Let a→, b→ and c→ be three vectors. Then, scalar triple product a→ b→ c→ is equal to :
b→ a→ c→
a→ c→ b→
c→ b→ a→
b→ c→ a→
∫tan-1x31 + x2dx is equa to :
3tan-1x2 + c
tan-1x44 + c
tan-1x4 + c
None of these
The area between the parabola y = x2 and the line y = x is :
16 sq unit
13 sq unit
12 sq unit
∫0πxdx1 + sinx is equal to :
- π
π2
π
The solution of the differential equation sec2(x)tan(y))dx + sec2(y)tan(x))dy = 0 is :
tanytanx = c
tan2x tany= c
Cosine of the angle between two diagonals of a cube is equal to :
26
13
12
a→, b→ and c→ are three vectors with magnitude a→ = 4, b→ = 4, c→ = 2 and such that a→ is perpendicular to b→ + c→, b→, is perpendicular to c→ + a→ and c→ is perpendicular to a→ + b→. It follows that a→ + b→ + c→ is equal to :
9
6
5
4
B.
Given, a→ = 4, b→ = 4, c→ = 2 ...iand a→ . b→ + c→ = 0, b→ . c→ + a→ = 0and c→ . a→ + b→ = 0⇒ a→ . b→ + a→ . c→ = 0 ...ii b→ . c→ + b→ . a→ = 0 ...iii⇒ c→ . a→ + c→ . b→ = 0 ...ivOn adding Eqs. (ii), (iii) and (iv)2a→ . b→ + b→ . c→ + c→ . a→ = 0Now, considera→ + b→ + c→2 = a→2 + b→2 + c→2 + 2a→ . b→ + b→ . c→ + c→ . a→ = 42 + 42 + 22 + 0 = 36⇒ a→ + b→ + c→ = 6
If a→, b→, c→ are three non-coplanar vectors, then
a→ + b→ + c→ . a→ + b→ × a→ + c→ is :
0
2a→ b→ c→
- a→ b→ c→
a→ b→ c→
∫dxxx5 + 1 is equal to :
15logx5x5 + 1 + c
15logx5 + 1x5 + c
∫x + sinx1 + cosxdx is equal to
xtanx2 + c
xsec2x2 + c
logcosx2 + c