The focal length of a mirror is given by . In finding the values of u and v, the errors are equal to ' 'p'. Then, the relative error in f is
p1u + 1v
p21u - 1v
p1u - 1v
B.
Given, equation is 2f = 1v - 1u iDifferentiating the given equation, we have- 2f2df = - 1v2dv + - 1u2du= - p1v - 1u1v + 1u ∵ dvv = duu = p= - 2pf1v + 1u using eq (i)∴ dff = p1v + 1u
If u = logx3 + y3 + z3 - 3xyz, then x + y + zux + uy +uz = ?
0
x - y + z
2
3
∫ ex2 + sin2x1 + cos2xdx = ?
excotx + C
2exsec2x +C
excos2x + C
extanx + C
If ∫ x - sinx1 + cosxdx = xtanx2 + plogsecx2 + C, then p = ?
- 4
4
- 2
If ∫dxxlogx - 2logx - 3 = I + C, then I =?
1xloglogx - 3logx - 2
loglogx - 3logx - 2
loglogx - 2logx - 3
If ∫0bdx1 + x2 = ∫b∞dx1 + x2, then b = ?
tan-113
32
1
The approximate value of ∫13dx2 + 3x using Simpson's rule and dividing the interval [1, 3] into two equal parts is
13log115
107110
22110
119440
An integrating factor of the equation 1 + y + x2ydx + x + x3dy = 0 is
ex
x2
1x
x
The solution of the differential equationdydx - 2ytan2x = exsec2x is
ysin2x = ex + C
ycos2x = ∫ex . 1dx + C
y = excos2x + C
ycos2x + ex = C
If 1 + ix - i2 + i + 1 + 2iy + i2 - i = 1, then x, y =?
73, - 715
73, 715
75, - 715
75, 715