Question

If f(x) = |x| + |x − 1|, then which of the following is correct?

पर्याय

• f(x) is both continuous and differentiable, at x = 0 and x = 1.

• f(x) is differentiable but not continuous, at x = 0 and x = 1.

• f(x) is continuous but not differentiable, at x = 0 and x = 1.

• f(x) is neither continuous nor differentiable, at x = 0 and x = 1.

Answer 1

f(x) is continuous but not differentiable, at x = 0 and x = 1.

Explanation:

Given: f(x) = |x| + |x − 1|

We may rewrite f(x) as

`f(x) = [(-2x + 1",", x ≤ 0),(                 1",", 0 ≤ x ≤ 1),(    2x - 1",", x ≥ 1):}`

For continuity at x = 0

L.H.L.  `lim_(x -> 0^-)f(x) = lim_(h -> 0) f(0 - h)`

`=> lim_(h -> 0)-2(-h) + 1`

= 1

R.H.L.  `lim_(x -> 0^+)f(x) lim_(h -> 0) f(0 + h)`

`=> lim_(h -> 0) 1`

= 1

f(0) = 1

L.H.L. = R.H.L. = f(0)

So, F(x) is continuous at x = 0

at x = 1

L.H.L.  `lim_(x -> 1^-)f(x) = lim_(h -> 0) f(1 - h) = 1`

R.H.L.  `lim_(x -> 1^+)f(x) = lim_(h -> 0) f(1 + h)`

`=> lim_(h -> 0) 2(1 + h) - 1` 

= 1

f(1) = 2(1) − 1

= 1

Now, LHL = RHL = f(1)

Hence, f(x) is continuous at x = 1

For differentiability at x = 0

L.H.D. `lim_(h -> 0) (f(0 - h) - f(0))/(-h)`

= `(2h + 1 - 1)/(-h)`

= `(2h)/-h`

= −2

R.H.D. `lim_(h -> 0) (f(0 + h) - f(0))/(h)`

= `lim_(h -> 0) (1 - 1)/(h)`

= 0

So, LHD ≠ RDH

Hence, f(x) is not differentiable at x = 0

at x = 1

LHD,  `lim_(h -> 0) (f(1 - h) - f(1))/(-h)  h_1`

= `lim_(h -> 0)(1 - 1)/(-h)`

= 0

RHD = `lim_(h -> 0)(f(1 + h) - f(1))/(-h)`

= `(2(1 + h) - 1 - 1)/(h)`

= 2

LHD ≠ RHD

Hence, f(x) is not differentiable at x = 1 and x = 0