JEE Main 2017 — Application Of Derivatives Question with Solution

We have

$$\frac{(y-y_2)}{(x-x_1)}=f^{\prime }(x_1)$$

$$\Rightarrow y-y_1=f^{\prime }(x_1)(x-x_1)$$

$$\bullet$$ When y = 0: $$\frac{-y_1}{f^{\prime }(x_1)}=x-x_1$$

$$\Rightarrow x=x_1-\frac{y_1}{f^{\prime }(x_1)}$$

Therefore, point A is $$A\left(x_1-\frac{y_1}{f^{\prime }(x_1)},0\right)$$

$$\bullet$$ When x = 0: $$y-y_1=f^{\prime }(x).(-x_1)$$

$$\Rightarrow y=y_1-x_1{f}^{\prime }(x_1)$$

Therefore, point B is $$B(0,y_1-x_1{f}^{\prime }(x_1))$$

Point P divides AB in the ratio 1 : 3.

$$x_1=\frac{\left[3\left(x_1-\frac{y_1}{f^{\prime }(x_1)}\right)\right]}{4}$$

$$y_1=\frac{y_1-x_1{f}^{\prime }(x_1)}{4}$$

Therefore, $$4y_1=y_1-x_1{f}^{\prime }(x_1)$$

$$\Rightarrow f^{\prime }(x_1)=\frac{-3y_1}{x_1}\Rightarrow f^{\prime }(x)=\frac{-3y}{x}$$

Now, $$\frac{dy}{dx}=\frac{-3y}{x}\Rightarrow \frac{dy}{y}=\frac{-3dx}{x}$$

On integrating, we get

$$\ln y=-3\ln x+C\Rightarrow y=k{x}^{-3}$$

$$y(1)=1\Rightarrow k=1$$

$$y=\frac{1}{x^3}$$

When we substitute the values from the given options, only option (c) satisfies the above equation.