A) \[y={{e}^{a}}^{(x-1)}\]
B) \[y={{e}^{a}}^{(1-x)}\]
C) \[y={{e}^{a/2}}^{(x-1)}\]
D) \[{{e}^{a/2}}^{(x+1)}\]
Correct Answer: A
Solution :
Slope of the normal at (1,1) is - 1 /a \[\Rightarrow \]Slope of the tangent at (1, 1) is 'a' i. e., \[{{\left. \frac{dy}{dx} \right]}_{(1,1)}}=a\] ......(1) We are given that \[\frac{dy}{dx}\propto y;\frac{dy}{dx}=ky,\] where k is some constant = k dx \[\log |y|=kx+c,\]where c is a constant\[|y|={{e}^{kx+c}}\]\[y=\pm {{e}^{c}}{{e}^{kx}}=\]\[A{{e}^{kx}},\]where A is a constant. Since the curve passes through (I,I), therefore, \[1=A{{e}^{k}}\Rightarrow A={{e}^{-k}}\] Therefore, \[y={{e}^{-k}}.{{e}^{kx}}={{e}^{k}}(x-1)\] \[\Rightarrow \]\[\frac{dy}{dx}=k{{e}^{k}}(x-1)\] \[\Rightarrow \]\[{{\left. \frac{dy}{dx} \right]}_{(1,1)}}=k\Rightarrow a=k\] [Using (1)] Thus, the required curve is \[y={{e}^{a}}\] (x - 1).You need to login to perform this action.
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