
If the function $f\left( x \right)$ defined by $f\left( x \right) = \dfrac{{{x^{100}}}}{{100}} + \dfrac{{{x^{99}}}}{{99}} + ..... + \dfrac{{{x^2}}}{2} + x + 1$ , then $f'\left( 0 \right) = $
A) $100f'\left( 0 \right)$
B) 1
C) 100
D) None of these
Answer
577.2k+ views
Hint:
The function $f\left( x \right)$ is defined by $f\left( x \right) = \dfrac{{{x^{100}}}}{{100}} + \dfrac{{{x^{99}}}}{{99}} + ..... + \dfrac{{{x^2}}}{2} + x + 1$.
To find the value of $f'\left( 0 \right)$ , firstly find the value of $f'\left( x \right)$.
After that, substitute \[x = {\text{0}}\] , in the value of $f'\left( x \right)$.
Thus, find the value of $f'\left( 0 \right)$.
Complete step by step solution:
Here, the function $f\left( x \right) = \dfrac{{{x^{100}}}}{{100}} + \dfrac{{{x^{99}}}}{{99}} + ..... + \dfrac{{{x^2}}}{2} + x + 1$ .
We are asked to find the value of $f'\left( 0 \right)$ .
To find the value of $f'\left( 0 \right)$ , we firstly need to find the value of $f'\left( x \right)$ .
\[
\Rightarrow \dfrac{d}{{dx}}f\left( x \right) = \dfrac{d}{{dx}}\left( {\dfrac{{{x^{100}}}}{{100}} + \dfrac{{{x^{99}}}}{{99}} + ..... + \dfrac{{{x^2}}}{2} + x + 1} \right) \\
\Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {\dfrac{{{x^{100}}}}{{100}}} \right) + \dfrac{d}{{dx}}\left( {\dfrac{{{x^{99}}}}{{99}}} \right) + ..... + \dfrac{d}{{dx}}\left( {\dfrac{{{x^2}}}{2}} \right) + \dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( 1 \right) \\
\Rightarrow f'\left( x \right) = \dfrac{{100{x^{100 - 1}}}}{{100}} + \dfrac{{99{x^{99 - 1}}}}{{99}} + ..... + \dfrac{{2{x^{2 - 1}}}}{2} + 1 + 0 \\
\Rightarrow f'\left( x \right) = {x^{99}} + {x^{98}} + ..... + x + 1 \\
\]
Thus, we get \[f'\left( x \right) = {x^{99}} + {x^{98}} + ..... + x + 1\] .
Now, to get the value of $f'\left( 0 \right)$ , by substituting \[x = 0\] , in $f'\left( x \right)$ .
$\Rightarrow f'\left( 0 \right) = {\left( 0 \right)^{99}} + {\left( 0 \right)^{98}} + ..... + 0 + 1$
$ = 0 + 0 + ...... + 0 + 1 \\
=1 $
Hence, $f'\left( 0 \right) = 1$ .
So, option (B) is correct.
Note:
Here, students should understand the question properly and then carry forward to solve it to avoid mistakes. The differentiation of each and every step must be done carefully and proceed step-wise.
Also, the substitution of \[x = 0\], must be done carefully in $f'\left( x \right)$, to get the required answer without any error in it.
The function $f\left( x \right)$ is defined by $f\left( x \right) = \dfrac{{{x^{100}}}}{{100}} + \dfrac{{{x^{99}}}}{{99}} + ..... + \dfrac{{{x^2}}}{2} + x + 1$.
To find the value of $f'\left( 0 \right)$ , firstly find the value of $f'\left( x \right)$.
After that, substitute \[x = {\text{0}}\] , in the value of $f'\left( x \right)$.
Thus, find the value of $f'\left( 0 \right)$.
Complete step by step solution:
Here, the function $f\left( x \right) = \dfrac{{{x^{100}}}}{{100}} + \dfrac{{{x^{99}}}}{{99}} + ..... + \dfrac{{{x^2}}}{2} + x + 1$ .
We are asked to find the value of $f'\left( 0 \right)$ .
To find the value of $f'\left( 0 \right)$ , we firstly need to find the value of $f'\left( x \right)$ .
\[
\Rightarrow \dfrac{d}{{dx}}f\left( x \right) = \dfrac{d}{{dx}}\left( {\dfrac{{{x^{100}}}}{{100}} + \dfrac{{{x^{99}}}}{{99}} + ..... + \dfrac{{{x^2}}}{2} + x + 1} \right) \\
\Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {\dfrac{{{x^{100}}}}{{100}}} \right) + \dfrac{d}{{dx}}\left( {\dfrac{{{x^{99}}}}{{99}}} \right) + ..... + \dfrac{d}{{dx}}\left( {\dfrac{{{x^2}}}{2}} \right) + \dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( 1 \right) \\
\Rightarrow f'\left( x \right) = \dfrac{{100{x^{100 - 1}}}}{{100}} + \dfrac{{99{x^{99 - 1}}}}{{99}} + ..... + \dfrac{{2{x^{2 - 1}}}}{2} + 1 + 0 \\
\Rightarrow f'\left( x \right) = {x^{99}} + {x^{98}} + ..... + x + 1 \\
\]
Thus, we get \[f'\left( x \right) = {x^{99}} + {x^{98}} + ..... + x + 1\] .
Now, to get the value of $f'\left( 0 \right)$ , by substituting \[x = 0\] , in $f'\left( x \right)$ .
$\Rightarrow f'\left( 0 \right) = {\left( 0 \right)^{99}} + {\left( 0 \right)^{98}} + ..... + 0 + 1$
$ = 0 + 0 + ...... + 0 + 1 \\
=1 $
Hence, $f'\left( 0 \right) = 1$ .
So, option (B) is correct.
Note:
Here, students should understand the question properly and then carry forward to solve it to avoid mistakes. The differentiation of each and every step must be done carefully and proceed step-wise.
Also, the substitution of \[x = 0\], must be done carefully in $f'\left( x \right)$, to get the required answer without any error in it.
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