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  • Consider a curve  in the first quadrant as shown in the figure. Let the area <img alt="\mathrm{A}_{1}" src="/la

Consider a curve \mathrm{ y=y(x)} in the first quadrant as shown in the figure. Let the area \mathrm{A}_{1} is twice the area \mathrm{A}_{2}. Then the normal to the curve perpendicular to the line \mathrm{2 x-12 y=15} does NOT pass through the point.

Option: 1

\left ( 6,21 \right )


Option: 2

\left ( 8,9 \right )


Option: 3

\left ( 10,-4 \right )


Option: 4

\left ( 12,-15 \right )


Answers (1)

best_answer

\begin{aligned} & \mathrm{A_{1}+A_{2}=x y-8} \\ & \mathrm{\Rightarrow A_{1}+\frac{A_{1}}{2}=x y-8 }\\ & \mathrm{ \Rightarrow \quad A_{1}=\frac{2}{3}(x y-8) }\\ & \mathrm{ \Rightarrow \int_{4}^{x} f(x) d x=\frac{2}{3} x \cdot f(x)-\frac{16}{3} }\\ & \text { Differentiate wir.t. } \mathrm{x} \\ &\mathrm{ f(x)=\frac{2}{3} f(x)+\frac{2}{3} x f^{\prime}(x)} \\ & \Rightarrow \quad\mathrm{ \frac{f(x)}{3}=\frac{2}{3} x f^{\prime}(x)} \\ & \Rightarrow \quad \mathrm{\frac{f^{\prime}(x)}{f(x)}=\frac{1}{2 x}} \\ & \Rightarrow \mathrm{2 \int \frac{f^{\prime}(x)}{f(x)} d x=\int \frac{d x}{x}} \end{aligned}

\therefore \text{option {C}}

Posted by

Pankaj

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