Please Wait... While Loading Full Video

Tricks & Tips of differentiability for NDA Lecture - 3

Click for Only Video

Differentiability of GIF Function

Q 2930512412

lf `f(x) = [x - 2],` then

(A)

`f'(2.5) = 1/2 ` and `f'(5) =3`

(B)

`f'(2.5) =0` and `f'(5) =3`

(C)

`f'(2.5) =0` and `f'(5)` does not exist

(D)

both `f' (2.5)` and `f' (5)` do not exist

Solution:

We have

`Lf'(2.5) = lim_(h->0) (f(2.5 -h)- f(2.5))/(-h)`

`=lim_(h->0) ([2.5 -h -2]-[2.5 -2])/(-h)`

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

and `R f'(2.5) = lim_(h->0) (f (2.5 +h)- f(2.5))/h`

`=lim_(h->0) ([2.5 +h -2]- [2.5 -2])/h`

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

`:. f'(2.5)=0`

Also, `Lf'(5) = lim_(h->0) (f (5-h) -f(5))/(-h)`

`=lim_(h->0) ([5-h-2]-[5-2])/(-h)`

`=lim_(h->0) (2-3)/(-h) -> oo`

Hence `f' (2 · 5) = 0` while `f' (5)` does not exist
Correct Answer is `=>` (C) `f'(2.5) =0` and `f'(5)` does not exist
Q 2519780610

The function `f(x) = [x^2] + [- x]^2`, where [ ] denotes the greatest integer function. is
BCECE Mains 2015
(A)

continuous and derivable at `x = 2`

(B)

neither continuous nor derivable at `x = 2`

(C)

continuous but not derivable at `x = 2`

(D)

None of the above

Solution:

We have,

` lim_(x -> 2^-) f(x) = lim_(h -> 0) f(2- h)`

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

`=> lim_(x -> 2^-) f(x) =3 + (- 2)^2 = 7`

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

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

` => lim_( x -> 2^+) f(x) = 4 + (- 3)^2 = 13`

Clearly, `lim_(x -> 2^-) f(x) != lim_(x -> 2^+) f(x)`.

So, `f( x)` is discontinuous at `x = 2`.

Consequently, it is not differentiable at `x = 2`.

Miscellaneous

Q 2930012812

If `f(x) = { tt (( 3 text(,), x <0),(2x+1 text(,), x ge 0))` then

(A)

both `f(x)` and `f(|x|)` are differentiable at `x = 0`

(B)

`f(x)` is differentiable but `f(|x|)` is not differentiable at `x = 0`

(C)

`f(|x|)` is differentiable but `f(x)` is not differentiable at `x = 0`

(D)

both `f(x)` and `f(|x|)` are not differentiable at `x = 0`

Solution:

We have,

`Lf'(0) = lim_(h->0) (f(0-h)-f(0))/(-h) = lim_(h->0) (3-1)/(-h) -> -oo`

`:. f(x)` is not differentiable at `x = 0`

Also, if `x < 0` or `x ge 0` then `| x | ge 0`

`:. f(|x|) = 2|x| +1` for all `x`.

`:. R f'(0) = lim_(h->0) (2 h+1-1)/h =2`

and `L f'(0) = lim_(h->0) (f(0-h) -f(0))/(-h)`

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

`:. f(|x|)` is differentiable at `x = 0`.

The correct option is (3)
Correct Answer is `=>` (C) `f(|x|)` is differentiable but `f(x)` is not differentiable at `x = 0`
Q 2920123011

Let ` f: R-> R` be any function. Define `g:R -> R` by `g(x)= |f(x)|` for all `x`. Then `g` is

(A)

onto if `f` is onto

(B)

one-one if `f` is one-one

(C)

continuous if `f` is continuous

(D)

differentiable if `f` is differentiable

Solution:

Let `h (x) =| x|` for all `x`. Clearly, `h (x)` is continuous for all `x`.

Then `g(x)= |f(x)| = h[f(x)] = (hof) (x)` for all `x`.

Since composition of two continuous functions is continuous,
therefore, `g` is continuous if `f` is continuous.

