wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

f(x)=2x+1x33x2+2x.

Open in App
Solution

f(x)= 2x+1x23x+2x=2x+1x(x2)(x1)
[f(x)]2=2x+1x(x2)(x1)=Ax+B(x2)+C(x1)
simplifying 2x+1=A(x2)(x1)+B(x1)x+C(x2)x
2x+1=At x2(A+B+C)+x(3AC2C)+A(2A+0+0)
comparing coefficient
2A=1 A=1/2A+B+C=0 B+C=1/2
2=.AB2C 2=32B2C B+2C=322=72
B+C=1/2 , B+2C=7/2
solving C=3 , B=5/2 , A=1/2
[f(x)]2=12x+52(x2)+3(x1)
differentiating w.r.t x
2f(x),f(x)=12x2+52(x2)2+3(x1)2
f(x)=122x+1x(x2)(x1)=⎢ ⎢ ⎢ ⎢12x252(x2)2+3(x1)2⎥ ⎥ ⎥ ⎥


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Sum of n Terms of an AP
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon