The value of limx→11 - x . tanπx2 will be
π2
2π
π
Let f(x) = x2 - 4x + 3x2 + 2x - 3, x ≠ 1k , x = 1 If f(x) is continuous at x = 1, then the value of k will be
1
12
- 1
- 12
Let f(x) = xn . sin1x, x ≠ 00 , x = 0 Then, f(x) is differentiable at x = 0, if
n ∈ 0, 1
n ∈ 1, 2
n ∈ 1, ∞
n ∈ - ∞, ∞
In which interval the function fx = log105x - x24 is defined ?
[1, 4]
[0, 5)
(0, 1)
(- 1, ∞)
limx→0x . 2x - x1 - cosx equals
log2
12log2
2log2
None of these
Let f(2) = 4 and f'(2)= 1. Then, limx→2xf2 - 2fxx - 2 is given by
2
- 2
- 4
3
A.
limx→2xf2 - 2fxx - 2Using L' Hospital's rule,limx→2f2 - 2f'x1 = f2 - 2f'2 = 4 - 2 = 2
If y = logsinxtanx, then dydxπ4 is equal to
4log2
- 4log2
If y = logexx - 22 for x ≠ 0, 2, then y'(3) is equal to
13
23
43
sin-11 + x22x is
continuous but not differentiable at x = 1
differentiable at x = 1
neither continuous nor differentiable at x = 1
continuous everywhere
If x2 + y2 = atan-1yx, a > 0, then d2ydx2 at x = 0 is
0
2ae- π2
- 2ae- π2