Question

Let f(x) = $2\sqrt{x},g\left(x\right)=3-1/x,h\left(x\right)=e^x+e^{-x}$ and u(x) = $2+x^2$ then

• A. $f\left(x\right)<g\left(x\right)forx>1$
• B. $f\left(x\right)\le g\left(x\right)forx>1$
• C. $h\left(x\right)>u\left(x\right)forx\ne 0$
• D. $h\left(x\right)\le u\left(x\right)forx\ne 0$

Answer 1

The correct option is C $h\left(x\right)>u\left(x\right)forx\ne 0$

Consider an arbitrary function
$m\left(x\right)=h\left(x\right)-u\left(x\right)$
Hence
$m^{\prime }\left(x\right)=h^{\prime }\left(x\right)-u^{\prime }\left(x\right)$...(i)
Now consider
$h\left(x\right)$
Applying $AM\ge GM$ we get
$\frac{e^x+e^{-x}}{2}\ge \sqrt{e^x.e^{-x}}$
Or
$\frac{h\left(x\right)}{2}\ge 1$
Or
$h\left(x\right)\ge 2$
$h^{\prime }\left(x\right)>0$ for $x>0$.
Hence $h\left(x\right)$ is an increasing function with a minimum value of 2.
Now
$u\left(x\right)=\left(x^2\right)+2$
Hence $u\left(x\right)>0$ for all x.
Also
$u^{\prime }\left(x\right)>0$ for all x>0.
And minimum value of $u\left(x\right)=2$.
Hence $u\left(x\right)$ and $h\left(x\right)$ are similar.
However, exponential, increase at a faster rate than quadratic.
Hence $h\left(x\right)$ increases at a faster rate than $u\left(x\right)$
Therefore $h\left(x\right)\ge u\left(x\right)$ for all x, where $h\left(x\right)=u\left(x\right)atx=0$