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Answer please! - Limit , continuity and differentiability - JEE Main-8

Let y = e^{x} then the equation of the tangent at (0, 1) will be?

  • Option 1)

    y = 2x + 3

  • Option 2)

    y = 2x + 1

  • Option 3)

    y = x + 1

  • Option 4)

    y - 3x = 1

 
Answers (1)
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As we have learnt,

 

Equation of the tangent -

To find the equation of the tangent we need either one slope + one point or two points.

\therefore \:\:(y-y_{\circ})=m(x_{\circ }-y_{\circ })
 

or\:\:(y-y_{2})=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{2})

- wherein

Where  (x_{\circ},y_{\circ})   is the point on the curve and M = MT  slope of tangent.

 

 \frac{\mathrm{d} y}{\mathrm{d} x} = e^{x} \Rightarrow\frac{\mathrm{d} y}{\mathrm{d} x} = e^{x} \Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} at (0, 1) = = e^{0} = 1 = slope of tangent

Therefore, equation of tangent = (y-1) = 1(x-0) \Rightarrow y = x + 1

 


Option 1)

y = 2x + 3

Option 2)

y = 2x + 1

Option 3)

y = x + 1

Option 4)

y - 3x = 1

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