Let f be any continuously differentiable function on [a, b] and twice differentiable on (a, b) such that f(a) = f"(a) = 0 and f(b) = 0. Then,
f''(a) = 0
f'(x) = 0 for some
Let f : be such that f(2x - 1) = f(x) for all . If f is continuous at x = 1 and f(1) = 1, then
f(2) = 1
f(2) = 2
f is continuous only at x = 1
f is continuous at all points
Let f(x) be a differentiable function in [2, 7]. If f(2) = 3 and f'(x) 5 for all x in (2, 7), then the maximum possible value of f(x) at x = 7 is
7
15
28
14
The function , where denotes the greatest integer x, is
continuous for all values of x
discontinuous at x =
not differentiable for some values of x
discontinuous at x = - 2
Let R be the set of all real numbers and f : [- 1, 1] R be defined by
Then,
f satisfies tile conditions of Rolle's theorem on [-1, 1]
f satisfies the conditions of Lagrange's mean value theorem on [-1, 1]
f satisfies the conditions of Rolle's theorem on [0, 1]
f satisfies the conditions of Lagrange's mean value theorem on [0, 1]
D.
f satisfies the conditions of Lagrange's mean value theorem on [0, 1]
Given,
Continuity at x = 0,
Hence, f(x) is continuous for all values of x.
Differentiability at x = 0
Thus, f( x) is not differentiable at x = 0.
Now, f(0) = 0 and f(1) = 1 sin(1)
, Rolle's theorem is not satisfy.
Hence, Lagrange's mean value theorem is satisfy on [0, 1].
Suppose that f(x) is a differentiable function such that f'(x) is continuous, f'(0) = 1 and f''(0) does not exist. Let g(x) = xf'(x), Then,
g'(0) does not exist
g'(0) = 0
g'(0) = 1
g'(0) = 2
Applying Lagrange's Mean Value Theorem for a suitable function f(x) in [0, h], we have f(h) = f(0) + hf'(), . Then, for f(x) = cos(x), the value of is
1
0
1/2
1/3