Find the derivative of `y = ln(2x^3 − x)^2`

Answer

`a rarr r; b rarr p,q; c rarr q,r; d rarr p,s`

Explanation

a. The radical axis of `x^(2)+y^(2)+2a_(1)x+b=0` and `x^(2)+y^(2)+2a_(2)x=b=0` is `(a_(1)-a_(2))x=0` or `x=0` . It must touch both the circles.
Solving it with one of the circles, we get
`y^(2)+b=0` or `b le 0`
b. The radical axis `x^(2)+y^(2)+2a_(1)x+b=0` and `x^(2)+y^(2)+2a_(2)y+b=0` is `a_(1)x-a_(2)y=0`
Solving it with one the circles, we have
`x^(2)=((a_(1))/(a_(2)))^(2)x^(2)+2a_(1)x+b=0`
This equation must have equal roots. Hence,
`4a_(1)^(2)-4b[1+((a_(1)^(2))/(a_(2)^(2)))]=0`
`a_(1)^(2)-b[1+(a_(1)^(2))/(a_(2)^(2))]=0`
Options p and q satisfy this condition.
c. If the straight line `a_(1)x-by+b^(2)=0` touches the circle `x^(2)+y^(2)=a_(2)x=by`, then
`(|a_(1)(a_(2))/(2)-b(b)/(2)+b^(2)|)/(sqrt(a_(1)^(2)+b^(2)))=sqrt((a_(2)^(2))/(4)+(b^(2))/(4))`
`(|a_(1)a_(2)+b^(2)|)/(sqrt(a_(1)^(2)+b^(2)))=sqrt(a_(2)^(2)+b^(2))` ltbgt or `a_(1)^(2)a_(2)^(2)+2b^(2)a_(1)a_(2)+b^(4)=a_(1)^(2)a_(2)^(2)+a_(1)^(2)b^(2)+a_(2)^(2)b^(2)+b^(4)`
i.e., `b^(2)=0` or `2a_(1)a_(2) =a_(1)^(2)+a_(2)^(2)`
Options q and r satisfy this condition.
d. The line `3x+4y-4=0` touches the circle
`(x-a_(1))^(2)+(y-a_(2))^(2)=b^(2)`. Then,
`(|3a_(1)+4a_(2)-4|)/(5)=b`