Skip to content
Question 126 of 148

Q.Find the points on the curve y=x3−3x2+x−2y=x^3-3x^2+x-2 at which the tangent is parallel to the line y=xy=x.

Puducherry TnboardTamil Nadu HSC (DGE) Board 2022Subjective· 2mImportance★★★★★
85% · 126/148 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Sets the derivative equal to the slope of y=x and solves for the x-coordinates, then finds the matching y-values on the curve.

  1. Given y=x3−3x2+x−2y=x^3-3x^2+x-2, differentiate: dydx=3x2−6x+1\dfrac{dy}{dx}=3x^2-6x+1.
  2. The line y=xy=x has slope 11; a tangent parallel to it must also have slope 11.
  3. Set 3x2−6x+1=1⇒3x2−6x=0⇒3x(x−2)=03x^2-6x+1=1 \Rightarrow 3x^2-6x=0 \Rightarrow 3x(x-2)=0.
  4. So x=0x=0 or x=2x=2. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.