Question

In the square $ABCD$ with side $AB=2$, two points $M$ and $N$ are on the adjacent sides of the square such that $MN$ is parallel to the diagonal $BD$. If $x$ is the distance of $MN$ from the vertex $A$ and $f(x)=$ Area $(△AMN)$, then range of $f(x)$ is :

• A. $(0,\sqrt{2}]$
• B. $(0,2]$
• C. $(0,2\sqrt{2}]$
• D. $(0,2\sqrt{3}]$

Answer 1

Answer: (B)

$AP=x;MN=y;BD=2\sqrt{2}$
hence, $\frac{y}{2\sqrt{2}}=\frac{2\sqrt{2}-x}{\sqrt{2}}\Rightarrow {\Delta }^{\prime }$ 's CNM and CDB are similar $y=2(2\sqrt{2}-x)$
$f(x)=\frac{xy}{2}=x(2\sqrt{2}-x)=2-(x-\sqrt{2}{)}^2$