Given `A (0, 0)` and `B (x,y)`with `x epsilon (0,1)` and `y>0`. Let the slope of the line AB equals `m_1` Point C lies on the line `x= 1` such that the slope of BC equals `m_2` where `0< m_2< m_1` If the area of the triangle ABC can expressed as `(m_1- m_2)f(x)`, then largest possible value of `f(x)` is:

Options

A. 1
B. `1//2`
C. `1//4`
D. `1//8`

Answer

D

Explanation

Let the coordinates of C be (1,c). Then,
`m_(2) = (c-y)/(1-x)`
`"or " m_(2) = (c-m_(1)x)/(1-x)`
`"or " m_(2)-m_(2)x = c-m_(1)x`
`"or " (m_(1)-m_(2))x = c-m_(2)`
`"or " c = (m_(1)-m_(2))x + m_(2) " " (1)`
Now, the area of `DeltaABC` is
`(1)/(2)||{:(0,0,1),(x,m_(1)x,1),(1,c,1):}|| = (1)/(2) |[cx-m_(1)x]|`
`=(1)/(2)|[{(m_(1)-m_(2))x+m_(2)}x-m_(1)x]|`
`=(1)/(2)|[(m_(1)-m_(2))x^(2)+m_(2)x-m_(1)x]|`
`=(1)/(2)(m_(1)-m_(2))(x-x^(2)) " " [because x gt x^(2) "in" (0,1)]`
`"Hence, " f(x) = (1)/(2)(x-x^(2))`
`f(x)_(max) = (1)/(8) "when " x =(1)/(2)`