InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
E is a _____(a) shifting operator (b) Displacement operator (c) 1 + ∆ (d) all of these (e) none of these |
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Answer» (d) all of these |
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| 2. |
For the given points (x0 , y0 ) and (x1, y1 ) the Lagrange’s formula is ______(a) y(x) = (x - x1)/(x0 - x1)y0 + (x - x0)/(x1 - x0)y1(b) y(x) = (x1 - x)/(x0 - x1)y0 + (x - x0)/(x1 - x0)y1(c) y(x) = (x - x1)/(x0 - x1)y1 + (x - x0)/(x1 - x0)y0(d) y(x) = (x1 - x)/(x0 - x1)y1 + (x - x0)/(x1 - x0)y0 |
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Answer» (a) y(x) = (x - x1)/(x0 - x1)y0 + (x - x0)/(x1 - x0)y1 |
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| 3. |
∇ f(a) = ______ (a) f(a) + f(a – h) (b) f(a) – f(a + h) (c) f(a) – f(a – h) (d) f(a) |
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Answer» (c) f(a) – f(a – h) |
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| 4. |
If f(x) = x2 + 2x + 2 and the interval of differencing is unity then ∆ f(x) _______ (a) 2x – 3 (b) 2x + 3 (c) x + 3 (d) x – 3 |
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Answer» (b) 2x + 3 f(x) = 2x2 + 2x + 2 h = 1 ∆f(x) = (x + 1)2 + 2(x + 1) + 2 – x2 – 2x – 2 = x2 + 2x + 1 +2x + 2 + 2 – x2 – 2x – 2 = 2x + 3 |
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| 5. |
If h = 1 then prove that (E-1 ∆) x3 = 3x2 – 3x + 1. |
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Answer» h = 1 To prove (E-1 ∆) x3 = 3×2 – 3x + 1 L.H.S = (E-1 ∆) x3 = E-1 (∆x3 ) = E-1 [(x + h)3 – x3 ] = E-1 ( x + h)3 – E-1 (x3) = (x – h + h)3 – (x – h)3 = x3 (x – h)3 But given h = 1 So(E-1 ∆) x3 = x3 - (x – 1)3 = x3 – [x3 - 3x2 + 3x – 1] = 3x2 – 3x + 1 = RHS |
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| 6. |
If c is a constant then ∆c = ______ (a) c (b) ∆ (c) ∆2(d) 0 |
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Answer» The correct answer is : (d) 0 |
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| 7. |
If h = 1, then ∆(x2) = ________ (a) 2x (b) 2x – 1 (c) 2x + 1(d) 1 |
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Answer» (c) 2x + 1 ∆(x2 ) = (x + h)2 – x2 = (x + 1)2 – x2 = 2x + 1 |
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| 8. |
∇ = _______ (a) 1 + E (b) 1 – E (c) 1 – E-1 (d) 1 + E-1 |
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Answer» The correct answer is : (c) 1 – E-1 |
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| 9. |
Evaluate (log ax) |
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Answer» \(\triangle\)log ax = log a a (x + h) - log ax = log [a(x + h)/ax] = log(1 + h/x) |
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| 10. |
E = ______ (a) 1 + ∆ (b) 1 – ∆ (c) 1 + ∇ (d) 1 – ∇ |
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Answer» (a) 1 + Δ E = 1 + ∆ |
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| 11. |
E f(x) = _______ (a) f(x – h) (b) f(x) (c) f(x + h) (d) f(x + 2h) |
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Answer» (c) f(x + h) |
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| 12. |
If m and n are positive integers then ∆m ∆n f(x) = _______ (a) ∆m + n f(x) (b) ∆m f(x)(c) ∆n f(x) (d) ∆m - n f(x) |
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Answer» The correct answer is : (a) ∆m + n f(x) |
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| 13. |
E-n f(x) is ______ (a) f(x + nh) (b) f(x – nh)(c) f(-nh) (d) f(x – n) |
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Answer» The correct answer is : (b) f(x – nh) |
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| 14. |
If f(x) = x2 + 3x then show that ∆f(x) = 2x + 4 |
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Answer» f(x) = x2 + 3 x ∆f(x) = f(x + h) – f(x) = (x + h)2 + 3(x + h) – x2 – 3x = x2 + 2xh + h2 + 3x + 3h – x2 – 3x = 2xh + 3h + h2 Put h = 1, ∆f(x) = 2x + 4 |
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| 15. |
∆f(x) = _______ (a) f(x + h) (b) f(x) – f(x + h) (c) f(x + h) – f(x) (d) f(x) – f(x – h) |
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Answer» (c) f(x + h) – f(x) ∆f(x) = f(x + h) – f(x) |
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| 16. |
If ‘n’ is a positive integer ∆n [∆-n f(x)] _______ (a) f(2x) (b) f(x + h) (c) f(x) (d) ∆ f(2x) |
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Answer» The correct answer is : (c) f(x) |
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| 17. |
Match the following(a) ∆f(x)(i) 2x + 1(b) E2 f(x)(ii) 1 + ∆(c) E(iii) f(x + h) – f(x)(d) ∆x2 , h = 1(iv) f(x + 2h) |
| Answer» (a) – (iii), (b) – (iv), (c) – (ii), (d) – (i) | |
| 18. |
∆4 y3 = _______ (a) (E – 1)4 y3 (b) (E3 – 1) y3 (c) (E – 1)3 y0(d) (E – 1)4 y0 |
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Answer» (a) (E – 1)4 y3 |
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| 19. |
Say true or false 1. ∇y2 = y1 – y0 2. ∇2 yn = ∇yn – ∇yn + 1 3. When 5 values are given, the polynomial which fits the data is of degree 4 4. E ∆ = ∆ E 5. f(2) + ∆f(2) = f(3) |
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Answer» 1. False 2. True 3. True 4. True 5. True |
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| 20. |
Fill in the blanks. 1. The two methods of interpolation are _______ and _______ 2. If values of x are not equidistant we use _______ method. 3. ∆(f(x) + g(x)) = ______ 4. ∆k yn = ______ 5. The first three terms in Newton’s method will give a ________ interpolation. |
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Answer» 1. graphical method, algebraic method 2. Lagrange’s method 3. ∆f(x) + ∆g(x) 4. ∆k-1 yn + 1 – ∆k - 1 yn 5. Parabolic |
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