AP EAMCET201920 Apr 2019Morning ShiftMathematicsFunctionsActual
Consider the following lists. ( array ll List I & List II (A) f(x)= |x+2| x+2 , x -2 & 1. [ 1 3 , 1 ] (B) g(x)= [x x R & 2. Z (C) h(x)=|x-[x]|, x R & 3. W (D) f(x)= 1 2- 3 x , x R & 4. [0,1) & 5. -1,1 array )
Options
- A( array llll A & B & C & D 5 & 3 & 2 & 1 array )
- B( array llll A & B & C & D 3 & 2 & 4 & 1 array )
- C( array llll A & B & C & D 5 & 3 & 4 & 1 array )
- D( array llll A & B & C & D 1 & 2 & 3 & 4 array )
Correct answer
C. ( array llll A & B & C & D 5 & 3 & 4 & 1 array )
Step-by-step solution
( aligned ( A ) f(x) & = |x+2| x+2 , x -2 & = array ll x+2 x+2 , & x > -2 - x+2 x+2 , & x -2 -1, & x < -2 cases . aligned ) So, range of (f(x) ) is ( -1,1 ). (B) ( ) (g(x)=|[x]|, x R ) As ([x] I |[x]| W ) So, range of (g(x) ) is (W ). (C) ( aligned & h(x)=|x-[x]|, x R=| x | [0,1) & [ x =x-[x] and x [0,1)] aligned ) So, range of (h(x) ) is ([0,1) ). ( array lrl (D) f(x)= & 1 2- 3 x , x R & & -1 3 x 1, x R & -1 - 3 x 1 & 2-1 2- 3 x 2+1 & 1 3 1 2- 3 x 1 1 array ) So, range of (f(x) ) is ( [ 1 3 , 1 ] ). Hence, option