Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The number of points on the curve <img alt="y=54 x^5-135 x^4-70 x^3+180 x^2+210 x" src="https://entrancecorner.oncodecogs.com/gif.latex?y%3D54%20x%5E5-135%20x%5E4-70%20x%5E3&plus;180%20x%5

The number of points on the curve y=54 x^5-135 x^4-70 x^3+180 x^2+210 x  at which the normal lines are parallel to x+90 y+2=0 is 

Option: 1

4


Option: 2

2


Option: 3

0


Option: 4

3


Answers (1)

Given curve is y=54 x^5-135 x^4-70 x^3+180 x^2+210 x

\frac{d y}{d x}=270 x^4-540 x^3-210 x^2+360 x+210

\because Normal is parallel to x+90 y+2=0

Then tangent is \perp^r \text { to } x+90 \mathrm{y}+2=0

Then \left(270 x^4-540 x^3-210 x^2+360 x+210\right)\left(\frac{-1}{90}\right)=-1

            \begin{aligned} & 270 x^4-540 x^3-210 x^2+360 x+120=0 \\ & \Rightarrow 9 x^4-18 x^3-7 x^2+12 x+4=0 \\ & \Rightarrow(x-1)(x-2)(3 x+1)(3 x+2)=0 \\ & \Rightarrow x=1,2,-\frac{1}{3},-\frac{2}{3} \end{aligned}

Number of points are 4

Posted by

Ramraj Saini

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

JEE Main 2025: BTech ECE cut-offs for female candidates go up at top NITs
anam-khan

JEE Main 2025 Session 2 City Slip (OUT) Live: Admit card likely on March 29 at jeemain.nta.nic.in; cut-offs
anam-khan

AESL expands Aakash Digital 2.0 to offer AI-powered coaching for NEET, JEE, Olympiads

Latest Question