The general solution of the differential equation(D2– 4D+ 4)y = 0 is of form
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Solution
Given (D2– 4D +4) y = 0
⇒f(D) y = 0 where,
f(D) = D2– 4
D + 4f(D) = 0
∴ D2 – 4D + 4 = 0
⇒(D – 2)2 = 0
∴ D = 2, 2
∴ yc= (c1+ c2x) e2x
The box 1 contains chips numbered 3, 6, 9, 12 and 15.The box 2 contain chips numbered 6, 11, 16, 21 and 26. Two chips, one frame each box are drawn at random.The numbers written on these chips are multiplied.The probability for the product to be an even number is
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Solution
n(s)=5c1+x5c1=25
Let E be the went of picking one chip from each box such that product of numbers on chips is even
∴ n (E) = (2 × 5) + (3 × 3)
= 19
∴ prob = \(\frac{19}{25}\)
The equation x3 – x2+ 4x – 4 = 0 is to be solved using Newton-Raphson method. If x = 2 taken as the initial approximation of the solution, then the next approximate using this method, will be
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Solution
QUESTIONS CARRY 2 MARKS EACH
If matrix X =\(\begin{bmatrix} a &1 \\ -a^{2}+a-1 & 1-a \end{bmatrix}\) and X2– X + I= 0, then inverse of X is
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Solution
The vehicle/carrier on which sensors are borne is known as
If the standard deviation of the spot speed of vehicles in a highway is 8.8 kmph and the mean speed of the vehicle is 3.3 kmph, the co-efficient of variation in speed is
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Solution
Given the sight distance as 120 m, the height of driver’s eye as 1.5m and height of object is 0.15m.Grade difference of international gradient 0.09. The required length of summit parabolic curve is
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Solution
A vehicle travelling on dry level pavement at 80 kmph had the brakes applied. The vehicle travelled 76.5 m before stopping. What is the co-efficient that has developed?
Aeration of water is done to remove