PARAGRAPH: Consider f ( x )= x ^2-6 x +4 x ^2+2 x +4 and g ( x )=(1+m) x ^2-2(1+3 m) x -2(1+m) where m is a parameter. QUESTION: If ( ⁻¹ f(x) )=k , has exactly two distinct real solutions then the integral value of ' k ' can be

Step-by-step solution

aligned & f(x)= x^2-6 x+4 x^2+2 x+4 =1- 8 x x^2+2 x+4 & f^ (x)=-8 [ (x^2+2 x+4 )-x(2 x+2) (x^2+2 x+4 )^2 ]=-8 [ -x^2+4 (x^2+2 x+4 )^2 ]= 8 (x^2-4 ) (x^2+2 x+4 )^2 & f^ (x)=0 x=2 or -2 & f(2)= 4-12+4 4+4+4 = -4 12 = -2 3 aligned aligned & Let y= 8 x x^2+2 x+4 & x^2 y+2(y-4) x+4 y=0 & x R D 0 & 4(y-4)^2-16 y^2 0 & (y-4)^2-(2 y)^2 0 & (3 y-4)(y+4) 0 aligned f(-2)= 4+12+4 4-4+4 =5 Hence range of f(x) is [- 1 3 , 5 ] the graph of y=f(x) is as shown Hence, - 1 3 f ( x ) 5 (i) obviously f is continuous and has y=1 as its