# Numerical Analysis Questions and Answers – Gauss’s Backward Interpolation Formula

This set of Numerical Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Gauss’s Backward Interpolation Formula”.

1. What will be the solution for the following table using Gauss’s backward interpolation formula, where x = 3?

 x f(x) 1 1 2 2 3 1

a) 1
b) 0.328
c) 0.327
d) 0.322

Explanation:
Gauss's backward difference interpolation method to find solution
h=2-1=1
Now the central difference table is

 x p=(x-2)/1 y Δy Δ2y 1 -1 1 1 2 0 2 -2 -1 3 1 1

Solution of Gauss's backward interpolation is y(3)=1.

2. What will be the solution for the following table using Gauss’s backward interpolation formula, where x = 3?

 x f(x) 1 2 2 3 3 4

a) 0.2452
b) 0.2864
c) 0.2862
d) 4

Explanation:
Gauss's backward method to find solution
h=2-1=1
Now the central difference table is

 x p=(x-2)/1 y Δy Δ2y 1 -1 2 1 2 0 3 0 1 3 1 4
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Solution of Gauss's backward interpolation is y(3)=4.

3. What will be the solution for the following table using Gauss’s backward interpolation formula, where x = 3?

 x f(x) 1 2 2 3 3 4 4 8

a) 11.152
b) 4
c) 11.122
d) 11.128

Explanation: Given,
Gauss's backward method to find solution
h=2-1=1
Now the central difference table is

 x p=(x-2)/1 y Δy Δ2y Δ3y 1 -1 2 1 2 0 3 0 1 3 3 1 4 3 4 4 2 8

Solution of Gauss's backward interpolation is y(3)=4.

4. What will be the solution for the following table using Gauss’s backward interpolation formula, where x = 3?

 x f(x) 1 2 2 6 3 4 4 8 5 10 6 12

a) 1
b) 4.1774
c) 4
d) 2

Explanation:
Gauss's backward difference interpolation method to find solution
h=2-1=1
Now the central difference table is

 x p=(x-3)/1 y Δy Δ2y Δ3y Δ4y Δ5y 1 -2 2 1 2 -1 3 0 1 3 3 0 4 3 -8 4 -5 15 4 1 8 -2 7 2 2 5 2 10 0 2 6 3 12

Solution of Gauss's backward interpolation is y(3)=4.

5. What will be the solution for the following table using Gauss’s backward interpolation formula, where x = 3?

 x f(x) 1 2 2 6 3 4 4 8 5 10 6 12 7 24 8 26 9 28

a) 4
b) 9
c) 7
d) 6

Explanation:
Gauss's backward method to find solution
h=2-1=1
Now the central difference table is

 x p=(x-5)/1 y Δy Δ2y Δ3y Δ4y Δ5y Δ6y Δ7y Δ8y 1 -4 2 1 2 -3 3 0 1 3 3 -2 4 3 -8 4 -5 15 4 -1 8 -2 7 -14 2 2 1 -25 5 0 10 0 8 -39 162 2 10 -38 137 6 1 12 10 -30 98 12 -20 60 7 2 24 -10 30 2 10 8 3 26 0 2 9 4 28

Solution of Gauss's backward interpolation is y(3)=4.

6. What will be the solution of the equation 4x-1 using Gauss’s backward interpolation formula where x1 = 2 and x2 = 4 and x = 2.1, step value (h) = 0.25?
a) 8.4
b) 7.4
c) 8
d) 5

Explanation:
Gauss's backward method to find solution
h=2.25-2=0.25
Now the central difference table is

 x p=(x-3)/0.25 y Δy Δ2y 2 -4 7 1 2.25 -3 8 0 1 2.5 -2 9 0 1 2.75 -1 10 0 1 3 0 11 0 1 3.25 1 12 0 1 3.5 2 13 0 1 3.75 3 14 0 1 4 4 15

Solution of Gauss's backward interpolation is y(2.1)=7.4.

7. What will be the solution of the equation -2x2 using Gauss’s backward interpolation formula, where x1 = 1 and x2 = 2 & x = 2 & step value (h) = 0.25?
a) 2
b) 7
c) 5
d) -8

Explanation:
Gauss's backward method to find solution
h=1.25-1=0.25

Now the central difference table is

 x p=(x-1.5)/0.25 y Δy Δ2y Δ3y 1 -2 -2 -1.125 1.25 -1 -3.125 -0.25 -1.375 0 1.5 0 -4.5 -0.25 -1.625 0 1.75 1 -6.125 -0.25 -1.875 2 2 -8

Solution of Gauss's backward interpolation is y(2)=-8.

8. What will be the solution of the equation 3x using Gauss’s backward interpolation formula, where x1 = 2 and x2 = 4 & x = 2 & step value (h) = 0.25?
a) 8
b) -6
c) 5
d) 1

Explanation:
Gauss's backward method to find solution
h=2.25-2=0.25
Now the central difference table is

 x p=(x-3)/0.25 y Δy Δ2y 2 -4 -6 -0.75 2.25 -3 -6.75 0 -0.75 2.5 -2 -7.5 0 -0.75 2.75 -1 -8.25 0 -0.75 3 0 -9 0 -0.75 3.25 1 -9.75 0 -0.75 3.5 2 -10.5 0 -0.75 3.75 3 -11.25 0 -0.75 4 4 -12

Solution of Gauss's backward interpolation is y(2)=-6.

