Paragraph: The sum of an infinitely decreasing geometric progression whose first term is a and common ratio r , is equal to least value of the quadratic trinomial P(x)=3 x^2-x+ 25 12 , in [0,2] . Also the first term of the geometric progression is equal to the square of its common ratio Question: If the minimum value of P(x) lies between roots of equation x^2+(k+1) x+a+2 3 =0 , then the maximum integral value of k is

Step-by-step solution

Given, P(x)=3 x^2-x+ 25 12 =3 [x^2- 1 3 x+ 25 36 ] ^3 ( (x- 1 6 )^2+ 2 3 ) . P(x) |_ min. =2 , occurs at x= 1 6 . Given, a 1-r =2 ......(1) and a = r ^2 ......(2) Solving (1) and (2), we get aligned & r^2 1-r =2 r^2=2-2 r r^2+2 r=2 & r=( 3 -1) [ As r (0,1)] & a=r^2=4-2 3 aligned Let f(x)=x^2+(k+1) x+a+2 3 Now, f(x)=0 a^2+(k+1) x+4=0 . Since 2 lies between the roots, hence f(2) 0 aligned & 4+2 k+2+4 0 & 2 k+10 0 k -5 aligned So, k _ max =-6 Let y= P(x) P^ (x) = 3 x^2-x+ 25 12 6 x-1 3 x^2-x+ 25 12 =6 x y-y 3 x^2-x(1+