Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Help me understand! - Limit , continuity and differentiability - JEE Main-10

Let y = x^3 + x^2 - x then the slope of the tangent at (2, 10) will be ?

  • Option 1)

    10

  • Option 2)

    15

  • Option 3)

    20

  • Option 4)

    25

 

Answers (1)

best_answer

As we have learnt,

 

Slope of the tangent -

Let y = f(x) is a curve then  dy / dx = f'(x) and at a particular point (h, k) it gives slope of tangent. From fig

M_{T}=\lim_{\delta x\rightarrow 0}\:\frac{(y+\delta y)-y}{(x+\delta x)-x}=\lim_{\delta x\rightarrow 0}\:\frac{\delta y}{\delta x}

- wherein

 

 \frac{\mathrm{d} y}{\mathrm{d} x} at (2,10) will b the required slope 

\\*\therefore \frac{\mathrm{d} y}{\mathrm{d} x} = 3x^2 +2x - 1 \\*\Rightarrow \frac{\mathrm{d} }{\mathrm{d} x} \; at\; (2, 10) = 3(4) + 2(2) - 1 =15

 


Option 1)

10

Option 2)

15

Option 3)

20

Option 4)

25

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

Latest Question