Paragraph: A line PQ parallel to the diagonal BD of a square A B C D with side length ' a ' unit is drawn at a distance x from the verter A , where x [0, 2 a| cuts the adjacent sides at P and Q . Let f(x) be the area of the segment of a square cut off by PQ. with A as one of the vertex. Question: For a=2 , the domain of the function (x)= f⁻¹(x)-f(x) is/are

Step-by-step solution

Two cases are possible for PQ w.r.L diagonal OA Case -1 : When x=A R AS i.e. x [0, a 2 ] Triangle ARP is an isosceles right angle triangle, Hence A P= 2 x=A Q Area of segment i.e. f(x)= 1 2 ( 2 x)^2=x^2, x [0, a 2 ] Case -II: When x= AR AS i.e. x [ a 2 , 2 a ] C P=C Q= 2 ( 2 a-x) Area of segment f(x)=a^2-( 2 a-x)^2 aligned & x [ a 2 , 2 a ] & i.c. f: [0, 2 a] |0, a^2 | aligned f(x)= cases x^2, & 0 x a 2 a^2-( 2 a-x)^2, & a 2 x a cases When 0 x a 2 , f(x)=x^2, 0 f(x) a^2 2 Put x=f⁻¹(x) (f⁻¹(x) )^2=x, 0 x a^2 2 , 0 f