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ASSIGNMENT
| SUMMER | 2014 |
| PROGRAM | BCA |
| SEMESTER | 1 |
| SUBJECT CODE & NAME | BCA 1030- BASIC MATHEMATICS |
| CREDIT | 4 |
| BK ID | B0950 |
| MAX. MARKS | 60 |
Q.1 (i) Let A = {1, 2, 3, 4, 5, 6} and B = {2, 4, 6, 8}. Find A – B and B – A.
Answer: A-B = {1, 3, 5} , B-A = {8}
(ii) In a group of 50 people, 35 speak Hindi, 25 speak both English and Hindi and all the people speak at least one of the two languages. How many people speak only English and not Hindi?
How many people speak English?
Answer: n (A U B )= people who speak in either Hindi and English.
Given people speak at least one of the languages.
n (A U B) = 50.
Q.2 (i) Express 7920 in radians and (7π/12) c in degrees.
(ii) Prove that (tan θ + sec θ – 1)/ (tan θ + sec θ +1) = Cos θ / (1-sin θ) = (1+sin θ)/ Cos θ
Answer: (i) The conversion is 180O= π radian
So 79200 = (7920*3.14)/180 = 138.247 radians
(7π/12) c in degrees:-
1radian = 57.29577795
Degrees = radians * (180/pi)
= (7π/12)*180/pi = 105
(ii). Solution:-
(tan θ + sec θ – 1)/ (tan θ + sec θ +1) =(1+sin θ)/ Cos θ
If (tan θ + sec θ – 1)/ (tan θ + sec θ +1) = (1/cosθ)+(sinθ/cosθ)
If (tan θ + sec θ – 1)/ (tan
Q.3 (i) Define continuity of a point
(ii) Test the continuity of the function f where f is defined by f(x) = {x-2/|x-2| if x ≠ 2, 7 if x = 2.
Answer: (i) Definition of Continuity
Let a be a point in the domain of the function f(x). Then f is continuous at x=a if and only if
lim f(x) = f(a)
x –> a
A function f(x) is continuous on a set
Q.4 Solve dy/dx = (y+x-2)/(y-x-4).
Answer:dy/dx = (y+x-2)/(y-x-4) ——————————– (i)
Put y = vx
Diff w.r.t “x”
dy/dx = v.1+x.dx/dx
dy/dx = v+xdy/dx
Q.5 (i) a bag contains two red balls, three blue balls and five green balls.
Three balls are drawn at random. Find the probability that
- a) The three balls are of different colors’.
- b) Two balls are of the same color.
Let nCk = number of ways to pick up k items from a set of n items.
Of course you should already know that nCk+=+n%21%2F%28k%21%2A%28n-k%29%21%29 (*)
Bag consists of 2 red balls (R), 3 blue
Q.6 Solve: 2x + 3y + 4z = 20, x + y + 2z = 9, 3x + 2y + z = 10.
Answer: These equations are written as
[2 3 4 [20
1 1 2 = 9
3 2 1] 10]
AX = B
Where A = [2 3 4 , 1 1 2 , 3 2 1 ] X =[ X,Y,Z] ,B= [20,9,30]
Therefore |A| = Determinant of |A| = 5
Now we have to find the value of Δ1. So replace first column of A with the values of B and find Determinant.
Therefore Δ1 = 5
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