Question

If $f:R\to R$ and $g:R\to R$ are functions defined by $f\left(x\right)=3x-1;g\left(x\right)=\sqrt{x+6}$, then the value of $(g\circ f^{-1})\left(2009\right)$ is

• A. $26$
• B. $29$
• C. $16$
• D. $15$

Answer 1

The correct option is A $26$

Given $f\left(x\right)=3x-1,g\left(x\right)=\sqrt{x+6}$

Let $f\left(x\right)=y$

$\Rightarrow y=3x-1\Rightarrow y+1-3x\Rightarrow x=\frac{y+1}{3}$

Now, $f^{-1}\left(y\right)=x=\frac{y+1}{3}$

$\Rightarrow f^{-1}\left(x\right)=\frac{x+1}{3}$ and $g\left(x\right)=\sqrt{x+6}$

Consider, $(g\circ f^{-1})\left(2009\right)=g\left[f^{-1}\left(2009\right)\right]$

$=g\left(\frac{2009+1}{3}\right)=g\left(\frac{2010}{3}\right)$

$=\sqrt{\frac{2010}{3}+6}=\sqrt{\frac{2028}{3}}=\sqrt{676}=26$

Hence, option $A$ is correct.