The focal length of a mirror is given by 2f = 1v - 1u. In finding the values of u and v, the errors are equal to ' 'p'. Then, the relative error in f is
p21u + 1v
p1u + 1v
p21u - 1v
p1u - 1v
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
A.
1 + ix - i2 + i + 1 + 2iy + i2 - i = 1⇒ 1 + ix - i2 - i4 - i2 + 1 + 2iy + i2 + i4 - i2 = 1⇒ 21 + ix - 2i - i1 + ix + i24 + 1 + 21 + 2iy + 2i + i1 + 2ix + i24 + 1 = 1⇒ 2 + 2i - i - i2x - 2i + i25 + 2 + 4i + i + 2i2y + 2i + i25 = 1⇒ 3 + ix - 2i - 1 + 5iy + 2i - 1 = 5⇒ 3 + ix + 5iy = 7⇒ 3x + ix + 5iy - 7 = 0⇒ 3x - 7 + x + 5yi = 0 + 0iOn compairing, we get3x - 7 = 0⇒ x = 73, x + 5y = 0⇒ y = - 715Hence, x, y = 73, - 715