(3,128)
(13,23)
(12,3)
(2,18)
(y−y1)(x−x1)=f′(x1)y−y1=f′(x1)(x−x1)y=0, −yf′(x1)=x−x1 ⇒x=x1−y1f′(x1)A=(x1−y1f′(x1),0)x=0, y−y1=f′(x1)(−x1) ⇒y=y1−x1f′(x1)B=(0,y1−x1f′(x1))P divides AB in ratio 1:3 ∴x1=3(x1−y1f′(x1))4y1=y1−x1f′(x1)44y1=y1−x1f′(x1)f′(x1)=−3y1x1f′(x)=−3yxdydx=−3yxdyy=−3xdxlny=−3lnx+clny=lnx−3+cy=kx−3y(1)=1 ⇒k=1y=1x3. Option D satisfies the above function.