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

Correct Answer is : [0,aa+1]≤ft[0,{a/a+1}][0,a+1a] ∪ (1 , ∞)

[0 , ∞) - ≤ft[-{a/a+1},\;{a/a+1}]

Correct Answer is : [0,aa+1]≤ft[0,{a/a+1}][0,a+1a] ∪ (1 , ∞)

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AP EAMCET Engineering 23 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 2 of 160

Marks: +1, -0

The range of f (x) =

\sqrt{\frac{a-|x|}{(a+1)-|x|}}

, (a > 0) is

Solution:

Consider the function,

f(x)=\sqrt{\frac{a-|x|}{(a+1)-|x|}},\;(a>0)

So,

f(x) \leq 0,\;\forall x\in

domain of

f(x)

Let,

\frac{a-|x|}{(a+1)-|x|}=y

Then,

a-|x|=y(a+1)-y|x|

(y-1)|x|=y(a+1)-a

|x| = \frac{y(a+1)}{y-1} \leq 0,\;\forall x \in

domain of

f(x)

This implies,

y \in \left(-\infty, \frac{a}{a+1}\right) \cup (1, \infty)

So, range of function is,

f(x)=\sqrt{y} \in \left[0,\sqrt{\frac{a}{a+1}}\right] \cup (1,\infty)

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