∫x - 1x + 1dx is equal to
2x2 + 1 + sin-1x + c
x2 - 1 - sin-1x + c
2x2 - 1 + sin-1x + c
x2 - 12 + sin-1x + c
The solution of the differential equation (1 + y) tan-1(x)dx + y(1 + x2)dy = 0 is
logtan-1xx + y1 + x2 = c
log1 + y2 + tan-1x2 = c
log1 + x2 + logtan-1y = c
tan-1x1 + y2 + c = 0
The value of ∫- 11logx - 1x + 1dx is
1
2
0
4
Considering four sub-intervals, the value of ∫042xdx by Simpson's rule is
648
653
6212
618
B.
Here, a = 0, b = 4, n = 4
h = b - an = 4 - 04 = 1
By Simpson's rule
∫022xdx = h3y0 + y4 + 4y1 + y3 + 2y2
= 131 + 16 + 42 + 8 + 24= 1317 + 40 + 8 = 653
If a→ = b→ = 1 and a→ + b→ = 3, then the value of 3a→ - 4b→ . 2a→ + 5b→ is
- 21
- 212
21
212
A unit vector in the plane of i^ + 2j^ + k^ and i^ + j^ + 2k^ and perpendicular to 2i^ + j^ + k^, is
j^ - k^
i^ + j^2
j^ + k^2
j^ - k^2
If a→ is perpendicular to b→ and c→, a→ = 2, b→ = 3, c→ = 4 and the angle between b→ and c→ is 2π3, then [a→ b→ c→] is equal to
43
63
123
183
If a→, b→ and c→ are perpendicular to b→ + c→, c→ + a→ and a→ + b→ respectively and if a→ + b→ = 6, b→ + c→ = 8 and c→ + a→ = 10 then a→ + b→ + c→ is equal to
52
50
102
10
If I1 = ∫sin-1xdx and I2 = ∫sin-11 - x2dx, then
I1 = I2
I2 = π2I1
I1 + I2 = π2x
I1 + I2 = π2
∫sinθ + cosθsin2θdθ is equal to
logcosθ - sinθ + sin2θ + c
logsinθ - cosθ + sin2θ + c
sin-1sinθ - cosθ + c
sin-1sinθ + cosθ + c