Question

Given $A(0,0)$ and $B(x,y)$ with $x\in (0,1)$ and $y>0$. Let the slope of the line $AB$ equal to $m_1$. Point $C$ lies on the line $x=1$ such that the slope of $BC$ equal to $m_2$ where $0<m_2<m_1$. If the area of the $△ABC$ can be expressed as $(m_1-m_2)f\left(x\right)$, then the largest possible value of $f\left(x\right)$ is

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{1}{8}$

Answer 1

Answer: (D)

$\frac{1}{8}$

Let the coordinates of $C$ be $(1,c)$.

Then, $m_2=\frac{c-y}{1-x}$

$\Rightarrow m_2=\frac{c-m_{1}x}{1-x}$
$\Rightarrow m_2-m_{2}x=c-m_{1}x$
$\Rightarrow (m_1-m_2)x=c-m_2$
$\Rightarrow c=(m_1-m_2)x+m_2$ ......... $\left(i\right)$
Now area of $\Delta$ $ABC$ is
$\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ x& m_{1}x& 1\\ 1& c& 1\end{array}\right|=\frac{1}{2}[cx-m_{1}x]$
$=\frac{1}{2}|\left[\right((m_1-m_2)x+m_2)x-m_{1}x]|$

$=\frac{1}{2}|\left[\right(m_1-m_2)x^2+m_{2}x-m_{1}x]|$

$=\frac{1}{2}(m_1-m_2)(x-x^2)$ .... [$\because x-x^{2}in(0,1)$]
Hence, $f\left(x\right)=\frac{1}{2}(x-x^2)$
$\Rightarrow f(x{)}_{max}=\frac{1}{8}$ when $x=\frac{1}{2}$.