Skip to content
EXERCISE 8.1 · Q13

Q.Test the continuity of the following function at the point indicated against it: f(x)=x2+8x−202x2−9x+10f(x) = \dfrac{x^2+8x-20}{2x^2-9x+10}, for 0<x<30 < x < 3, x≠2x \ne 2, =12= 12, for x=2x = 2, =2−2x−x2x−4= \dfrac{2-2x-x^2}{x-4}, for 3≤x<43 \le x < 4, at x=2x = 2.

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est
18% · 13/73 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 →

On 0<x<3, x≠20<x<3,\,x\ne2: f(x)=x2+8x−202x2−9x+10f(x)=\dfrac{x^2+8x-20}{2x^2-9x+10}, and separately f(2)=12f(2)=12.

Factor: x2+8x−20=(x+10)(x−2)x^2+8x-20=(x+10)(x-2), and 2x2−9x+10=(x−2)(2x−5)2x^2-9x+10=(x-2)(2x-5) (checking: (x−2)(2x−5)=2x2−5x−4x+10=2x2−9x+10(x-2)(2x-5)=2x^2-5x-4x+10=2x^2-9x+10, correct). So for x≠2x\ne2,

f(x)=(x+10)(x−2)(x−2)(2x−5)=x+102x−5.f(x)=\frac{(x+10)(x-2)}{(x-2)(2x-5)}=\frac{x+10}{2x-5}.

lim⁡x→2f(x)=2+102(2)−5=12−1=−12.\lim_{x\to2} f(x)=\frac{2+10}{2(2)-5}=\frac{12}{-1}=-12. …

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.