Question

Let $f(x)=(x+2)e^{\ln (x+2)}$ and $g(x)=\frac{e^{\left(\frac{1}{-{\log }_{x}e}\right)}}{x}-\frac{2x}{e^{-\ln 2}}$. 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, $h(x)=f(x)+g(x)=(x+2)e^{\ln (x+2)}+\frac{e^{\left(\frac{1}{-{\log }_{x}c}\right)}}{x}-\frac{2x}{e^{-\ln 2}}$
$=(x+2{)}^2+\frac{1}{x^2}-2x\cdot e^{\ln 2}=x^2+4x+4+\frac{1}{x^2}-4x=x^2+\frac{1}{42}+4$
$\Rightarrow h(x)>6\forall x\in R^{+}-\{1\}$
[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.]