Question

The least integral value of $f(x)=\frac{(x-1{)}^7+3(x-1{)}^6+(x-1{)}^5+1}{(x-1{)}^5}\forall x>1$, is equal to

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (D)

Let $x-1=t>0$
$$\begin{gathered} \therefore f(x)=\frac{t^7+3t^6+t^5+1}{t^5}=t^2+3t+1+\frac{1}{t^5} \\ <br/>\Theta \text{ A.M. }\ge GM. \end{gathered}$$
$\Rightarrow \frac{t^2+t+t+t+1+\frac{1}{t^5}}{6}\ge \sqrt[6]{t^2\cdot t\cdot t\cdot t\cdot 1\cdot \frac{1}{t^5}}]$
$\Rightarrow f(x)\ge 6$
$\therefore \text{ least integral value }=6$