Engineering Mathematics Miscellaneous
- The probability that a student knows the correct answer to a multiple choice question is 2/3. If the student does not known the answer, then the student guesses the answer. The probability of the guessed answer being correct is 1/4. Given that the student has answered the question correctly, the conditional probability that the student known the correct answer is
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Probability = {(2 / 3) × 1} = 8 {(2 / 3) × 1} + 
1 × 1 
9 3 4
Correct Option: D
Probability = {(2 / 3) × 1} = 8 {(2 / 3) × 1} + 1 × 1 9 3 4
- The accuracy of Simpson's rule quadrature for a step size h is
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NA
Correct Option: C
NA
- An explicit forward Euler method is used to numerically integrate the differential equation
dy = y dt
using a time step of 0.1. With the initial condition y(0) = 1, the value of y(10) computed by this method is ______ (correct to two decimal places).
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General formula
yn + 1 = yn + hf (tn , yn)
For n = 0, y1 = y0 + hf (t0, y0)
= y0 + hy0
= 1 + 0.1 (1)
y1 = 1.1
For n = 1, y2 = y1 + hf (t1, y1)
= y1 + hy1
= 1.1 + 0.1 (1.1)
y2 = 1.21
For n = 2, y3 = y2 + hf (t2, y2)
= y2 + hy2
= 1.21 + 0.1 × 1.21
y3 = 1.331
For n = 3, y4 = y3 + hf (t3, y3)
= y3 + hy3
= 1.331 + 0.1 × 1.331
y4 = 1.4641
For n = 4, y5 = y4 + hf (t4, y4)
= y4 + hy4
= 1.4641 + 0.1 × (1.4641)
y5 = 1.61051
For n = 5, y6 = y5 + hf (t5, y5)
= y5 + hy5
= 1.61051 + 0.1 × 1.61051
y6 = 1.771561
For n = 6, y7 = y6 + hf (t6, y6)
= y6 + hy6
= 1.771561 + 0.1 × 1.771561 = 1.9487
For n = 7, y8 = y7 + hf (t7, y7)
= y7 + hy7
= 1.9487 + 0.1 × (1.9487)
y8 = 2.14357
For n = 8, y9 = y8 + hf (t8, y8)
= y8 + hy8
= 2.14357 + 0.1 × 2.14357
y9 = 2.3579
For n = 9, y10 = y9 + hf (t9, y9)
= y9 + hy9
= 2.3579 + 0.1 × (2.3579)
y10 = 2.5937Correct Option: C
General formula
yn + 1 = yn + hf (tn , yn)
For n = 0, y1 = y0 + hf (t0, y0)
= y0 + hy0
= 1 + 0.1 (1)
y1 = 1.1
For n = 1, y2 = y1 + hf (t1, y1)
= y1 + hy1
= 1.1 + 0.1 (1.1)
y2 = 1.21
For n = 2, y3 = y2 + hf (t2, y2)
= y2 + hy2
= 1.21 + 0.1 × 1.21
y3 = 1.331
For n = 3, y4 = y3 + hf (t3, y3)
= y3 + hy3
= 1.331 + 0.1 × 1.331
y4 = 1.4641
For n = 4, y5 = y4 + hf (t4, y4)
= y4 + hy4
= 1.4641 + 0.1 × (1.4641)
y5 = 1.61051
For n = 5, y6 = y5 + hf (t5, y5)
= y5 + hy5
= 1.61051 + 0.1 × 1.61051
y6 = 1.771561
For n = 6, y7 = y6 + hf (t6, y6)
= y6 + hy6
= 1.771561 + 0.1 × 1.771561 = 1.9487
For n = 7, y8 = y7 + hf (t7, y7)
= y7 + hy7
= 1.9487 + 0.1 × (1.9487)
y8 = 2.14357
For n = 8, y9 = y8 + hf (t8, y8)
= y8 + hy8
= 2.14357 + 0.1 × 2.14357
y9 = 2.3579
For n = 9, y10 = y9 + hf (t9, y9)
= y9 + hy9
= 2.3579 + 0.1 × (2.3579)
y10 = 2.5937
- Gauss seidel method is used to solve the following equations (as per the given order):
x1 + 2x2 + 3x3 = 1
2x1 + 3x2 + x3 = 1
3x1 + 2x2 + x3 = 1
Assuming initial guess as x1 = x2 = x3 = 0, the value of x3 after the first iteration is ______.
