Let d be the number of solutions of the equation ( x-1)=( 2 -1) x in [0,2 ] . If d lies between the roots of the equation x^2+(k-1) x+k^2+k-11=0 , then k can be:

Step-by-step solution

1- x x =( 2 -1) x x 1- x x =( 2 -1) 2 ^2 x 2 2 x 2 x 2 = 2 -1 either x 2 =0 i.e., x=2 n x=0,2 or x 2 = 2 -1 x 2 =n + 8 x=2 n + 4 only x= 4 Hence, d=3 i.e., 0,2 , / 4 Now, d=3 lies between the roots of the equation x^2+(k-1) x+k^2+k-11=0 f(3) < 0 aligned & 9+3 k-3+k^2+k-11 < 0 & k^2+4 k-5 < 0 & (k+5)(k-1) < 0 aligned k (-5,1) a , b , c