I AM CERTAIN THAT NO ONE WILL ANSWER THIS QUESTION...
Suppose that l be a line through the points (0,3)and (5,-2) and tangent to the graph of the function \[F(x)=\frac{ x^2-p }{ 3x+1 }\] where ‘p’ is real number. What is the equation of a line parallel to l and tangent to the graph of F.
A) 4x+4y+7 = 0 B) 9x+9y+37 = 0 C) x+y-3 = 0 D) 4x+6y-37 = 0
solve y=3-x=F(x)=....
u get two roots of x, equate them because the line is tangent and can have only one intersection point.
from this i got p=-7.
now answer can only be either A or B
so solving both of them with F(x), gave only B) option with one intersection point, A) doesn't even intersect.
First of all what you have to do is finding l so as to get p \[l:\frac{ y-3 }{ x-0 }=\frac{ -2-3 }{ 5-0 }\] \[l:y=-x+3\]thus \[F(x)=\frac{ x^2-p }{ 3x+1} =-x+3=y\]simplifying the above expression \[4x^{2}-8x-p-3=0\]Now since l is tangent to F(x), the quadratic equation above has only one solution. That means \[b^{2}-4ac=0\] \[(-8)^{2}-4(4)(-p-3)=0\]thus \[(-8)^{2}-4(4)(-p-3)=0\]\[p=-7\]Now we get F(x) to be \[F(x)=\frac{ x^2+7 }{ 3x+1 }\]Therefore lets solve F’(x)=-1 so as to get the coordinate of the unknown line. \[F'(x)=\frac{ 2x(3x+1)-3(x^{2}+7) }{ (3x+1)^2 }=-1\] \[3x^{2}+2x-21=-9x^{2}-6x-1\] \[12x^{2}+8x-20=0\]thus \[x=1 or x=\frac{ -5 }{ 3 }\]and \[F(1)=2 and F(\frac{ -5 }{ 3 })=\frac{ -22 }{ 9 }\]Thus we got this two points (1,2) and (-5/3,-22/9). The two possible lines therefore are \[\frac{ y-2 }{ x-1 }=-1{\rightarrow}y=-x+3\]w/c is the given line l. And \[\frac{ y-(-\frac{ 22 }{9 }) }{ x-(-\frac{ 5 }{ 3 })}=-1{\rightarrow}9y+9x+37=0\]w/c is the answer.