A) \[-\frac{1}{3}\]
B) \[\frac{1}{13}\]
C) \[\frac{1}{3}\]
D) \[-\frac{1}{13}\]
Correct Answer: C
Solution :
\[f(x)=y={{x}^{2}}-x+5\] \[{{x}^{2}}-x+\frac{1}{4}-\frac{1}{4}+5=y\] \[{{\left( x-\frac{1}{2} \right)}^{2}}+\frac{19}{4}=y\] \[{{\left( x-\frac{1}{2} \right)}^{2}}y-\frac{19}{4}\] \[x-\frac{1}{2}=\pm \sqrt{y-\frac{19}{4}}\] \[x=\frac{1}{2}\pm \sqrt{y-\frac{19}{4}}\] As\[x>\frac{1}{2}\] \[x=\frac{1}{2}+\sqrt{y-\frac{19}{4}}\] \[g(x)=\frac{1}{2}+\sqrt{x-\frac{19}{4}}\] \[g'(x)=\frac{1}{2\sqrt{x-\frac{19}{4}}}\] \[g'(7)=\frac{1}{2\sqrt{7-\frac{19}{4}}}=\frac{1}{2\frac{\sqrt{28-19}}{2}}=\frac{1}{3}\]
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