. Then
f(x) is continuous at x = 1
f(x) is not continuous at x = 1
f(x) is differentiable at x = 1
f(x) is not differentiable at x = 1
If f(x) =
then
does not exist
f is not continuous at x = 2
f is continuous but not differentiable at x = 2
f is continuous and differentiable at x = 2
If f is a real-valued differentiable function such that f(x)f' (x) < 0 for all real x, then
f(x) must be an increasing function
f(x) must be a decreasing function
must be an increasing function
must be a decreasing function
Rolle's theorem is applicable in the interval [- 2, 2] for the function
f(x) = x3
f(x) = 4x4
f(x) = 2x3 + 3
f(x) =
If f(x) and g(x) are twice differentiable functions on (0, 3) satisfying f"(x) = g''(c), f'(1) = 4g'(D) = 6, f(2) = 3, g(2) = 9, then f(1) - g(1) is
4
- 4
0
- 2
B.
- 4
According to question,
f"(x) = g"(x)
Integrating w.r.t. x, we get
f'(x) = g'(x) + C1
Put x = 1 ⇒ f'(1) = g'(1) + C1
4 = 6 + C,
C1 = - 2
f'(x) = g'(x) - 2
Again, integrating w.r.t.x, we get
For function , Rolle's theorem is
applicable, when
applicable, when
applicable, when
applicable, when
f(x) = the function f(x) is
increasing when x
strictly increasing when x > 0
strictly increasing at x = 0
not continuous at x = 0 and so it is not increasing when x > 0