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AP EAMCET201923 Apr 2019Morning ShiftMathematicsFunctionsActual

The range of (f(x)= a-|x| (a+1)-|x| ,(a>0) ) is

Options

  1. A([0, a] )
  2. B([0, )- [- a a+1 , a a+1 ] )
  3. C( [0, a a+1 ] (1, ) )
  4. D( [0, a a+1 +1 ] )

Correct answer

C. ( [0, a a+1 ] (1, ) )

Step-by-step solution

Given function is (f(x)= a-|x| (a+1)-|x| ,(a > 0) ) ( f(x) 0, x ) domain of (f(x) ). Now, let ( a-|x| (a+1)-|x| =y ) ( aligned & a-|x|=y(a+1)-y|x| [ assuming |x| a+1] & (y-1)|x|=y(a+1)-a & |x|= y(a+1)-a y-1 0, x domain of f(x) & y (- , a a+1 ] (1, ) ( as a > 0) aligned ) So, range of (f(x)= y [0, a a+1 ] (1, ) ) ( [as y 0] ) Hence, option (c) is correct.

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