Answer :

It is given that


Now, f’(x) =0



The points x =1 and x = -1 divide the real line in three disjoint intervals


(-∞,-1),(-1,1) and (1,∞)


Now in interval, (-1,1)


it is clear that -1 < x < 1


x2 < 1




Therefore, f’(x) = 1- < 0 (-1,1) ~ {0}


Therefore, f is strictly decreasing on (-1,1) ~ {0}


x < -1 or 1 < x


x2 > 1




Therefore, f’(x) = 1- > 0 (-∞, -1) and (1,∞)


Therefore, f is strictly increasing in interval I disjoint from (-1,1)


Hence Proved.


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