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 :
C. ( P ) (5) ( Q ) (3) ( R ) (2) ( S ) (4)
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 )= \ array cll x | x | 1 x & ; & x 0 \\ 0 & ; & x=0 array . \\ & RHD = _ x 0 x^2 1 x -0 x =0 \\ & LHD = _ x 0 -x^2 1 x -0 x =0 aligned Hence h ( x ) is differentiable on R (R) aligned & a =0, ~b =0, c =1, ~d =0 \\ & h(x)=x-g(x)= \ array ccc 3 x -1 & ; & 0 x 1 2 \\ 0 & ; & otherwise array . aligned h ( x ) is ONTO (S) aligned & a =0, ~b =0, c =0, ~d =1 \\ & ~h ( x )= g ( x )= \ array ccl 1-2 x & ; & 0 x 1 2 \\ 0 & ; & otherwise array . aligned Range of h(x) is [0,1]
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