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Numerical interpolation using Langrange's formula
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Option :
1.  X 300 304 305 307 f(x) 2.4771 2.4829 2.4843 2.4871
and x=301
2.  X 2 2.5 3 f(x) 0.69315 0.91629 1.09861
and x=2.7
3.  X 0 1 3 4 f(x) -12 0 12 24
and x=5
4.  X -1 0 3 6 7 f(x) 3 -6 39 822 1611
and x=8
5.  X 0.2 0.3 0.5 0.7 0.8 f(x) 1.1487 1.23114 1.41421 1.62451 1.7411
and x=0.40
 f(x) = x1 = and x2 = x = Step value (h) =  OR  Inverval (N) = Option : 1. Value f(2) 2. Interpolation table 3. Equation f(x) 4. Equation f(x) then value f(2) f(x)=2x^3-4x+1x1 = 2 and x2 = 4x = 3.8Step value (h) = 0.5 or N = 5f(x)=2x^3-4x+1x1 = 2 and x2 = 4x = 2.1Step value (h) = 0.25 or N = 8f(x)=x^3-x+1x1 = 2 and x2 = 4x = 3.8Step value (h) = 0.5 or N = 5f(x)=x^3-x+1x1 = 2 and x2 = 4x = 2.1Step value (h) = 0.25 or N = 8f(x)=x^3+x+2x1 = 2 and x2 = 4x = 3.8Step value (h) = 0.5 or N = 5f(x)=x^3+x+2x1 = 2 and x2 = 4x = 2.1Step value (h) = 0.25 or N = 8f(x)=sin(x)x1 = 0 and x2 = 1.57x = 2Step value (h) = 0.25 or N = 8f(x)=cos(x)x1 = 0 and x2 = 1.57x = 2Step value (h) = 0.25 or N = 8
 Decimal Place = 0 1 2 3 4 5 6 7 8
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