Question

If n is such that $36\le n\le 72,$ then $X=\frac{n^2+2\sqrt{n}(n+4)+16}{n+4\sqrt{n}+4}$ satisfies?

• A. 20 < x < 54
• B. 23 < x < 58
• C. 25 < x < 64
• D. 28 < x < 59

Answer 1

The correct option is D 28 < x < 59

$x=\frac{n^2+2\sqrt{n}(n+4)+16}{n+4\sqrt{n}+4}$
Let $\sqrt{n}=t$
$\Rightarrow x=\frac{t^4+2t(t^2+4)+16}{t^2+4t+4}$
$=\frac{(t+2)(t^3+8)}{(t+2{)}^2}=\frac{t^3+8}{t+2}$
$\Rightarrow x=t^2-2t+4$ .......(1)
For $t=6tot=6\sqrt{2}$
$(40-12)<x<(72+4-12\sqrt{2})$
[Putting in equation (1)]
$\Rightarrow 28<x<76-12\sqrt{2}or28<x<59$