Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Confused! kindly explain, - Limit , continuity and differentiability - JEE Main-12

Let y = x^3 + x^2 - x then the equation of tangent at (1, 1) on the curve will be ?

  • Option 1)

    4x -y = 3

  • Option 2)

    4x +y = 5

  • Option 3)

    4x -y = 1

  • Option 4)

    4x + 10y = 14

 

Answers (1)

best_answer

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

Therefore, using slope form of line we have,

y -1 = 4(x-1) \Rightarrow 4x -y = 3

 


Option 1)

4x -y = 3

Option 2)

4x +y = 5

Option 3)

4x -y = 1

Option 4)

4x + 10y = 14

Posted by

Himanshu

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today
anam-khan

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs
anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question