Question

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

• A. 2
• B. 3
• C. 6
• D. 7

Answer 1

Answer: (D)

Clearly, domain of $h ( x )= x \in R ^{+}-\{1\}$.
We have, $h(x)=f(x)+g(x)=(x+2) e^{\ln (x+2)}+\frac{e^{\left(\frac{1}{-\log _x c}\right)}}{x}-\frac{2 x}{e^{-\ln 2}}$
$=(x+2)^2+\frac{1}{x^2}-2 x \cdot e^{\ln 2}=x^2+4 x+4+\frac{1}{x^2}-4 x=x^2+\frac{1}{42}+4$
$\Rightarrow h ( x )>6 \forall x \in R ^{+}-\{1\} $
[Domain of $h(x)$ is $x \in R ^{+}-\{1\}$ ]
$\Rightarrow h ( x )>6 \forall x \in R ^{+}-\{1\}$
So, smallest integral value in the range of $h ( x )$ is 7 . Ans.]