Let f(x) = x3 - x + p (0 ≤ x ≤ 2) where p is a constant. The value c of mean value theorem is
32
63
33
233
D.
Given, f(x) = x3 - x + p, 0 ≤ x ≤ 2
Here, a = 0, b = 2
f'c = fb - fab - a ...i⇒ f'c = f2 - f02 - 0∴ 3c2 - 1 = 23 - 2 + p - 02 - 0 + p2 - 0⇒ 3c2 - 1 = 62 = 3 ⇒ 3c2 = 4⇒ c = ± 23 = ± 233∴ c = 233 ∵ - 233 ∉ 0, 2
If the function fx = x2 - k + 2x + 2kx - 2 for x ≠ 22 for x = 2 is continuous at x = 2, then k is equal to
- 12
- 1
0
12
If xexy + ye- xy = sin2x, then dydx at x = 0 is
2y2 - 1
2y
y2 - y
y2 - 1
If y = tan-12x - 11 + x - x2, then dydx at x = 1 is equal to
23
1
If f(x) = cos-12cosx + 3sinx13, then [f'(x)]2 is equal to
1 + x
1 + 2x
2
If u = tan-11 - x2 - 1x and v = sin-1x, then dudv is equal to
1 - x2
- x
If y = 11 + x + x2, then dydx is equal to
y2(2 + 2x)
- 1 + 2xy2
1 + 2xy2
- y2(1 + 2x)
If g(x) is the inverse of f(x) and f'(x) = 11 + x3, then g'(x) is equal to
g(x)
1 + g(x)
1 + {g(x)}3
11 + gx3
If y = f(x2 + 2) and f'(3) = 5, then dydx at x = 1 is
5
25
15
10
Let, f(x) = x2 + bx + 7. If f'(5) = 2f'72, then the value of b is
4
3
- 4
- 3