( by using L’Hospital Rule )

which does not exist at x = 0 .

here f(x)=|x−1| x∈R. So f(x) is not derivable when x−1=0 i.e. at x=1

therefore , f is neither continuous nor differentiable at x = 0

If f is strictly increasing function , then f ‘ (x) can be 0 also . For example , f(x) = x3 is strictly increasing , but its derivative is 0 at x = 0. As another example , take f(x) = x + cosx ; here f ‘(x) = 1 – sinx , which is either +ve or 0 and the function x + cos x is strictly increasing.

f(x) = x - is derivable at all x R – I , and f ‘(x) = 1 for all x R – I

Let f (x + y) = f(x) + f(y) x, y Suppose that f (6) = 5 and f ‘ (0) = 1, then f ‘ (6) is equal to

If x sin (a + y) = sin y, then is equal to