Engineering Mathematics Miscellaneous
- Match the items in columns I and II.
Column I
P. Gauss-Seidel method
Q. Forward Newton-Gauss method
R. Runge-Kutta method
S. Trapezoidal Rule
Column II
1. Interpolation
2. Non-linear differential equations
3. Numerical integration
4. Linear algebraic equations
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NA
Correct Option: D
NA
- The derivative of f(x) = cos(x) can be estimated using the approximation
f '(x) = f(x + h) - f(x - h) 2h
The percentage error is calculated asExact value - Approximation value × 100 Exact value
The percentage error in the derivative of f(x) at x = π / 6 radian, choosing h = 0.1 radian, is
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f(x) = cos(x)
at x = π rad = π × 180 = 30° 6 6 π [f '(x)]2 = -sin30° = -1 .......(1) 2 f '(x)= cos(x + h) - cos(x - h) 2h ∵ x = π = 30° 6 h = 0.1 = 28° π [f '(x)]2 = cosx . cosh - sinx . sinh - cosx.cosh - sinx.sinh 2h = - 2sinx . sinh = - 2sin30° × sin(18/ π) = -0.499 2h 2 × 0.1 % error = [f '(x)]1 - [f '(x)]2 [f '(x)]1
= 0.166% (> 0.1% and < 1%)Correct Option: C
f(x) = cos(x)
at x = π rad = π × 180 = 30° 6 6 π [f '(x)]2 = -sin30° = -1 .......(1) 2 f '(x)= cos(x + h) - cos(x - h) 2h ∵ x = π = 30° 6 h = 0.1 = 28° π [f '(x)]2 = cosx . cosh - sinx . sinh - cosx.cosh - sinx.sinh 2h = - 2sinx . sinh = - 2sin30° × sin(18/ π) = -0.499 2h 2 × 0.1 % error = [f '(x)]1 - [f '(x)]2 [f '(x)]1
= 0.166% (> 0.1% and < 1%)
- Three cards were drawn from a pack of 52 cards. The probability that they are a king, a queen, and a jack is
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ρ = 4C1 × 4C1 × 4C1 52C3 ρ = 16 5525
Correct Option: A
ρ = 4C1 × 4C1 × 4C1 52C3 ρ = 16 5525
- The probability that a screw manufactured by a company is defective is 0.1. The company sells screws in packets containing 5 screws and gives a guarantee of replacement if one or more screws in the packet are found to be defective. The probability that a packet would have to be replaced is ___.
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we require P(x ≥ 1) = 1 – P(x = 0)
= 1 – 5C0 (0.1)0 (0.9)5
= 0.4095 ≈ 0.41Correct Option: C
we require P(x ≥ 1) = 1 – P(x = 0)
= 1 – 5C0 (0.1)0 (0.9)5
= 0.4095 ≈ 0.41
- The probability of obtaining at least two "SIX" in throwing a fair dice 4 times is
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Let P be the probability that six happens in a fair dice,
∴ p = 1 , q = 5 6 6
Let X, be the number of times ‘six’ happens Probability of obtaining atleast two ‘six’ in throwing a fair dice 4 times is
= 1 – {P (X = O) + P (X = 1)}
= 1 – {4C0 P°q4 + 4C1, P’q3}Correct Option: B
Let P be the probability that six happens in a fair dice,
∴ p = 1 , q = 5 6 6
Let X, be the number of times ‘six’ happens Probability of obtaining atleast two ‘six’ in throwing a fair dice 4 times is
= 1 – {P (X = O) + P (X = 1)}
= 1 – {4C0 P°q4 + 4C1, P’q3}