Question

If $\mathrm{f}:\mathrm{R}\to \mathrm{R}$ and $\mathrm{g}:\mathrm{R}\to \mathrm{R}$ are defined by $\mathrm{f}\left(\mathrm{x}\right)=2\mathrm{x}+3$ and $\mathrm{g}\left(\mathrm{x}\right)={\mathrm{x}}^2+7$, then the value of $x$ for which $\mathrm{g}\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}=8$ are

• A. $0,-6$
• B. $-1,-2$
• C. $1,-1$
• D. $0,6$
• E. $0,2$

Answer 1

The correct option is B

$-1,-2$

Explanation for the correct answer:

Given functions $\mathrm{f}\left(\mathrm{x}\right)=2\mathrm{x}+3\mathrm{and}\mathrm{g}\left(\mathrm{x}\right)={\mathrm{x}}^2+7$

Finding the value of $\mathrm{g}\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}$

$$\begin{gathered} \therefore \mathrm{g}\left\{\mathrm{f}\left(\mathrm{x}\right)\right\}=8 \\ \Rightarrow \begin{array}{rcl}& & \mathrm{g}(2\mathrm{x}+3)\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & 8\end{array} \\ \begin{array}{rcl}& \Rightarrow & \end{array}\begin{array}{rcl}& & (2\mathrm{x}+3)\end{array}\begin{array}{rcl}& & {}^2\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & 7\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & 8\end{array}\begin{array}{rcl}& & \left[\because \mathrm{g}\left(\mathrm{x}\right)={\mathrm{x}}^2+7\right]\end{array} \\ \begin{array}{rcl}& \Rightarrow & {(2\mathrm{x}+3)}^2=8-7\end{array} \\ \begin{array}{rcl}& \Rightarrow & {(2\mathrm{x}+3)}^2=1\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \end{array} \\ \begin{array}{rcl}& \Rightarrow & 2\mathrm{x}+3=\pm 1\end{array} \\ \mathrm{Now},\begin{array}{rcl}& & 2\end{array}\begin{array}{rcl}& & \mathrm{x}\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & 2\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & o\end{array}\begin{array}{rcl}& & r\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & 2\end{array}\begin{array}{rcl}& & \mathrm{x}\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & 4\end{array}\begin{array}{rcl}& & \end{array} \\ \begin{array}{rcl}& & \mathrm{x}\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & 1\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & o\end{array}\begin{array}{rcl}& & r\end{array}\begin{array}{rcl}& & \end{array}\begin{array}{rcl}& & \mathrm{x}\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & 2\end{array} \end{gathered}$$

Hence, the correct answer is an option (B).