The map f (x) = A0 cos |x| + A1 sin |x| + A2 |x|^3 is differentiable at x = 0 iff ...?

(A) a1 = 0 and a2 = 0;


(B) a0 = 0 and a1 = 0;


(C) a1 = 0;


(D) a0 , a1 , a2 can take any real value;


Kindly explain your answer....

The map f (x) = A0 cos |x| + A1 sin |x| + A2 |x|^3 is differentiable at x = 0 iff ...?
(C) A1 = 0





A1sin l x l = A1 sin x for x %26gt; 0 and = - A1 sin x for x %26lt; 0


Hence, at x = 0,


left-hand derivative = - A1 cos x and its value = -A1


right-hand derivative = A1 cos x and its value = A1


For derivative to exist, both should be equal which is possible if A1 = 0.





A0 cos l x l and A2 l x^3 l are derivable at x = 0 for any values of A0 and A2.
Reply:C





f (x) = A0 cos |x| + A1 sin |x| + A2 |x|^3





|x| =x (x%26gt;=0)


|x| =-x (x%26lt;0)





if y=x^3


y'=3x^2


at x=0


y'=0





if y=(-x)^3=-x^3


y'=-3x^2


at x=0


y'=0





so yes we are good





|x|^3 is differentiable at 0, hence so is A2 |x|^3








if y=cos x


y'=-sinx


(at x=0)


y'=-sin0


=0





if y=cos (-x)


y'=-sin(-x)


(at x=0)


y'=-sin0


=0








if y=sin x


y'=cosx


(at x=0)


y'=cos0


=1





if y=sin (-x)


y'=-cos(-x)


(at x=0)


y'=-cos0


=-1





so sin|x| is NOT differentiable at 0!!!


C is the correct answer