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)={(x-4)/(|x-4|)+a ,x 4 Then f(x) is continous at x=4 when a=0,b=0 b. a=1,b=1 c. a=-1,b=1 d. a=-1,b=-1

The solution of (dy)/(dx)=(a x+h)/(b y+k) represent a parabola when (a) (a) a=0,b!=0 (b) a!=0,b!=0 (c) b=0,a!=0 (d) a=0,b in R

Let f(x)=a x^2+b x+c. Consider the following diagram. Then Fig c 0 a+b-c >0 a b c<0

Let f(x)=(1+b^2)x^2+2b x+1 and let m(b) the minimum value of f(x) as b varies, the range of m(b) is (A) [0,1] (B) (0,1/2] (C) [1/2,1] (D) (0,1]

if f(x) =ax +b and g(x) = cx +d , then f[g(x) ] - g[f(x) ] is equivalent to

Let f(x)=ax^(2)+bx+c , a ne 0 , a , b , c in I . Suppose that f(1)=0 , 50 lt f(7) lt 60 and 70 lt f(8) lt 80 . Number of integral values of x for which f(x) lt 0 is

Let f(x)=(g(x))/x w h e nx!=0 and f(0)=0. If g(0)=g^(prime)(0)=0a n dg^(0)=17 then f(0)= 3//4 b. -1//2 c. 17//3 d. 17//2

If f(x) is a twice differentiable function such that f(a)=0, f(b)=2, f(c)=-1,f(d)=2, f(e)=0 where a < b < c < d e, then the minimum number of zeroes of g(x) = f'(x)^2+f''(x)f(x) in the interval [a, e] is

Let f(x) be a quadratic expression such that f(-1)+f(2)=0 . If one root of f(x)=0 is 3 , then the other root of f(x)=0 lies in (A) (-oo,-3) (B) (-3,oo) (C) (0,5) (D) (5,oo)

Let alpha,beta be the roots of the equation (x-a)(x-b)=c ,c!=0. Then the roots of the equation (x-alpha)(x-beta)+c=0 are a , c b. b , c c. a , b d. a+c , b+c