Q1. Let f(x) = a[x] + belxl + clxl2, where a,b,c are real constants. If f(x) is differentiable at x=0, then:
(a) b=0, c=0 a=R
(b) a=c=0, b=R
(c) a=b=0, c=R
(d) None of these
1wel just applied logic..................
c cannot be zero as |x|2 is differentiable...........
so ans is either C or D
and i kno that [x] is not differentaile at 0
so a=0
just had a doubt whether b=0....then thought that at lim x=0(minus) it wud be 1/ex....so
even b=0
so ans is C
106This is my working:
for differentiable at x=0
f(x) should be continuous at x=0
f(0) = b , f(0+)=b and f(0-)= -a+b
These three shud be equal
So, -a+b=b
a=0
Now for differentiablity,
f'(0+) = lim(h→0) f(h)-bh = lim(h→0)(beh+ch2)-bh
=lim(h→0)b(eh-1)+ch2h
= b
f'(0-) = lim(h→0) f(-h)-b-h
= lim(h→0)(be-h+ch2)-b-h
=lim(-h→0)b(e-h-1)+ch2-h = b
So im not getting any condition for b
Only a=0 i think
1hey, the function is f(x) = a[x] + be|xl + clxl2
so, while taking LHD,
f'(-0)=\lim_{h\rightarrow 0}\frac{(be^{h}+ch^{2})-b}{-h}
which gives, f'(-0)= - b
so, b=-b theregore b=0