Paragraph: Let f(x)=(c-1) x^2+2 c x+c+4 and g(x)=c x^2+2(c+1) x+(c+1) , where c R . Question: If g(x) is always negative x (0,1) , then the number of integral values of c in the interval [-5,5] is:

Step-by-step solution

(i) c>0 not possible, think! If c D 0, g(0) 0 and -b 2 a 0 Solving we get c=-1 . (ii) Case 1: When both f(x) and g(x) are concave up i.e., c>1 Possible graph to have f(x) 0 and g(x) 0 to have unique solution. Hence, D_f=0 4 c^2-4(c-1)(c+4)=0 Hence, c= 4 3 At c= 4 3 , f(x)=(x+4)^2 and g(-4)>0 This case is possible. Hence, c= 4 3 Case 2: Both f(x) and g(x) are concave down i.e., c Here f(x) and g(x) have a common root. In this case, the value of c is -3 4 . Case-3: When one is concave up and another is concave down i