Question:
There exist a function f (x), satisfying $f (0) = 1, f'(0) = -1, f (x) > 0$ for all x, and
There exist a function f (x), satisfying $f (0) = 1, f'(0) = -1, f (x) > 0$ for all x, and
Updated On: Jun 14, 2022
- f ''(x) > 0 for all x
- -1 < f " (x) < 0 for all x
- - 2 $\le$ f "(x) $\le$ -1 for all
- f " (x) < -2 for all x
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The Correct Option is A
Solution and Explanation
$f\left(x\right) =e^{-x}$ is one such function.
Here $ f\left(0\right) = 1 ,f'\left(0\right) = - 1, f\left(x\right) > 0, \forall x $
$\therefore f" > 0 \forall x $
Here $ f\left(0\right) = 1 ,f'\left(0\right) = - 1, f\left(x\right) > 0, \forall x $
$\therefore f" > 0 \forall x $
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Concepts Used:
Continuity
A function is said to be continuous at a point x = a, if
limx→a
f(x) Exists, and
limx→a
f(x) = f(a)
It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is undefined or does not exist, then we say that the function is discontinuous.
Conditions for continuity of a function: For any function to be continuous, it must meet the following conditions:
- The function f(x) specified at x = a, is continuous only if f(a) belongs to real number.
- The limit of the function as x approaches a, exists.
- The limit of the function as x approaches a, must be equal to the function value at x = a.