A) \[y-2x=6+log\text{ }2\]
B) \[y+2x=6+log2\]
C) \[y+2x=6-log\text{ }2\]
D) \[y+2x=-6+log\text{ }2\]
E) \[y-2x=-6+log\text{ }2\]
Correct Answer: B
Solution :
Given, \[g(x)=\left\{ \begin{matrix} 2e, & if\,x\le 1 \\ \log (x-1), & if\,x>1 \\ \end{matrix} \right.\] For the point\[(3,\text{ }log2),\]we take \[y=g(x)=\log (x-1)\] \[\frac{dy}{dx}=\frac{1}{(x-1)}\] \[\Rightarrow \] \[{{\left( \frac{dy}{dx} \right)}_{(3,\log 2)}}=\frac{1}{3-1}=\frac{1}{2}\] \[\therefore \]Equation of normal is \[y-log2=-2(x-3)\] \[\Rightarrow \] \[y+2x=6+log2\]You need to login to perform this action.
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