Question

If $f\left(x\right)={\left(\frac{x}{2}\right)}^{10}$, then $f\left(1\right)+\frac{f^{\prime }\left(1\right)}{1!}+\frac{f^{\prime \prime }\left(1\right)}{2!}+....+\frac{f^{\left(10\right)}\left(1\right)}{10!}$ is equal to

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

Answer 1

Answer: (A)

$1$

Given, $f\left(x\right)={\left(\frac{x}{2}\right)}^{10}=\frac{1}{2^{10}}.x^{10}$

On differentiating w.r.t $x$ we get

$f^{\prime }\left(x\right)=\frac{1}{2^{10}}.10x^9$

again differentiating w.r.t $x$ we get

$f^{\prime \prime }\left(x\right)=\frac{1}{2^{10}}.9x^8$

similarly

$f^{10}\left(x\right)=\frac{1}{2^{10}}10!$

The given sum

$=\frac{1}{2^{10}}[1+\frac{10}{1!}+\frac{10.9}{2!}+\frac{\mathrm{10.9.8}}{3!}+.....+\frac{10!}{10!}]$

$=\frac{1}{2^{10}}[{{}^{10}C}_0+{{}^{10}C}_1+{{}^{10}C}_2+...+{{}^{10}C}_{10}]=\frac{2^{10}}{2^{10}}=1$