Let g(x)= (x-1)^n ^m(x-1) ; 0 0 and let p be the left hand derivative of |x-1| at x=1 . If _ x 1⁺ g(x)=p , then

Step-by-step solution

Given, g(x)= (x-1)^n ^m(x-1) ; 0 < x < 2, m 0, n are integers and |x-1|= array l x-1 ; x 1 1-x ; x < 1 array . The left hand derivative of |x-1| at x=1 is p=-1 . Also, aligned & _ h 0 (1+h-1)^n ^m(1+h-1) =-1 _ h 0 h^n ^m h =-1 & _ h 0 h^n m h =-1 aligned [Using L' Hospital rule] array ll & _ h 0 n h^ n-1 m 1 h (- h) =-1 & _ h 0 (- n m ) h^ n-2 ( h h ) =-1 & ( n m ) _ h 0 h^ n-2 ( h h ) =1 & n=2 and n m =1 & m=n=2 array