If x2-5x+7x-13=Ax-1+Bx-12+Cx-13 then A+B-C= _____

x2-5x+7(x-1)3x-1=y ⇒ x=y+1 (y+1)2-5(y+1)+7y3 =y2-3y+3y3 =1y-3y2+3y3A=1, B=-3, C=3A+B-C=1-3-3=-5

If x2-5x+7x-13=Ax-1+Bx-12+Cx-13 then A+B-C= _____

x2-5x+7(x-1)3x-1=y ⇒ x=y+1 (y+1)2-5(y+1)+7y3 =y2-3y+3y3 =1y-3y2+3y3A=1, B=-3, C=3A+B-C=1-3-3=-5

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

$If \frac{x^{2} - 5 x + 7}{{(x - 1)}^{3}} = \frac{A}{x - 1} + \frac{B}{{(x - 1)}^{2}} + \frac{C}{{(x - 1)}^{3}} then A + B - C =_____$

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

-5

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{rcl} \frac{x^{2} - 5 x + 7}{{(x - 1)}^{3}} \\ x - 1 & = & y \Rightarrow x = y + 1 \\ \frac{{(y + 1)}^{2} - 5 (y + 1) + 7}{y^{3}} \\ = & \frac{y^{2} - 3 y + 3}{y^{3}} \\ = & \frac{1}{y} - \frac{3}{y^{2}} + \frac{3}{y^{3}} \\ A & = & 1, B = - 3, C = 3 \\ A + B - C & = & 1 - 3 - 3 = - 5 \end{array}$