Q1: Let, f(x, y, z) = 4x2 + 7xy + 3xz2. The direction in which the function f(x, y, z) increases most rapidly at point P = (1, 0, 2) is (2022)
(a) 
(b) 
(c) 
(d) 
Ans: (b)
Sol: 
The directional derivative at point P is given by
at point (1, 0, 2)
Q2: Let 
The value of 
equals _________. (Give the answer up to three decimal places) (SET-2 (2017))
(a) 3.732
(b) 4.734
(c) 5.734
(d) 6.732
Ans: (c)
Sol: 
Q3: Let x and y be integers satisfying the following equations
2x2 + y2 = 34
x + 2y = 11
The value of (x + y) is ________. (SET-2 (2017))
(a) 4
(b) 3
(c) 7
(d) 10
Ans: (c)
Sol: 
Q4: Only one of the real roots of f(x) = x6 − x − 1 lies in the interval 1 ≤ x ≤ 2 and bisection method is used to find its value. For achieving an accuracy of 0.001, the required minimum number of iterations is _____. (SET-1 (2017))
(a) 1000
(b) 100
(c) 10
(d) 1
Ans: (c)
Sol: 
The minimum number of iterations by Bisection method is given by
Q5: A differential equation (di/dt) - 0.2i = 0 is applicable over −10 < t < 10. If i(4) = 10, then i(-5) is _____. (SET-2 (2015))
(a) 1
(b) 0.5
(c) 1.6
(d) 2.4
Ans: (c)
Sol: 
Q6: Two coins R and S are tossed. The 4 joint events HRHS, TRTS, hRHS, TRHS have probabilities 0.28, 0.18, 0.30, 0.24, respectively, where H represents head and T represents tail. Which one of the following is TRUE? (SET-2 (2015))
(a) The coin tosses are independent.
(b) R is fair, S is not.
(c) S is fair, R is not
(d) The coin tosses are dependent
Ans: (d)
Sol: From the given information, we can create a joint probability table as follows:
From the table, we can get
P(HR) = 0.58, P(TR) = 0.42, P(HS) = 0.52, P(TS) = 0.48
So, Coins R and S are biased (not fair). So choises (B) and (C) are both false.
The coin tosses are not independent since their probability of heads and tails is not 0.5.
R and S are dependent.
If R and S were independent then all the joint probabilities will be equal to the product of the marginal probabilities.
For example
So R and S are not independent.
i.e. R and S are dependent. So, Choise (A) is also false and choise (D) is true.
Q7: The function f(x) = ex − 1 is to be solved using Newton-Raphson method. If the initial value of x0 is taken as 1.0, then the absolute error observed at 2nd iteration is _____. (SET-3 (2014))
(a) 0.01
(b) 0.03
(c) 0.06
(d) 0.09
Ans: (c)
Sol: 
In Newton-Raphson method, we have
Putting the values, we get:
Putting the value, we get:
Q8: When the Newton-Raphson method is applied to solve the equation f(x) = x3 + 2x − 1 = 0, the solution at the end of the first iteration with the initial value as x0 = 1.2 is (2013)
(a) -0.82
(b) 0.49
(c) 0.705
(d) 1.69
Ans: (c)
Sol: 
Q9: Solution of the variables x1 and x2 for the following equations is to be obtained by employing the Newton-Raphson iterative method.
equation (1) 10x2sinx1 − 0.8 = 0
equation (2) 
− 10x2cosx1 − 0.6 = 0
Assuming the initial values ar x1 = 0.0 and x2 = 1.0, the jacobian matrix is (2011)
(a) 
(b) 
(c) 
(d) 
Ans: (b)
Sol: 
The Jacobian matrix is
Q10: Let x2 − 117 = 0. The iterative steps for the solution using Newton-Raphon's method is given by (2009)
(a) 
(b) 
(c) 
(d) 
Ans: (a)
Sol: 
Q11: Equation ex − 1 = 0 is required to be solved using Newton's method with an initial guess x0 = −1. Then, after one step of Newton's method, estimate x1 of the solution will be given by (2008)
(a) 0.71828
(b) 0.36784
(c) 0.20587
(d) 0
Ans: (a)
Sol: Here,
The newton raphson iterative equation is
Q12: The differential equation 
is discretised using Euler's numerical integration method with a time step ΔT > 0. What is the maximum permissible value of ΔT to ensure stability of the solution of the corresponding discrete time equation ? (2007)
(a) 1
(b) τ/2
(c) τ
(d) 2τ
Ans: (d)
