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
B.
Given, f : - 2, 2 → R fx = 1 + cx - 1 - cxx for - 2 ≤ x ≤ 0x + 3x + 1 for 0 ≤ x ≤ 2Now, LHL = limx→0 -fx= limh→0 1 - ch - 1 + ch - h × 1 - ch + 1 + ch1 - ch +1 + ch= limh→0 1 - ch - 1 + ch- h1 - 0 + 1 + 0= limh→0 - 2ch- h1 + 1 = cand RHL = limx→0 +fx= limh→0 f0 + h = limh→0 0 + h + 30 + h + 1limh→0 h + 3h + 1 = 0 + 30 + 1 = 3Since, f is continuous ar x = 0∴LHL = RHL ⇒ c = 3
If fx = xtan-1x, then limx→1fx - f1x - 1 = ?
π + 34
π4
π + 14
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