A) (1, 0)
B) \[(-1,0)\]
C) \[\left( -\frac{1}{2},0 \right)\]
D) None of these
Correct Answer: C
Solution :
Curve \[y={{e}^{2x}}\] Gradient\[{{\left( \frac{dy}{dx} \right)}_{(0,1)}}={{(2{{e}^{2x}})}_{(0,1)}}=2{{e}^{0}}=2\] \[\therefore \]Equation of tangent at the point (0, 1) \[y-1=2(x-0)\] \[\Rightarrow \] \[y-1=2x\] \[\Rightarrow \] \[y=2x+1\] Since, tangents meets at\[x-\]axis. So, putting\[y=0,\] \[2x+1=0\] \[\Rightarrow \] \[x=-\frac{1}{2}\]
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