Question

The function $f:\mathrm{ℝ}\to \mathrm{ℝ}$ defined by $f\left(x\right)=6^x+6^{\left|x\right|}$ is

• A. One-One and Onto
• B. Many-One and Onto
• C. One-One and Into
• D. Many-One and Into

Answer 1

The correct option is C

One-One and Into

The explanation of the correct option

Given function: $f\left(x\right)=6^x+6^{\left|x\right|}$.

The given function can be rewritten as, $f\left(x\right)=6^x+6^{\left|x\right|}=\left\{\begin{array}{ll}{6}^x+6^x& \mathrm{for}\mathit{}x\mathit{\ge }0\\ 6^x+6^{-x}& \mathrm{for}x<0\end{array}\right.$

$\Rightarrow f\left(x\right)=\left\{\begin{array}{ll}2·6^x& \mathrm{for}\mathit{}x\mathit{\ge }0\\ 6^x+6^{-x}& \mathrm{for}x<0\end{array}\right.$

Differentiate the given function with respect to $x$.

$$\begin{gathered} \frac{d}{dx}\left[f\left(x\right)\right]=\left\{\begin{array}{ll}\frac{d}{dx}\left(2·6^x\right)& \mathrm{for}\mathit{}x\mathit{\ge }0\\ \frac{d}{dx}\left(6^x+6^{-x}\right)& \mathrm{for}x<0\end{array}\right. \\ \Rightarrow f\text{'}\left(x\right)=\left\{\begin{array}{ll}2\ln \left(6\right)6^x& \mathrm{for}x\ge 0\\ \ln \left(6\right)6^x-\ln \left(6\right)6^x& \mathrm{for}x<0\end{array}\right. \\ \Rightarrow f\text{'}\left(x\right)=\left\{\begin{array}{ll}\ln \left(36\right)6^x& \mathrm{for}x\ge 0\\ 0& \mathrm{for}x<0\end{array}\right. \end{gathered}$$

As $\ln \left(36\right)6^x>0$ thus $f\text{'}\left(x\right)\ge 0$ for all values of $x$.

So, the given function is always increasing.

Thus, the given function is a One-One function.

Now consider the given function, $\Rightarrow f\left(x\right)=\left\{\begin{array}{ll}2·6^x& \mathrm{for}\mathit{}x\mathit{\ge }0\\ 6^x+6^{-x}& \mathrm{for}x<0\end{array}\right.$.

As $6^x$ and $6^{-x}$ both are always positive. thus $f\left(x\right)$ can not have any negative values.

So, the range of the function is not equal to the given co-domain $\left(\mathrm{ℝ}\right)$.

So, the given function is an Into function.

Hence, option C is correct .