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
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
Question 3 Explanation:
The bisection method is applied to compute a zero of the function f(x) = x4 – x3 – x2 – 4 in the interval [1,9]. The method converges to a solution after ––––– iterations
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
Given i=√-1, what will be the evaluation of the integral ?
Question 5 Explanation:
Newton-Raphson method is used to compute a root of the equation x2-13=0 with 3.5 as the initial value. The approximation after one iteration is
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) = x2-13 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
What is the value of Limn->∞(1-1/n)2n ?
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.
Two alternative packages A and B are available for processing a database having 10k records.Package A requires 0.0001n2 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?
Question 8 Explanation:
Since, 10nlog10n ≤ 0.0001n2 Given n = 10k records. Therefore, ⟹10×(10k)log1010k ≤ 0.0001(10k)2 ⟹10k+1k ≤ 0.0001 × 102k ⟹k ≤ 102k−k−1−4 ⟹k ≤ 10k−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.
1/2 ln 2
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.
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