If $f (x) = (x + 2) e^{ln (x + 2)}$ and $g (x) = \frac{e^{(\frac{1}{- {log}_{x} e})}}{x} - \frac{2 x}{e^{- ln 2}},$ then the smallest positive integer in the range of $h (x) = f (x) + g (x)$ is

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Solution

The correct option is D 7
$h (x) = f (x) + g (x) = (x + 2) e^{ln (x + 2)} + \frac{e^{(\frac{1}{- {log}_{x} e})}}{x} - \frac{2 x}{e^{- ln 2}}$
$= (x + 2)^{2} + \frac{1}{x^{2}} - 2 x e^{l n 2}$
$= x^{2} + 4 x + 4 + \frac{1}{x^{2}} - 4 x$
$= x^{2} + \frac{1}{x^{2}} + 4$
We know that $x^{2} + \frac{1}{x^{2}} \geq 2$
And, since the domain of $h (x)$ is $x ϵ R^{+} - {1}$
So, $x^{2} + \frac{1}{x^{2}} > 2$
$\Rightarrow h (x) > 6 \forall x ϵ R^{+} - {1}$
$∴$ Least integral value in the range of $h (x)$ is 7.

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