Ajay Verma
Last Activity: 10 Years ago
solution:
a= min (x^2+ 4x +5)
( using dy/dx= 0 at critical points)
y= x^2+ 4x +5 ; dy/dx = 2x+4 = 0 so x = -2
d2y/dx2 = 2(-2) + 4 = 1 so minimum..
so a = 1
b = lim y --> 0 = (1- cos 2y)/ y2 : 0/0 form so use DL hospi low..
b= lim y-->0 = 2 sin2y / 2y ; lim y-->0 sin 2x/ 2x = 1
soo b= 2
a-b/ a+b = 1-2 / 1+2 = -1/3
Thanks and Regards,
Ajay verma,
askIITians faculty,
IIT HYDERABAD
