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Tricks & Tips of differentiability for NDA Lecture - 2

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Differentiability of Modulus function

By Rule that |x-a| is not differentiable at x= a

Q 2821112921

Discuss the continuity and differentiability of `f(x) = |x + 1|+ |x| + |x - 1|. AA x in R`; also draw the graph
of `f(x)`.

Solution:

Continuous ` AA x in R`, differentiable for `x in R- {- 1, 0, 1}`
Q 2844212153

Consider the following statements

I. `f (x) = | x- 3 |` is continuous at x = 0.

II. `f (x) = | x- 3 |` is differentiable at x = 0.

Which of the above statement(s) is/ate correct?

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

` :. f(x) = | x-3 | = { tt ( (x-3 , x ge 3), (3-x , x < 3) )`

`:. LHL= lim_(x->0^-) f(x) = lim_(b->0) f ( 0- b)`

`= lim_(b ->0 ) (3 +b) =3`

and `RHL = lim_(x -> 0^+) f(x) = lim_(b->0) f (0+b)`

`lim_(b->0) (3- b) = 3`

`=> LHL = RHL`

So, f( x) is continuous at x = 0.
Now, LHD

`= f' (0^-) = lim_(b->0) (f(0) - f (0-b) )/b`

`= lim_(b->0) ( 3- (3- b) )/b =1`

and RHD

`= f' (0^+) = lim_(b->0) ( f (0+b) -f (0) )/b `



`= lim_(b ->0) (3+b -3)/b = 1`

`=> LHD = RHD`

`:. f(x)` is differentiable at x = 0 .
Hence, both Statements I and II are
correct.
Correct Answer is `=>` (C) Both I and II

Differentiability of Modulus function function By converting it into piecewise function

Q 2106101978

Consider the equation `x+ | y |= 2y.`

Which of the following statements are not correct?
1. y as a function of x is not defined for all real x.
2. y as a function of x is not continuous at x = 0.
3. y as a function of x is differentiable for all x.
NDA Paper 1 2016
(A)

1 and 2

(B)

2 and 3

(C)

1 and 3

(D)

1, 2 and 3

Solution:

Given, `x +| y | = 2y`

`x+ y =2y, ` for `y > 0`

`=> y = x`

and `x- y = 2y`, for `y < 0`

`=> y= 1/3 x`

Graph of quad `x + | y | = 2y`

From the graph

`y` as a function `x` defined for all real `x`

and `y` as function of `x` is not differentiable at `x = 0`

`:.` Statements `1` and `3` are incorrect.
Correct Answer is `=>` (C) 1 and 3
Q 2400245118

Which one of the following function is
differentiable for all real values of `x`?
NDA Paper 1 2012
(A)

`x/(|x|)`

(B)

`x | x |`

(C)

`1/(|x|)`

(D)

`1/x`

Solution:

Let us take the function `f(x) = x | x |`

Redefine this function,

`f(X) = { tt ( (x^2 , text (if) x ge 0 ), (-x^2 , text (if) x < 0 ) )`

`Lf' (0) = lim_(h->0) (f(0-h) -f(0) )/(-h) = lim_(h->0) (- (-h)^2 - 0)/(-h)`

`= lim_(h->0) (-h^2)/(-h) = lim_(h->0) (+h) =0`

`Rf'(0) = lim_(h->0) (f (0+h) -f (0) )/h = lim_(h->0) (h^2 - 0)/h`

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

`:. Lf' (0) = Rf' (0)`

Hence, `f(x)` is differentiable for all real values of `x .`
Correct Answer is `=>` (B) `x | x |`
Q 2940812713

If `4x + 3| y|= 5y`, then `y` as a function of `x` is

(A)

not continuous at `x = 0`

(B)

not defined for all real `x`

(C)

`(dy)/(dx)= 1/2` for `x < 0`

(D)

derivable at `x = 0`

Solution:

We have, `4x + 3 | y | = 5y`

`=> 4x +3y = 5y` if `y ge 0`

and `4x - 3y = 5y` if `y < 0 => y = { tt (( 2x text(,), x ge 0),( 1/2 x text(,), x < 0))`

Clearly, `y` is continuous at `x = 0` but not differentiable at `x = 0`.

Also, `(dy)/(dx) = { tt(( 2 text(,), x ge 0),( 1/2 text(,), x <0))`
Correct Answer is `=>` (C) `(dy)/(dx)= 1/2` for `x < 0`
Q 2410356219

What is the set of all points, where the function

`f(x) = x/(1+ |x|)` is differentiable?
NDA Paper 1 2009
(A)

Only `(-oo , oo)`

(B)

Only `(0, oo)`

(C)

Only `(-oo , 0) cup (0, oo)`

(D)

Only `(-oo ,0)`

Solution:

`:. f(x) = x/(1+ |x|)`

Redefined the function,

`f(x) = { tt ((x/(1-x) , x < 0 ), (x/(1+x) , x ge 0) )`

`:. LHD = f'(0^-) = lim_(h->0) (f(0-h) -f(0) )/(-h)`

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

and `RHD= f' (0^+)`

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

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

`:. LHD = RHD`

So, `f(x)` is differentiable at `x = 0`.

Hence, `f(x)` is differentiable in `( -oo, oo)`.
Correct Answer is `=>` (A) Only `(-oo , oo)`
Q 2106101978

Consider the equation `x+ | y |= 2y.`

Which of the following statements are not correct?
1. y as a function of x is not defined for all real x.
2. y as a function of x is not continuous at x = 0.
3. y as a function of x is differentiable for all x.
NDA Paper 1 2016
(A)

1 and 2

(B)

2 and 3

(C)

1 and 3

(D)

1, 2 and 3

Solution:

Given, `x +| y | = 2y`

`x+ y =2y, ` for `y > 0`

`=> y = x`

and `x- y = 2y`, for `y < 0`

`=> y= 1/3 x`

Graph of quad `x + | y | = 2y`

From the graph

`y` as a function `x` defined for all real `x`

and `y` as function of `x` is not differentiable at `x = 0`

`:.` Statements `1` and `3` are incorrect.
Correct Answer is `=>` (C) 1 and 3

Differentiability of Modulus function function By Graph

Q 2513701640

Let `f(x) = ||x| -1|`, then points were f(x) is not differentiable is are
JEE Prelims 2005
(A)

0, ±1

(B)

±1

(C)

0

(D)

1

Solution:

It is given that

From the graph shown here, it is obvious that f(x) is not differentiable at `x =-1, 0, 1,` that is, `0, ±1`.
Correct Answer is `=>` (A) 0, ±1
 
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