Let f: R R and g: R R be functions defined by f(x)= array ll x|x| ( 1 x ), & x 0, 0, & x=0, array . and g(x)= cases 1-2 x, & 0 x 1 2 0, & otherwise cases Let a, b, c, d R . Define the function h: R R by h(x)=a f(x)+b (g(x)+g ( 1 2 -x ) )+c(x-g(x))+d g(x), x R Match each entry in List-I to the correct entry in List-II. The correct option is :

Step-by-step solution

f(x)= array ccc x|x| 1 x & ; & x 0 0 & ; & x=0 array g(x)= array ccc 1-2 x & ; & 0 x 1 2 0 & ; & otherwise array . .g ( 1 2 -x )= array ccc 2 x & ; & 0 1 2 -x 1 2 0 & ; & otherwise array = array ccc 2 x & ; & 0 x 1 2 0 & ; & otherwise array .g(x)+g ( 1 2 -x )= array lll 1 & ; & 0 x 1 2 0 & ; & otherwise array (P) Now a =0, ~b =1, c =0, ~d =0 h ( x )= g ( x )+ g ( 1 2 - x )= array lll 1 & ; & 0 x 1 2 0 & ; & otherwise array . Hence Range of h ( x ) is 0,1 (Q) aligned & a =1, ~b =0, c =0, ~d =0 & ~h ( x )= f ( x )= a