Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let us consider a curve, passing through the point and the slope of the tangent to

Let us consider a curve, y=f(x) passing through the point (-2,2) and the slope of the tangent to the curve at any point (x, f(x)) is given by f(x)+x f^{\prime}(x)=x^{2}. Then:
Option: 1 x^{2}+2 x f(x)-12=0
Option: 2 x^{2}+2 x f(x)-12=0
Option: 3 x^{2}+2 x f(x)-12=0
Option: 4 x^{2}+2 x f(x)-12=0
Option: 5 x^{3}-3 x f(x)-4=0
Option: 6 x^{3}-3 x f(x)-4=0
Option: 7 x^{3}-3 x f(x)-4=0
Option: 8 x^{3}-3 x f(x)-4=0
Option: 9 x^{3}+x f(x)+12=0
Option: 10 x^{3}+x f(x)+12=0
Option: 11 x^{3}+x f(x)+12=0
Option: 12 x^{3}+x f(x)+12=0
Option: 13 x^{2}+2 x f(x)+4=0
Option: 14 x^{2}+2 x f(x)+4=0
Option: 15 x^{2}+2 x f(x)+4=0
Option: 16 x^{2}+2 x f(x)+4=0

Answers (1)

best_answer

Let\, y= f\left ( x \right )
x\frac{dy}{dx}+y= x^{2}
\Rightarrow \frac{dy}{dx}+\frac{1}{x}\cdot y= x
Integrating\, Factor\: IF= e^{\int \frac{1}{x}}dx= e^{lnx}= x
\Rightarrow y\cdot x= \int x^{2}dx= \frac{x^{3}}{3}+C
Passes through (-2,2)
\Rightarrow 2\cdot \left ( -2 \right )= \frac{\left ( -2 \right )^{3}}{3}+C\Rightarrow C= \frac{8}{3}-4= \frac{-4}{3}
\Rightarrow y= \frac{x^{2}}{3}-\frac{4}{3x}= f\left ( x \right )
\Rightarrow x^{3}-3xf\left ( x \right )-4= 0
option (2)

Posted by

Kuldeep Maurya

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