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The equations are
x1 + 2x2 + 3x3 = 5
2x1 + 3x2 + x3 = 1
3x1 + 2x2 + x3 = 3
By pivoting we getx1 = 3 - 2x2 - x3 .....(1) 3 x2 = 1 - 2x1 - x3 .....(2) 3 x3 = 5 - x1 - 2x2 .....(3) 3
When we put x2 = 0, x3 = 0 in equaiton (1), x1 = 1
put x1 = 1, x3 = 0 in equation (3), x2 = – .333
put x1 = 1, x2 = – .333 in equation (3), x3 = 1.555Correct Option: A
The equations are
x1 + 2x2 + 3x3 = 5
2x1 + 3x2 + x3 = 1
3x1 + 2x2 + x3 = 3
By pivoting we getx1 = 3 - 2x2 - x3 .....(1) 3 x2 = 1 - 2x1 - x3 .....(2) 3 x3 = 5 - x1 - 2x2 .....(3) 3
When we put x2 = 0, x3 = 0 in equaiton (1), x1 = 1
put x1 = 1, x3 = 0 in equation (3), x2 = – .333
put x1 = 1, x2 = – .333 in equation (3), x3 = 1.555
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Consider an ordinary differential equation dw = 0 - iz dz
If x = x0 at t = 0 , the increment in x calculated using Runge-Kutta fourth order multi-step method with a step size of ∆t = 0.2 is
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Given , dx = 4t + 4 dt
x = x0 at t = 0
n = 0.2
Calculate x(0.2) value
K1 = f(t0, x0) = f(0, x 0) = 4K2 = f t0 + h , x0 + 1 K1h 2 2
= f(0 + 0.1, x0 + 0.4)
= f(0.1, x0 + 0.4) = 4(0.1) + 4 = 4.4K3 = f x0 + h , x0 + 1 K2h 2 2
= f[t0 + 0.1, x0 + (2.2)(0.2)]
= f(0.1, x0 + 0.44) = 4(0.1) + 4 = 4.4
K4 = f(t0 + h, x0 + K3 h)
= f(0 + 0.2, x0 + 0.88)
= f(0.2, x0 + 0.88)
= 4(0.2) + 4 = 4.8x(0.2) = x1 = x0 + h (K1 + 2K2 + 2K3 + K4) 6 = x0 + 0.2 [ 4 + 2(4.4) + 2(4.4) + (4.8) ] 6 = x0 + 0.2 (4 + 8.8 + 8.8 + 4.8) 6
= x0 + 0.88
Increment as x = x 1 – x0 = x0 + 0.88 – x0 = 0.88Correct Option: D
Given , dx = 4t + 4 dt
x = x0 at t = 0
n = 0.2
Calculate x(0.2) value
K1 = f(t0, x0) = f(0, x 0) = 4K2 = f t0 + h , x0 + 1 K1h 2 2
= f(0 + 0.1, x0 + 0.4)
= f(0.1, x0 + 0.4) = 4(0.1) + 4 = 4.4K3 = f x0 + h , x0 + 1 K2h 2 2
= f[t0 + 0.1, x0 + (2.2)(0.2)]
= f(0.1, x0 + 0.44) = 4(0.1) + 4 = 4.4
K4 = f(t0 + h, x0 + K3 h)
= f(0 + 0.2, x0 + 0.88)
= f(0.2, x0 + 0.88)
= 4(0.2) + 4 = 4.8x(0.2) = x1 = x0 + h (K1 + 2K2 + 2K3 + K4) 6 = x0 + 0.2 [ 4 + 2(4.4) + 2(4.4) + (4.8) ] 6 = x0 + 0.2 (4 + 8.8 + 8.8 + 4.8) 6
= x0 + 0.88
Increment as x = x 1 – x0 = x0 + 0.88 – x0 = 0.88
