Let `f(x) = (e^x)/x^2` `x in R - {0}`. If f(x) = k has two distinct real roots, then the range of k, is

Let f (x) =x ^(2) + bx + c AA in R, (b,c, in R) attains its least value at x =-1 and the graph of f (x) cuts y-axis at y =2. If f (x) =a has two distinct real roots, then comlete set of values of a is :

Let f (x) =ax ^(2) + bx+ c,a gt = and f (2-x) =f (2+x) AA x in R and f (x) =0 has 2 distinct real roots, then which of the following is true ?

If the equation sin ^(2) x - k sin x - 3 = 0 has exactly two distinct real roots in [0, pi] , then find the values of k .

Let f(x)=2+x, x>=0 and f(x)=4-x , x < 0 if f(f(x)) =k has at least one solution, then smallest value of k is

Let f(x)=x+2|x+1|+2|x-1|* If f(x)=k has exactly one real solution,then the value of k is (a) 3 (b) 0(c)1(d)2

Let f(x)=(sqrt(log(x^(2)+kx+k+1)))/(x^(2)+k) If f(x) is continuous for all x in R then the range of k is

If the equation x^(2) + 4x - k = 0 has real and distinct roots, then

If the equation x^(2)+4x+k=0 has real and distinct roots, then find the value of 'k'.

Let f(x)=x^(2)+ax+b, where a,b in R. If f(x)=0 has all its roots imaginary then the roots of f(x)+f'(x)+f(x)=0 are

Let f(x)=min{e^(x), (3)/(2), 1+e^(-x)} if f(x)=k has at least 2 solutions then k can not be