Mathematics - Functions Question with Solution | TestHub

Let $t=x^2+1$, then $(t>1)$ and $g(t)=\frac{t}{\log t}\Rightarrow g^{\prime }(t)=\frac{\log t-1}{{\log }^{2}t}$.

Thus $\mathrm{g}(\mathrm{t})$ decreases for $t\in [1,e]$ and

increases for $t\in [e,\infty ),g(e)=\frac{e}{\log e}=e$.

We observe that ${\lim }_{t\to 1^{+}}g(t)\to \infty$ and ${\lim }_{t\to \infty }g(t)\to \infty$

Thus range of g is $[e,\infty )$. Hence range of f is $[e,\infty )$.