Show that the function f(x) = |x – 3|, x ∈ |R, is continuous but not differentiable at x = 3.

OR

If x = a sin t and find

Given:

Let c be a real number

Case1: Let c < 3. Then

f(c) = 3 – c

Since, , f is continuous at all negative real numbers

Case2: Let c = 3. Then

f(c) = 3 – 3 = 0

Since, , f is continuous at x = 3

Case3: Let c > 3. Then

f(c) = c – 3

Since, , f is continuous at all positive real numbers

∴ f(x) is continuous function

Condition 2:

Now, we have to show that f(x) = |x – 3|, x Є R is not differentiable at x = 3

Now, let us consider the differentiability of f(x) at x = 3

LHD

[∵ h < 0 ⇒ |h| = -h]

RHD

[∵ h > 0 ⇒ |h| = h]

∵ LHD ≠ RHD

∴ f(x) is not differentiable at x = 3

OR

Given: x = a sin t

and

Consider x = a sin t

Differentiating with respect to t, we get

…(i)

Now, consider

Differentiating with respect to t, we get

…(ii)

Now,

[from eq. (i) and (ii)]

Again differentiating with respect to x, we get

[from(i)]