Question #9318c

1 Answer
Sep 2, 2017

a=-5/2
b=-3/2

Explanation:

We have
f (x)=(x (1+acosx)-bsinx)/x^3
Given f (0)=1
And as f (x) is continuous so
lim f (x)=1
xrarr0
Now applying L'hopitals rule we get by differentiating numerator and denominator separately
limx→0 f (x)=(x×(0+a (-sinx))+(1+acosx)-bcosx)/(3x^2)
=>limx→0f (x)=(1-axsinx+(a-b)cosx)/(3x^2)
Now as xrarr0 denominator approaches zero hence numerator must also approach zero.
:.1+a-b=0
=>b-a=1 ........(1)
Now again applying L'hospital rule we get
limxrarr0f (x)=(-axcosx-asinx+(a-b)(-sinx))/(6x)
=(-axcosx-(2a-b)sinx)/(6x)
Now using limxrarro(sinx/x)=1 as xrarr0 we have
limxrarr0 f (x)=(-ax/xcosx-(2a-b)sinx/x)/6=-(3a-b)/6
Now -(3a-b)/6=1
=>3a-b=-6 .........(2)
From equation (1) and (2)
a=-5/2
b=-3/2