If `f(x)=(ax+b)/(cx+d)` `ad-bc!=0` is inverse of itself then 1. ad+bc=0 2. ac=bd 3.ac+bd=0 4.a+d=0

Let f(x)=(ax + b )/(cx+d) . Then the fof (x)=x , provided that : (a!=0, b!= 0, c!=0,d!=0)

Let f(x)=(ax + b )/(cx+d) . Then the fof (x)=x , provided that : (a!=0, b!= 0, c!=0,d!=0)

If {:A=[(a,b),(c,d)]:} such that ad - bc ne 0 , then A^(-1) , is

If (a^2+b^2)x^2+2(ac+bd)x+c^2+d^2=0 has no real root

Statement-1: If ad-bc ne 0 , then f(x)=(ax+b)/(cx+d) cannot attain the value {(a)/(c )} . Statement-2: The domain of the function g(x)=(b-dx)/(cx-a) is R-{(a)/( c)}

a, b, c, d are integers such that ad + bc divides each of a, b, c and d.Prove that ad + bc = pm1

The function f: C -> C defined by f(x) = (ax+b)/(cx+d) for x in C where bd != 0 reduces to a constant function if

If the lines ax+by+c=0, bx+cy+a=0 and cx+ay+b=0 (a, b,c being distinct) are concurrent, then (A) a+b+c=0 (B) a+b+c=1 (C) ab+bc+ca=1 (D) ab+bc+ca=0

The system of linear equations ax+by=0,cx+dy=0 has a non trivial solution if (A) ad+bc=0 (B) ad-bc=0 (C) ad-bc,0 (D) ad-bc.0

Discuss the nature of roots of equation x^(2) +ax +b =0 and x^(2) +cx + d=0, where a,b,c, d are real numbers and ac = 2( b +d)