Solution of the differential equation dydx + yx = sinx is :
xy + cosx = sinx + c
xy - cosx = sinx + c
xycosx = sinx + c
xy - cosx = cosx + c
The projection of the vector 2i^ + j^ - 3k^ on the vector i^ - 2j^ + k^ is
- 314
314
- 32
32
An unit vector perpendicular to the plane containing the vectors i^ - j^ + k^ and - i^ + j^ + k^ is
± i^ - j^2
i^ + k^2
± j^ - k^2
∓ i^ + j^2
D.
Let a = i^ - j^ + k^ and b = - i^ + j^ + k^And unit vector perpendicular to the plane a and b = ± a × ba × bNow, a × b = i^j^k^1- 11- 111 = i^- 1 - 1 - j^1 + 1 + k^1 - 1 = - 2i^ - 2j^ a × b = 22 + 22 = 22∴ a × ba × b = ∓ 2i^ + j^22 = ∓ i^ + j^2
If a, b and c are mutually perpendicular unit vectors, then a + b + c is equal to
3
a2 + b2 + c2/3
1
∫sin2x1 + cos2xdx =
- 12log1 + cos2x + c
2log1 + cos2x + c
12log1 + cos2x + c
- log1 + cos2x + c
∫ex1 + sinx1 + cosxdx =
extanx2 + c
extanx + c
ex1 + sinx1 - cosx + c
∫1 + tanxe- xcosxdx is equal to
e- xtanx + c
e- xsecx + c
exsecx + c
∫π4π2csc2xdx is equal to
- 1
0
12
∫0π4log1 + tanxdx is equal to
π8loge2
π4log2e
π4loge2
π8loge12
The area bounded by the parabola y2 = 4ax and the line x = a and x = 4a is
35a23
4a23
7a23
56a23