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  • I have a doubt, kindly clarify. - Limit , continuity and differentiability - JEE Main-11

The angle formed by the  tangent drawn at (1, -2) to the curve y = x^3 + x^2 - 4x with positive x-axis is?

  • Option 1)

    \frac{\pi}{6}

  • Option 2)

    \frac{\pi}{4}

  • Option 3)

    \frac{\pi}{3}

  • Option 4)

    \frac{3\pi}{4}

 

Answers (1)

best_answer

As we have learnt,

 

Geometrical interpretation of dy / dx -

Slope of tangent line is tan\theta  where  \theta is the angle made by the line with the  +ve direction of  x  axis.

\therefore \:\:\frac{dy}{dx}=tan\theta

-

 

 Slope of tangent at (1, -2) = \frac{\mathrm{d} y}{\mathrm{d} x} at (1, -2)

Now, 

\frac{\mathrm{d} y}{\mathrm{d} x} = 3x^2 + 2x -4

\therefore slope at (1, -2) = 3 + 2 -4 = 1

\Rightarrow \tan\theta = 1 \Rightarrow \theta = \frac{\pi}{4}

 


Option 1)

\frac{\pi}{6}

Option 2)

\frac{\pi}{4}

Option 3)

\frac{\pi}{3}

Option 4)

\frac{3\pi}{4}

Posted by

Himanshu

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