# Numerical Methods and Calculus

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Question 1 |

A | |

B | |

C | |

D |

**GATE CS 2013**

**Numerical Methods and Calculus**

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Question 1 Explanation:

A function is continuous at some point c,
Value of f(x) defined for x > c = Value of f(x) defined for x < c = Value of f(x) defined for x = c
All values are 2 in option A

Question 2 |

8.983 | |

9.003 | |

9.017 | |

9.045 |

**GATE CS 2013**

**Numerical Methods and Calculus**

**Discuss it**

Question 2 Explanation:

Since the intervals are uniform, apply the uniform grid formula of trapezoidal rule.
This solution is contributed by

**Anil Saikrishna Devarasetty**Question 3 |

Consider the function f(x) = sin(x) in the interval [π/4, 7π/4]. The number and location(s) of the local minima of this function are

One, at π/2 | |

One, at 3π/2 | |

Two, at π/2 and 3π/2 | |

Two, at π/4 and 3π/2 |

**GATE CS 2012**

**Numerical Methods and Calculus**

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Question 3 Explanation:

Question 4 |

The bisection method is applied to compute a zero of the function f(x) = x

^{4}– x^{3}– x^{2}– 4 in the interval [1,9]. The method converges to a solution after ––––– iterations1 | |

3 | |

5 | |

7 |

**GATE CS 2012**

**Numerical Methods and Calculus**

**Discuss it**

Question 4 Explanation:

In bisection method, we calculate the values at extreme points of given interval, if signs of values are opposite, then we find the middle point. Whatever sign we get at middle point, we take the corner point of opposite sign and repeat the process till we get 0.
f(1) < 0 and f(9) > 0
mid = (1 + 9)/2 = 5
f(5) > 0, so zero value lies in [1, 5]
mid = (1+5)/2 = 3
f(3) > 0, so zero value lies in [1, 3]
mid = (1+3)/2 = 2
f(2) = 0

Question 5 |

Given i=√-1, what will be the evaluation of the integral ?

0 | |

2 | |

-i | |

i |

**GATE CS 2011**

**Numerical Methods and Calculus**

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Question 5 Explanation:

Question 6 |

Newton-Raphson method is used to compute a root of the equation x

^{2}-13=0 with 3.5 as the initial value. The approximation after one iteration is3.575 | |

3.676 | |

3.667 | |

3.607 |

**GATE CS 2010**

**Numerical Methods and Calculus**

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Question 6 Explanation:

In Newton-Raphson's method, We use the following formula to get the next value of f(x). f'(x) is derivative of f(x).

f(x) = x^{2}-13 f'(x) = 2x Applying the above formula, we get Next x = 3.5 - (3.5*3.5 - 13)/2*3.5 Next x = 3.607

Question 7 |

What is the value of Lim

_{n->∞}(1-1/n)^{2n}?0 | |

e ^{-2} | |

e ^{-1/2} | |

1 |

**GATE CS 2010**

**Numerical Methods and Calculus**

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Question 7 Explanation:

The value of e (mathematical constant) can be written as following
And the value of 1/e can be written as following.

Question 8 |

Two alternative packages A and B are available for processing a database having 10k records.Package A requires 0.0001n

^{2}time units and package B requires 10nlog10n time units to process n records. What is the smallest value of k for which package B will be preferred over A?12 | |

10 | |

6 | |

5 |

**GATE CS 2010**

**Numerical Methods and Calculus**

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Question 8 Explanation:

Since,
10nlog

_{10}n ≤ 0.0001n^{2}Given n = 10^{k}records. Therefore, ⟹10×(10^{k})log_{10}10^{k}≤ 0.0001(10^{k})^{2}⟹10^{k+1}k ≤ 0.0001 × 10^{2k}⟹k ≤ 10^{2k−k−1−4}⟹k ≤ 10^{k−5}Hence, value 5 does not satisfy but value 6 satisfies. 6 is the smallest value of k for which package B will be preferred over A. Option (C) is correct.Question 9 |

0 | |

1 | |

ln 2 | |

1/2 ln 2 |

**GATE-CS-2009**

**Numerical Methods and Calculus**

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Question 9 Explanation:

(1-tanx)/(1+tanx) = (cosx - sinx)/(cosx + sinx)
Let cosx + sinx = t
(-sinx + cosx)dx = dt
(1/t)dt = ln t => ln(sinx + cosx)
=> ln(sin Π/4 + cos Π/4)
=> ln(1/√2 + 1/√2)
=> 1/2 ln 2

There are 91 questions to complete.