The following code is supposed to return n!, for positive n.  int factorial(int n){   if (n == 0)     return 1;   else     return (n + factorial(n - 1)); } An analysis of the code using our "Three Question" approach reveals that:  Group of answer choices it fails the base-case question. it fails the smaller-caller question. it fails the general-case question. it passes on all three questions and is a valid algorithm. None of these is correct.

Q1 The periodic function sin⁡(2x) has multiple roots between x values of -5π and 5π. If xL = -15 and xU = 15, which of the following statements is true using a bracketed method? Select one: a. All roots will be returned b. The middle root will be returned c. The chosen bracket is invalid for bracketed methods d. A single root will be returned e. The algorithm will be stuck in an infinite loop     Q2 Consider x and y to represent data points (xi,yi), where i = 1, 2, 3, … n. What is the length of pafter running the following command? p = polyval(x,y) Select one: a. n b. n - 1 c. n + 1 d. Empty variable e. 1   Q3 Consider a system of linear equations in the form of AX = B, where X is the unknown vector. Which of the following can be used to solve for X? Select one: a. X = A\B b. X = B./A c. X = inv(B)*A d. X = inv(A)./B e. X = B\A

One remarkably simple formula for calculating the value of p is the so-called Madhava-Leibniz series: p4 = 1-13+15-17+19-.... Consider the recursive function below to calculate the first n terms of this formula: double computePI(int n) { if (n <= 1) { return 1.0;} int oddnum = 2 * n - 1; if ((n % 2) == 0 { } return -1.0 oddnum + computePI(n − 1); } else { } return 1.0 / oddnum + computePI (n - 1); Which statements about the run-time performance of this function are true? 1.Each time this function is called it will invoke at least two more recursive calls II.The number of recursive calls this function will make is approximately equal to the value of the parameter variable n III.Not counting overhead, this function will be about as efficient as an iterative implementation of the same formula

Make a code using Recursion The countSubstring function will take two strings as parameters and will return an integer that is the count of how many times the substring (the second parameter) appears in the first string without overlapping with itself. This method will be case insensitive.   For example: countSubstring(“catwoman loves cats”, “cat”) would return 2 countSubstring(“aaa nice”, “aa”) would return 1 because “aa” only appears once without overlapping itself.     public static int countSubstring(String s, String x) { if (s.length() == 0 || x.length() == 0) return 1;   if (s.length() == 1 || x.length() == 1){ if (s.substring(0,1).equals(x.substring(0,1))){ s.replaceFirst((x), " "); return 1 + countSubstring(s.substring(1), x); } else { return 0 + countSubstring(s.substring(1), x); } } return countSubstring(s.substring(0,1), x) + countSubstring(s.substring(1), x); }   public class Main { public static void main(String[] args) {…