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TG EAMCET 2 May 2025 Shift 1 Paper

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Question : 1 of 160

Marks: +1, -0

If

f(x)=\tan\;\left(\;\frac{\pi}{\sqrt{x+1}+4}\right)

is a real valued function, then the range of

f

$[-1,1]$
$(0,1]$
$[-1, \infty)$
$R$

Solution:

Given,

f(x)=\tan\;\left(\;\frac{\pi}{\sqrt{x+1}+4}\right)

For

\sqrt{x+1}

to be defined,

we have

x+1 \ge 0 \Rightarrow x \ge -1

\therefore \; \sqrt{x+1}+4 \ge 0+4=4

\Rightarrow 0< \;\frac{1}{\sqrt{x+1}+4} \le \;\frac{1}{4}

\Rightarrow 0< \;\frac{\pi}{\sqrt{x+1}+4} \le \;\frac{\pi}{4}

\therefore \; f(x)=\tan\;\left(\;\frac{\pi}{\sqrt{x+1}}+4\right) , \;\text{ where }\;

\;0< \;\frac{\pi}{\sqrt{x+1}+4} \le \;\frac{\pi}{4}

So, the function

f(x)

is strictly increasing in the interval

(0, \;\frac{\pi}{4}]

And the range of the given function is

(\tan (0), \tan\left(\;\frac{\pi}{4}\right) ]

, which is

(0,1]

\therefore

The range of

f(x)

is

(0,1]

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