Question

Find the range of the real function defined as:-
${f(x) = \sqrt{2 - x} - \dfrac{ \sqrt{25 - {x}^{2} } }{ \sqrt{ {x}^{2} - 4 } } + \dfrac{x + 2}{ |x|}}$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:{f(x) = \sqrt{2 - x} - \dfrac{ \sqrt{25 - {x}^{2} } }{ \sqrt{ {x}^{2} - 4 } } + \dfrac{x + 2}{ |x|}}$

Let assume that,

$\rm :\longmapsto\:g(x) = \sqrt{2 - x}$

$\rm :\longmapsto\:h(x) = \dfrac{ \sqrt{25 - {x}^{2} } }{ \sqrt{ {x}^{2} - 4 } }$

$\rm :\longmapsto\:p(x) = \dfrac{x + 2}{ |x|}$

So,

Now, Let find the domain of each function.

Consider,

$\rm :\longmapsto\:g(x) = \sqrt{2 - x}$

Now, g(x) is defined, if

$\rm :\longmapsto\:2 - x \geqslant 0$

$\rm :\longmapsto\: - x \geqslant - 2$

$\rm :\longmapsto\: x \leqslant 2$

$\bf\implies \:x \: \in \: ( - \infty ,2]$

Now, Consider,

$\rm :\longmapsto\:h(x) = \dfrac{ \sqrt{25 - {x}^{2} } }{ \sqrt{ {x}^{2} - 4 } }$

So, h(x) is defined when

$\rm :\longmapsto\:25 - {x}^{2} \geqslant 0 \: \: and \: {x}^{2} - 4 > 0$

$\rm :\longmapsto\:{x}^{2} - 25\leqslant 0 \: \: and \: {x}^{2} - 4 > 0$

$\rm :\longmapsto\:(x - 5)(x + 5) \leqslant 0 \: \: and \: \: (x - 2)(x + 2) > 0$

$\rm :\implies\: - 5 \leqslant x \leqslant 5 \: \: and \: \: x < - 2 \: \: or \: \: x > 2$

$\bf\implies \:x \: \in \: [ - 5, - 2) \: \cup \: (2,5] - - - (2)$

Now, Consider

$\rm :\longmapsto\:p(x) = \dfrac{x + 2}{ |x|}$

So, p(x) is defined when

$\bf\implies \:x \: \in \: ( - \infty ,0) \: \cup \: (0, \infty ) - - - (3)$

So, from equation (1), (2) and (3), we concluded that

Domain of

$\rm :\longmapsto\:{f(x) = \sqrt{2 - x} - \dfrac{ \sqrt{25 - {x}^{2} } }{ \sqrt{ {x}^{2} - 4 } } + \dfrac{x + 2}{ |x|}}$

is

$\bf\implies \:x \: \in \: [ - 5, - 2)$

and

Thus,

$\bf\implies \: |x| = - x$

Now, we check the monotonocity of f(x) in [ - 5, - 2)

f(x) can be rewritten as

$\rm :\longmapsto\:{f(x) = \sqrt{2 - x} - \dfrac{ \sqrt{25 - {x}^{2} } }{ \sqrt{ {x}^{2} - 4 } } - \dfrac{x + 2}{ x}}$

$\rm :\longmapsto\:{f(x) = \sqrt{2 - x} - \dfrac{ \sqrt{25 - {x}^{2} } }{ \sqrt{ {x}^{2} - 4 } } -1 - \dfrac{2}{ x}}$

So, Differentiate f(x) w. r. t. x, we get

$\rm :\longmapsto\:f'(x) = - \dfrac{1}{2 \sqrt{2 - x} } - \dfrac{1}{2 \sqrt{\dfrac{25 - {x}^{2} }{ {x}^{2} - 4 } } }\dfrac{( {x}^{2} - 4)( - 2x) - (25 - {x}^{2})(2x) }{ {x}^{2} - 4} - 0 + \dfrac{1}{ {x}^{2} }$

$\rm :\longmapsto\:f'(x) = - \dfrac{1}{2 \sqrt{2 - x} } + \dfrac{1}{ \sqrt{\dfrac{25 - {x}^{2} }{ {x}^{2} - 4 } } }\dfrac{(21x ) }{ {x}^{2} - 4} + \dfrac{1}{ {x}^{2} }$

$\rm :\longmapsto\:As \:x \: \in \: [ - 5, - 2)$

$\bf\implies \:f'(x) < 0 \: \forall \: x \in \: [ - 5, - 2)$

$\bf\implies \:f(x) \: has \: maximum \: value \: at \: x = - 5$

So,

$\rm :\longmapsto\:{f( - 5) = \sqrt{2 + 5} - \dfrac{ \sqrt{25 - {5}^{2} } }{ \sqrt{ {5}^{2} - 4 } } + \dfrac{ - 5 + 2}{5}}$

$\rm :\longmapsto\:{f( - 5) = \sqrt{7} + \dfrac{ - 3}{5}}$

$\rm :\longmapsto\:{f( - 5) = - \dfrac{3}{5}} + \sqrt{7}$

So,

Range of f(x) is

$\bf\implies \:f(x) \: \in \: ( - \infty , \: - \: \dfrac{3}{5} + \sqrt{7} \bigg]$