Question:
The functions $f , g$ and $h$ satisfy the relations $f ^{'}\left(x\right)=g\left(x+1\right)$. Then $f ^{"}\left(2x\right)$ is equal to
The functions $f , g$ and $h$ satisfy the relations $f ^{'}\left(x\right)=g\left(x+1\right)$. Then $f ^{"}\left(2x\right)$ is equal to
Updated On: Jun 6, 2022
- $h\left(2x\right)$
- $4h\left(2x\right)$
- $h\left(2x-1\right)$
- $h\left(2x+1\right)$
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The Correct Option is A
Solution and Explanation
We have,
$ f^{\prime}(x)= g(x+1) $
$ \Rightarrow f^{\prime \prime}(x) =g^{\prime}(x+1) $
$\text { But } g^{\prime}(x)=h(x-1) $
$ \Rightarrow g^{\prime}(x+1) =h(x+1-1) $
$ =h(x) $
$ \therefore f^{\prime \prime}(x)=h(x) $
$ \Rightarrow f^{\prime \prime}(2 x) =h(2 x)$
$ f^{\prime}(x)= g(x+1) $
$ \Rightarrow f^{\prime \prime}(x) =g^{\prime}(x+1) $
$\text { But } g^{\prime}(x)=h(x-1) $
$ \Rightarrow g^{\prime}(x+1) =h(x+1-1) $
$ =h(x) $
$ \therefore f^{\prime \prime}(x)=h(x) $
$ \Rightarrow f^{\prime \prime}(2 x) =h(2 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.