The correct option is (3)
Correct Answer is `=>` (C) continuous if `f` is continuous
Q 2623478341








Column IColumn II
(A) Number of points of discontinuity of `f(x) = tan^2 x - sec^2 x` in
`(0, 2 pi )` is 		(P) `4`
(B) Number of points at which `f(x) = sin^(-1) x + tan^(-1) x + cot^(-1) x` is
non-differentiable in `(-1, 1)` is 		(Q) `3`
(C) Number of points of discontinuity of `y =[sin x], x in [0, 2 pi)`
where `[*]`represents greatest integer function 		(R) `2`
(D) Number of points where `y = | (x- 1 )^3| + | (x- 2)^5 | + | x- 3|` is
non-differentiable 		(S) `1`
(T) `0`


(A)

(A)-> (s), (B)-> (t), (C)-> (r), (D)-> (q)

(B)

(A)-> (t), (B)-> (t), (C)-> (r), (D)-> (q)

(C)

(A)-> (r), (B)-> (t), (C)-> (r), (D)-> (q)

(D)

(A)-> (r), (B)-> (p), (C)-> (r), (D)-> (q)

Solution:

(A) ` tan^2 x` is discontinuous at `x = pi/2 , (3 pi )/2`

`sec^2 x` is discontinuous at `x = pi/2 , (3 pi)/2`

`=>` Number of discontinuities `= 2`

(B) Since `f(x) = sin^(-1) x + tan^(-1) x + cot^(-1) x = sin^(-1) x + pi/2`

`:. f(x)` is differentiable in `(- 1, 1) => ` no. of points of non-diff. `= 0`

(C) ` y = [ sin x ] = { tt ( (0, 0 le x < pi/2 ) , ( 1 , x = pi/2 ) , ( 0 , pi/2 < x le pi ) , ( -1 , pi < x < 2 pi ) , ( 0 , x = 2 pi ) )`

`:.` points of discontinuity are `pi/2 , pi`

(D) `y = | (x-1)^3 | + | (x-2)^5 | + | x-3 |` is non differentiable at `x = 3` only
Correct Answer is `=>` (C) (A)-> (r), (B)-> (t), (C)-> (r), (D)-> (q)
Q 2653378244

`S_1 :` If `f` is continuous and `g` is discontinuous at `x = a`, then `f(x)* g(x)` is disqontinuous at `x = a`.

`S_2 :` `f(x) = sqrt (2-x) + sqrt (x-2)` is not continuous at `x = 2`.

`S_3 :` `e^(- |x| )` is differentiable at `x = 0`.

`S_4 :` If `f(x)` is differentiable every where, then `| f |^2` is differentiable every where(

(A)

TTFF

(B)

TTFT

(C)

FTFT

(D)

FFFT

Solution:

`S_1 :` False (take `f(x) = 0, x in R`

`S_2 :` Domain of `f(x)` is `{ 2}`

`:. f(x)` is not continuous at `x = 2`

`S_3 :` `e^(- |x| )` not differentiable at `x = 0`

`S_4 :` Derivative of `| f|^2` is `0` where ever `f(x)` is `0` and the derivative of `| f|^2` is

`2 | f(x) | * (f(x) )/( | f(x) |) = 2f (x)` where ever

`f(x) ne 0`
Correct Answer is `=>` (C) FTFT
Q 2663478345








Column IColumn II
(A) `f(x) = | x^3 |` is (P) continuous in `(-1, 1)`
(B) `f(x) = sqrt ( |x| )` is (Q) discontinuous in `(-1, 1)`
(C) `f(x) = | sin^(-1) x| ` is (R) differentiable in `(0, 1)`
(D) `f(x) = cos^(-1) |x|` is (S) not differentiable atleast at one point in
`(-1, 1)`
(T) differentiable in `(-1, 1)`


(A)

(A) -> (p, t, r), (B) -> (p, r, s), (C)-> (p, r, s), (D) -> (p, r, s)

(B)

(A) -> (p, s, t), (B) -> (p, r, s), (C)-> (p, r, s), (D) -> (p, r, s)

(C)

(A) -> (p, t, r), (B) -> (p, r, s), (C)-> (p, t s), (D) -> (p, r, s)

(D)

(A) -> (p, t, r), (B) -> (p, r, s), (C)-> (p, q, t), (D) -> (p, r, s)

Solution:

(A) `f(x) = | x^3 |` is continuous and differentiable

(B) `f(x) = sqrt (|x|)` is continuous

`f' (x) = 1/(2 sqrt (|x|) ) * x/(|x|)` {does not exist at `x = 0`}

(C) `f(x) = |sin^(-1) x |` is continuous

`f '(x) = (sin^(-1) x)/( | sin^(-1) x | ) * 1/(sqrt (1-x^2) )` {does not exist at `x = 0`}

(D) `f(x) = cos^(-1) |x|` is continuous

`f '(x) = (-1)/(sqrt (1-x^2) ) * x/(|x|)` {does not exist at `x = 0`}
Correct Answer is `=>` (A) (A) -> (p, t, r), (B) -> (p, r, s), (C)-> (p, r, s), (D) -> (p, r, s)
 
SiteLock