limx→0 tanx - sinxx2 = ?
0
1
12
- 12
If f(x) = x + sinx for x ∈ - π2, π2, then its left hand derivative at x = 0 is
- 1
- 2
- 3
If u = ux, y = siny + ax - y + ax2, then it implies
uxx = a2 . uyy
uyy = a2uxx
uxx = - a2 . uyy
uyy = - a2uxx
limx→∞x + 6x + 1x + 4 = ?
e4
e6
e5
e
The coordinates of the point P on the curve x = aθ + sinθ, y = a1 - cosθ, where the tangent is inclined at an angle π4 to x-axis, are
aπ4 - 1, a
aπ2 + 1, a
aπ2, a
(a, a)
If f : - 2, 2 → R is defined by fx = 1 + cx - 1 - cxx for - 2 ≤ x ≤ 0x + 3x + 1 for 0 ≤ x ≤ 2continuous on - 2, 2, then c is equal to
23
3
32
If fx = xtan-1x, then limx→1fx - f1x - 1 = ?
π + 34
π4
π + 14
π + 24
D.
Given, f(x) = xtan-1x thenlimx→1 fx - f1x - 1 00 form= limx→1 f'x - 01Using L'Hospital rule= limx→1x1 + x2 + tan-1x= 112 + 12 + tan-11 = 12 + π4= 2 + π4 or π + 24
The value of c in the Lagrange's mean value theorem for f(x) = x - 2 in the interval [2, 6] is
92
52
4
If gx = xx for x> 2, then limx→2 gx - g2x - 2 = ?
1/2
limx→π22x - πcosx = ?
5