Mathematical Methods Mid - I, January - 2015

1.∇2 f (2) =
  • f(2) – f(0)
  • f(2) +2f (1) + f( 0)
  • f(2) -2 f(10+ f(0)
  • f(2) +2f (1) + f( 0)
Answer: C
2.∇E1/2 =
  • E
  • δ
  • E2
  • μ
Answer: B
3.If h = 1, δμ(x2 +3x) =
  • 0
  • 2x
  • 2x +3
  • 2x – 1
Answer: C
4.If y(0) = 9, y(1) = 18, y(2) = 20 and y(3) = 24. Assuming that the third order differences are constant, then, y(4) =
  • 30
  • 33
  • 35
  • 39
Answer: D
5. f(2.5) using Newtons forward formula
  • 15.625
  • 35.4
  • 48.75
  • 20.12
Answer: A
6. f(3.5) by Gauss forward difference formula is
  • 4.0563
  • 3.9512
  • 3.7523
  • 2.75
Answer: A
7. f(25) = by Gauss backward interpolation formula is
  • 47.12
  • 42.23
  • 39.23
  • 38.44
Answer: B
8.If y = a0 + a1x then a0 by least squares method is
Answer: B
9.If n = 5, Σxi = 10, Σyi = 16.9, Σxiyi = 47.1, Σxi2 = 30 and y = a + bx then y =
  • .72 + 1.33 x
  • 1 + x
  • 1 - 2x
  • .9 - 1.5 x
Answer: A
10.If y = axb, then by least squares method the first normal equation is
  • Σ log yi = n log a + b Σ log xi
  • Σ log yi = na + b Σ log xi
  • Σ log yi = n log a + b Σ xi
  • Σ log xi yi = n log a + b Σ log xi2
Answer: A
11.Δy1 + Δ2y1 = ………………………………
Answer: Δy2
12.μy5/2 = ………………………..
Answer: 1/2 (y2 + y3)
13.If h = 1, Δ3x(2) = ………………………………………………….
Answer: 0
14.The solution of the difference equation (E2 + 4E + 4) y = 0 ………………………….
Answer: (A+ Bn)(−2)n
15.Newton’s Backward interpolation formula is ………………….
Answer:
16.If y(5) = 87.9, y(8) = 81.3, then by lagranges formula y(9) = …………….
Answer: 81.1
17.x2 + 5x in terms os factorials is………………
Answer: x(2) + 6x(1)
18.n = 7, Σxi = 28, Σyi = 153, Σxiyi = 848.6, Σxi2 = 140 Σxi2 yi = 848.6, Σxi3 = 2401, Σxi4 = 4676 and y = a +bx + cx2 then the third normal equation is ……………………………….
Answer: 848.6 = 140a + 2401b + 4676c
19.If y = aex, then by least squares method a = ……………………..
Answer:  or 
20. and y = a + bx, the first normal equation by least squares method is …………………….
Answer: 44 = 3a + 15 b