if x2+13x+15x-1x2+2=Ax-1+Bx+Cx2+2, then the descending order of A, B, C is

x2+13x+15(x-1) (x2+2)=Ax-1+Bx+cx2+2 x2+13x+15=A(x2+2)+(Bx+c) (x-1) x=1 ⇒ 29=3A ⇒ A=293 x2 cpeff ⇒ 1=A+B ⇒ B=1-293=-263 const ⇒ 15=2A-C ⇒ C=2A-15=583 -15=133Descending order A, C, B

if x2+13x+15x-1x2+2=Ax-1+Bx+Cx2+2, then the descending order of A, B, C is

x2+13x+15(x-1) (x2+2)=Ax-1+Bx+cx2+2 x2+13x+15=A(x2+2)+(Bx+c) (x-1) x=1 ⇒ 29=3A ⇒ A=293 x2 cpeff ⇒ 1=A+B ⇒ B=1-293=-263 const ⇒ 15=2A-C ⇒ C=2A-15=583 -15=133Descending order A, C, B

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$i f \frac{x^{2} + 13 x + 15}{(x - 1) (x^{2} + 2)} = \frac{A}{x - 1} + \frac{Bx + C}{x^{2} + 2}, then the descending order of A, B, C is$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

A, B, C

B, C, A

C, A, B

A, C, B

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\frac{x^{2} + 13 x + 15}{(x - 1) (x^{2} + 2)} = \frac{A}{x - 1} + \frac{Bx + c}{x^{2} + 2} x^{2} + 13 x + 15 = A (x^{2} + 2) + (Bx + c) (x - 1) x = 1 \Rightarrow 29 = 3 A \Rightarrow A = \frac{29}{3} x^{2} cpeff \Rightarrow 1 = A + B \Rightarrow B = 1 - \frac{29}{3} = \frac{- 26}{3} const \Rightarrow 15 = 2 A - C \Rightarrow C = 2 A - 15 = \frac{58}{3} - 15 = \frac{13}{3}$