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  • Answer please! - Limit , continuity and differentiability - JEE Main-9

Let y = x^3 + x^2 -x then the slope of normal at (1, 1) is 

  • Option 1)

    \frac{-1}{2}

  • Option 2)

    \frac{-1}{3}

  • Option 3)

    \frac{-1}{4}

  • Option 4)

    \frac{-1}{5}

 

Answers (1)

best_answer

As we have learnt,

 

Slope of Normal -

M_{T}\times M_{N}=-1
 

\therefore \:M_{N}=\frac{-1}{MT}
 

M_{N}=\frac{-1}{\frac{dy}{dx}_{(x_{1},y_{1})}}

- wherein

Where ( x, y ) is the point on the curve.

 

 \frac{\mathrm{d} y}{\mathrm{d} x} at (1,1) = 3x^2 + 2x - 1( at x =1) = 4 = slope of tangent 

Therefore, slope of normal = -\frac{1}{}4

 


Option 1)

\frac{-1}{2}

Option 2)

\frac{-1}{3}

Option 3)

\frac{-1}{4}

Option 4)

\frac{-1}{5}

Posted by

Himanshu

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