৪৯তম বিসিএস ⎯ ফলিত গণিত [৫৬১]
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৪৯তম বিসিএস ⎯ ফলিত গণিত [৫৬১]
৪৯তম বিসিএস ⎯ ফলিত গণিত [৫৬১] · তারিখ অনির্ধারিত · ৪০ প্রশ্ন
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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.
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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].
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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.
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Convergence is guaranteed by the Intermediate Value Theorem, which requires continuity and a sign change. Differentiability is not necessary.
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The error reduces by half in each step, so convergence is linear (slow) compared to Newton-Raphson (quadratic).
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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).
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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
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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.
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(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.
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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.
x : 1, 2, 3 f(x) : 1, 8, 27
Estimate f(1.5) using Newton’s Forward Interpolation.
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Here f(x) = x3 At x = 1.5, exact value is 1.5 = 3.3751.
Forward difference method gives same result.
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With 3 points, polynomial is quadratic. Any method produces the same quadratic with almost equal effort.
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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.
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Both forward and backward formulas require equal spacing. Lagrange works for unequal spacing.
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Central difference interpolation (Gauss forward/backward) is designed for approximations near the middle.
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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.
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Forward interpolation uses forward differences, suitable when xxx is near the first tabulated point.
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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.
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সঠিক উত্তর: 5.25
অপশনে সঠিক উত্তর না থাকায় প্রশ্নটি বাতিল করা হলো।
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Find f(1.5) using Lagrange interpolation.
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Polynomial fitting points is f(x)=x2+x+1.
At x=1.5: f(1.5)=2.25+1.5+1=4.75
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Iterative methods converge gradually, but Gaussian elimination (a direct method) solves the system exactly in finite steps.
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Jacobi updates all variables simultaneously using only the previous iteration’s values, unlike Gauss–Seidel which reuses updated values.
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If the coefficient matrix is diagonally dominant (∣aii∣>∑∣aij∣), iterative methods converge reliably.
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Direct methods (Gaussian, LU, Gauss–Jordan) give the exact solution in finite steps. Jacobi is iterative.
Solve using Cramer’s Rule.
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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.
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Sparse systems benefit from iterative approaches because they avoid heavy matrix factorizations.
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Near the root, Newton–Raphson converges quadratically (error roughly squares each step).
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Gauss–Seidel uses freshly updated values inside the same iteration for faster convergence.
The Jacobian matrix J is:
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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
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Newton–Raphson converges quadratically near the root, but a poor initial guess may lead to divergence or convergence to the wrong root.