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ASSIGNMENT
| PROGRAM | BCA |
| SEMESTER | 1 |
| SUBJECT CODE & NAME | BCA 1030- BASIC MATHEMATICS |
| CREDIT | 4 |
| BK ID | B0947 |
| MAX.MARKS | 60 |
Note: Answer all questions. Kindly note that answers for 10 marks questions should be approximately of 400 words. Each question is followed by evaluation scheme.
Q.1 (i) Express 7920 in radians and (7π/12) c in degrees.
Answer: (i) The conversion is 180O= π radian
So 79200 = (7920*3.14)/180 = 138.247 radians
(7π/12) c
(ii) Prove that (tan θ + sec θ – 1)/ (tan θ + sec θ +1) = Cos θ / (1-sin θ) = (1+sin θ)/ Cos θ)
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
Q.2 (i) y= xm/n, m,n being integers, n>0 find dy/dx
Solution: Let y= xm/n
Let dy be the increment in y corresponding to the increment dx in x.
(ii) Differentiate log (2x+3) from first principle.
Solution: First principles is also known as “delta method”, since many texts use Δx (for “change in x) and Δy (for “change in y”). This makes the algebra appear more difficult, so here we use h for Δx instead. We still call it “delta method”.
We can approximate this value by taking a point somewhere near to P(x, f(x)), say Q(x + h, f(x + h)).
Putting this together, we can write the slope of the tangent at P as:
Q.3 Evaluate ò2cosx+3sinx/4cosx+5sinx dx = I
Solution:
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
Q.5 (i) If a = cos q + i sin q, 0<q<2P prove that 1+a/1-a = i cot q/2
Solution: Given a = cos q + i sin q, 0<q<2P
a = sin(P/2-q)+cos(P/2-q)
(ii) If x+iy = Öa+ib/c+id prove that (x2 + y2) = a2+b2/c2+d2
Solution:
Given value is x+iy = Öa+ib/c+id i = iota
Squaring both sides we get i2 = -1 and i4 = 1
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
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