∫exx2 + 1x + 12dx is equal to
exx + 1 + C
- exx + 1 + C
exx - 1x + 1 + C
exx + 1x - 1 + C
∫x + 1x1 + xexdx is equal to
log1 + xexxex + C
logxex1 + xex + C
log1 + xex + C
∫fxg'x - f'xgxfxgxloggx - logfxdx is equal to
loggxfx + C
12loggxfx2 + C
gxfxloggxfx + C
loggxfx - gxfx + C
∫0π4sinx + cosx3 + sin2xdx = ?
12log3
log2
log(3)
14log3
∫- 111 + x + x2 - 1 - x + x21 + x + x2 + 1 - x + x2dx = ?
3π2
π2
0
- 1
The area of the region described by(x, y)/x2 + y ≤ 1 and y ≤ 1 - x is
π2 - 23
π2 + 23
π2 + 43
π2 - 43
The solution of dydx + 1x = eyx2 is
2x = 1 +Cx2 ey
x = 1 +Cx2 ey
2x2 = 1 +Cx2 e - y
x2 = 1 +Cx2 e - y
The differential equation dydx = 1ax + by + c ,where a, b, c are all non-zero real numbers, is
linear in y
linear in x
linear in both x and y
homogeneous equation
Let D be the domain of a twice differentiable function f.For all x ∈ D, f"(x) + f(x) = 0 and f(x)= ∫g(x)dx + constant.If h(x) = f(x)2 + g(x)2 and h(0) = 5, then h(2015) - h(2014) is equal to
5
3
1
The foci of the ellipse x216 + y2b2 = 1 and the hyperbola x2144 - y281 = 125 coincide. Then, the value of b2 is
7
9
B.
We have, x216 + y2b2 = 1 ⇒ a = 4, b = b∴ e = 1 - b216 = 16 - b24Hence, the focus will be ± 16 - b2, 0Also, x2144 - y281 = 125⇒x214425 - y28125 = 1⇒ x21252 - y2952 = 1⇒ a = 125, b = 95∴ e = 1 + b2a2 = 1 + 812514425 = ± 1512Hence, the focus will be± 125 × 1512, 0 ie ±3, 0On compairing, we getb2 = 7