Question

If f(x)=$\left(\frac{x}{2}\right)^{10}, then\, f \left(1\right)+\frac{f '\left(1\right)}{\lfloor1}+\frac{f \left(1\right)}{\lfloor2}+\frac{f '\left(1\right)}{\lfloor3}+\ldots+\frac{f ^{\left(10\right)}\left(1\right)}{\lfloor10}$ is equal to

• A. 1
• B. 10
• C. 11
• D. 512

Answer 1

Answer: (A)

1

Answer 2

Given, $f(x)=\left(\frac{x}{2}\right)^{10}=\frac{1}{2^{10}} \cdot x^{10}$
On differentiating w.r.t. $x$, we get
$f^{\prime}(x)=\frac{1}{2^{10}} 10 x^{9}$
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 $=\frac{1}{2^{10}}\left[1+\frac{10}{1 !}+\frac{10 \cdot 9}{2 !}+\frac{10 \cdot 9 \cdot 8}{3 !}+\ldots+\frac{10 !}{10 !}\right]$W $=\frac{1}{2^{10}}\left[{ }^{10} C_{0}+{ }^{10} C_{1}+{ }^{10} C_{2}+\ldots+{ }^{10} C_{10}\right] $ $=\frac{2^{10}}{2^{10}}=1 \left[\because{ }^{n} C_{0}+{ }^{n} C_{1}+\ldots+C_{n}=2\right.$