9. What will be the solution of the equation x – 1 using Gauss’s backward interpolation formula, where x1 = 1 and x2 = 2 & x = 1 & step value (h) = 0.25?
a) 1
b) 3
c) 0
d) 4

Explanation:
Gauss's backward method to find solution
h=1.25-1=0.25
Now the central difference table is

 x p=(x-1.5)/0.25 y Δy Δ2y 1 -2 0 0.25 1.25 -1 0.25 0 0.25 1.5 0 0.5 0 0.25 1.75 1 0.75 0 0.25 2 2 1

Solution of Gauss's backward interpolation is y(1)=0.

10. What will be the solution of the equation -x using Gauss’s backward interpolation formula, where x1 = 1 and x2 = 2 & x = 1 & step value (h) = 0.25?
a) 4
b) 2
c) 3
d) 1

Explanation:
Gauss's backward method to find solution
h=1.25-1=0.25
Now the central difference table is

 x p=(x-1.5)/0.25 y Δy Δ2y 1 -2 -1 -0.25 1.25 -1 -1.25 0 -0.25 1.5 0 -1.5 0 -0.25 1.75 1 -1.75 0 -0.25 2 2 -2

Solution of Gauss's backward interpolation is y(1)=-1.

11. What will be the solution of the equation -sin(x) using Gauss’s backward interpolation formula, where x1 = 1 and x2 = 2 & x = 1 & step value (h) = 0.25?
a) 4
b) 2
c) 3
d) -0.8415

Explanation:
Gauss's backward method to find solution
h=1.25-1=0.25
Now the central difference table is

 x p=(x-1.5)/0.25 y Δy Δ2y Δ3y Δ4y 1 -2 -0.8415 -0.1075 1.25 -1 -0.949 0.059 -0.0485 0.003 1.5 0 -0.9975 0.062 -0.0039 0.0135 -0.0008 1.75 1 -0.984 0.0612 0.0747 2 2 -0.9093

Solution of Gauss's backward interpolation is y(1)=-0.8415.

12. What will be the solution of the equation -cos(x) using Gauss’s backward interpolation formula, where x1 = 1 and x2 = 2 & x = 1 & step value (h) = 0.25?
a) -0.5403
b) 2
c) 3
d) 0.0167

Explanation:
Gauss's backward method to find solution
h=1.25-1=0.25
Now the central difference table is

 x p=(x-1.5)/0.25 y Δy Δ2y Δ3y Δ4y 1 -2 -0.5403 0.225 1.25 -1 -0.3153 0.0196 0.2446 -0.0152 1.5 0 -0.0707 0.0044 -0.0003 0.249 -0.0155 1.75 1 0.1782 -0.0111 0.2379 2 2 0.4161

Solution of Gauss's backward interpolation is y(1)=-0.5403.

13. What will be the solution of the equation -tan(x) using Gauss’s backward interpolation formula, where x1 = 1 and x2 = 2 & x = 1 & step value (h) = 0.25?
a) 0.9496
b) -1.5574
c) 3
d) 0.0167

Explanation:

Gauss's backward method to find solution
h=1.25-1=0.25
Now the central difference table is

 x p=(x-1.5)/0.25 y Δy Δ2y Δ3y Δ4y 1 -2 -1.5574 -1.4522 1.25 -1 -3.0096 -9.6397 -11.0919 40.3533 1.5 0 -14.1014 30.7137 -94.0241 19.6218 -53.6708 1.75 1 5.5204 -22.9571 -3.3353 2 2 2.185

Solution of Gauss's backward interpolation is y(1)=-1.5574.

14. What will be the solution of the equation cot(x) using Gauss’s backward interpolation formula, where x1 = 1 and x2 = 2 & x = 1 & step value (h) = 0.25?
a) 0.9496
b) 4190.0569
c) -0.6421
d) 0.0167

Explanation:
Gauss's backward method to find solution

h=1.25-1=0.25
Now the central difference table is

 x p=(x-1.5)/0.25 y Δy Δ2y Δ3y Δ4y 1 -2 -0.6421 0.3098 1.25 -1 -0.3323 -0.0485 0.2614 0.0392 1.5 0 -0.0709 -0.0093 -0.0054 0.2521 0.0337 1.75 1 0.1811 0.0244 0.2765 2 2 0.4577

Solution of Gauss's backward interpolation is y(1)=-0.6421.

15. What will be the solution of the equation -sec(x) using Gauss’s backward interpolation formula, where x1 = 1 and x2 = 2 & x = 1 & step value (h) = 0.25?
a) -1.8508
b) 4190.0569
c) 3.3389
d) 0.0167

Explanation:
Gauss's backward method to find solution
h=1.25-1=0.25

Now the central difference table is

 x p=(x-1.5)/0.25 y Δy Δ2y Δ3y Δ4y 1 -2 -1.8508 -1.3205 1.25 -1 -3.1714 -9.6449 -10.9655 40.3575 1.5 0 -14.1368 30.7125 -94.0243 19.7471 -53.6668 1.75 1 5.6102 -22.9543 -3.2072 2 2 2.403

Solution of Gauss's backward interpolation is y(1)=-1.8508.

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