Question

If f(x)=${\left(\frac{x}{2}\right)}^{10},thenf\left(1\right)+\frac{f^{\prime }\left(1\right)}{\lfloor 1}+\frac{f\left(1\right)}{\lfloor 2}+\frac{f^{\prime }\left(1\right)}{\lfloor 3}+\dots +\frac{f^{\left(10\right)}\left(1\right)}{\lfloor 10}$ is equal to

• A. 1
• B. 10
• C. 11
• D. 512
• E. 1024

Answer 1

Answer: (A)

Given, $f(x)={\left(\frac{x}{2}\right)}^{10}=\frac{1}{2^{10}}\cdot x^{10}$
On differentiating w.r.t. $x$, we get
$f^{\prime }(x)=\frac{1}{2^{10}}10x^9$
Again, differentiating, we get
f^{\prime \prime}(x)=\frac{1}{2^{10}} 10 \cdot 9 x^{8}

Similar, $f^{(10)}=\frac{1}{2^{10}}10!$
$\therefore$ The given sum
$=\frac{1}{2^{10}}\left[1+\frac{10}{1!}+\frac{10\cdot 9}{2!}+\frac{10\cdot 9\cdot 8}{3!}+\dots +\frac{10!}{10!}\right]$W
$=\frac{1}{2^{10}}\left[{}^{10}{C}_0+{}^{10}{C}_1+{}^{10}{C}_2+\dots +{}^{10}{C}_{10}\right]$
$=\frac{2^{10}}{2^{10}}=1[\because {}^n{C}_0+{}^n{C}_1+\dots +C_n=2$