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Summer 2015 ASSIGNMENT
PROGRAM-BCA
SEMESTER- I
CREDIT-4
BK ID-B0947
MAX. MARKS-60
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
Get fully solved assignment. Buy online from website
online store
or
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we will revert you within 2-3 hour or immediate
Charges rs
125/subject and rs 500/semester only.
if urgent then call us on 08791490301, 08273413412
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