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  • Help me answer: - Limit , continuity and differentiability - JEE Main-13

The angle formed by tangent drawn at (0, 0) to the curve y^3 + y + x = 0 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

-

 

 y^3 + y + x = 0 \Rightarrow 3y^2\frac{\mathrm{d} y}{\mathrm{d} x} + \frac{\mathrm{d} y}{\mathrm{d} x} + 1 = 0 \Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-1}{1+ 3y^2}

Now, slope of the tangent at (0,0) = \frac{\mathrm{d} y}{\mathrm{d} x} at (0,0) = -1

\tan\theta = -1 \Rightarrow \theta = \frac{3\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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