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Given \mathrm{A}(0,0) and \mathrm{B}(\mathrm{x}, \mathrm{y}) with \mathrm{x} \in(0,1) and \mathrm{y}>0.Let the slope of the line \mathrm{AB} equals \mathrm{m}_{1}.Point \mathrm{C} lies on the line \mathrm{x}=1 such that the slope of \mathrm{BC} equals \mathrm{m}_{2} where 0<\mathrm{m}_{2}<\mathrm{m}_{1}.If the area of the triangle \mathrm{ABC} can be expressed as \left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) f(\mathrm{x}), then the largest possible value of f(\mathrm{x}) is

Option: 1

1


Option: 2

1/2


Option: 3

1/4


Option: 4

1/8


Answers (1)

best_answer

Let the coordinates of \mathrm{C} be (1, \mathrm{c})



\mathrm{m}_{2}=\frac{\mathrm{c}-\mathrm{y}}{1-\mathrm{x}} ; \quad \mathrm{m}_{2}=\frac{\mathrm{c}-\mathrm{m}_{1} \mathrm{x}}{1-\mathrm{x}}
\mathrm{m}_{2}-\mathrm{m}_{2} \mathrm{x}=\mathrm{c}-\mathrm{m}_{1} \mathrm{x}
\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) \mathrm{x}=\mathrm{c}-\mathrm{m}_{2}
\mathrm{c}=\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) \mathrm{x}+\mathrm{m}_{2}\quad\cdots(1)

\mathrm{\text{now area of}\, \triangle \mathrm{ABC}=\frac{1}{2}\left|\begin{array}{ccc}0 & 0 & 1 \\ \mathrm{x} & \mathrm{m}_{1} \mathrm{x} & 1 \\ 1 & \mathrm{c} & 1\end{array}\right|=\frac{1}{2}\left[\mathrm{cx}-\mathrm{m}_{1} \mathrm{x}\right]=\frac{1}{2}\left|\left[\left(\left(\mathrm{~m}_{1}-\mathrm{m}_{2}\right) \mathrm{x}+\mathrm{m}_{2}\right) \mathrm{x}-\mathrm{m}_{1} \mathrm{x}\right]\right|}
                                            =\frac{1}{2}\left|\left[\left(\mathrm{~m}_{1}-\mathrm{m}_{2}\right) \mathrm{x}^{2}+\mathrm{m}_{2} \mathrm{x}-\mathrm{m}_{1} \mathrm{x}\right]\right|=\frac{1}{2}\left(\mathrm{~m}_{1}-\mathrm{m}_{2}\right)\left(\mathrm{x}-\mathrm{x}^{2}\right) \quad\left(\mathrm{x}>\mathrm{x}^{2} \text { in }(0,1)\right.

\text{Hence, }\: \left.\mathrm{f}(\mathrm{x})=\frac{1}{2}\left(\mathrm{x}-\mathrm{x}^{2}\right) ; \quad \mathrm{f}(\mathrm{x})\right]_{\max }=\frac{1}{8}$ when $\left.\mathrm{x}=\frac{1}{2}\right]


                                          
 

Posted by

Devendra Khairwa

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