Question

(a) (4 marks) Consider the function S(x) = x-cos(x). 1) Prove that S has at least one zero in the interval [0, ) f(0)f(x) <0. (O)f(x) > 0 and is continuous f(0)f(x) < 0 and is continuous. ii) Prove that S has at most one zero in the interval [0, x) f' <0 on (0,#] so that is strictly increasing on (0,r)- 1'>0 on (0,#] so that f is strictly increasing on (0, #) 1'>on (0,r) so that is strictly decreasing on (0,8).

Answer 1

í f(x)= 23 - COS X f(0) = 03-COS(O) - 0-1 = -1 f(1) = 173- Cos (IT) - 173-(-1) = 1+1 at x=0 W > n = π U 7 is aris once a Since f lies below aris and at f lies above it de will intersect surely pass the x aris between f(x) is continuous. f(x) = x² - coin is continuous function since has two terms 73 and and both are separately continuous in [0,11]. f has at least one zero in interval [o,in flo)f(uco and f is continuous Conn : ; floJfC30 ne E Co, T] is f(z) is. f'(): 32-(-sin) f'(x) = 3x² + sinx : 3x2 > 0 in and sin a>0 in ne [0, 1] :: f'(x) = 3x2 + sink is greater than 0 in ac(0,1] strictly interval f'(x) >0 in in breasing that interval since f(x) >O in ne in v FCO,11] · fra strictly increases in (0,17] f has almost one zero in the interval [0,] since t'o on (0,1] so that f is ncreasing (0.11] in an if and Answer :- on Strictly in