Square root

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What is the inverse of $g\left(x\right)=\sqrt{x-1} $ 
$Og\left(x\right)=\sqrt{x+1} $ 
$+1$ $g^{'}\left(x\right)=5-$ $y_{9}$ $g\sqrt{ef\left(x\left(nght\right)} =\left(f_{tac\left(x+1\right)0sq\pi \left(x\right)}$ 
$Og^{'}\left(x\right)=x^{2}+1$ 
$○g\left(x\right)=x^{2}-1$ 
« Previous Ne 

Q.4 $1$ $\left(x^{5}\longdiv{3}$ $\left(16x^{3}\longdiv{3}$ $2$ $\left(\dfrac {1} {4}x^{\dfrac {4} {9}}\longdiv{}3/2$ $=$ 
$x^{2}$ $\times 3$ 

$\left(\dfrac {\sqrt{p9} -\sqrt [3] {9} } {3\sqrt{p} -6\sqrt{p9} }\longdiv{}6$ $=$ 
$-$ $\left(3\right)$ $9$ $\left(4\right)$ $\left(1$ 
$\left(1\right)$ $p$ $9$ $\left(2\right)$ $9$ $P$ 


$1Bj$ If the pth and qth terms of a G.P. are $9$ and $p$ respectively, show that $0=$ 
term is $1$ $\left(\dfrac {q^{p}} {p^{9}}\longdiv{p-9}$ 


The rationalizing factor of \sqrt{23} is 
$°$ $Options^{°}$ $0$ 
A 24 
23 
C \sqrt{23} 
D None of these 

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