(x, y) = x - y - 1 = 0 The Jacobian matrix is:
      1. None
      সঠিক উত্তর:
      উত্তর
      সঠিক উত্তর:
      ব্যাখ্যা

      ৩৯.
      Using one Gauss–Seidel iteration with x0 = 0, y0 = 0, solve: 2x + y = 5, x + 3y = 9. After 1st iteration:
      1. x1 = 1.67, y1 = 2.44
      2. x1 = 2 y1 = 2.33
      3. x1 = 2.5, y1 = 2.17
      4. x1 = 2.5, y1 = 3
      সঠিক উত্তর:
      x1 = 2.5, y1 = 2.17
      উত্তর
      সঠিক উত্তর:
      x1 = 2.5, y1 = 2.17
      ব্যাখ্যা

      Rewrite: x=(5−y)/2,  y=(9−x)/3
      Gauss–Seidel first iteration:
      x1 = (5−0)/2 = 2.5,
      y1 = (9−x1)/3 = (9−2.5)/3 = 2.17

      ৪০.
      Which of the following is a disadvantage of Newton–Raphson method for nonlinear systems?
      1. Quadratic convergence
      2. Requires a good initial guess
      3. Works for linear systems only
      4. Always convergent
      সঠিক উত্তর:
      Requires a good initial guess
      উত্তর
      সঠিক উত্তর:
      Requires a good initial guess
      ব্যাখ্যা

      Newton–Raphson converges quadratically near the root, but a poor initial guess may lead to divergence or convergence to the wrong root.