Solution:

`:. f(x) = x/(1+ |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 differentiablc at x = 0.

Hence, f(x) is differentiable in `( -oo, oo)`.

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Solution:

Given, `f' (0) = 1` and `f' '(0) ` does not exist.

Also, given `g(x)=x f'(x)`

Put `x = 0`, we get

`g'(0) =0 f''(0) + f'(0)`

`=0+1=1`

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Solution:

We know, `| x | - | x-1 | { tt ( (-x +x -1 ,x < 0 ) , ( x+x -1 , 0 le x < 1) , ( x- (x+1) , 1 le x) )`

`=> |x| - |x-1| = { tt ( (-1 , x < 0), ( 2x-1 , 0 le x < 1), ( 1, 1 le x) )`

`:. f(x) = { |x| - |x -1 | }^2`

`= { tt ( (1, text (when) 0 < x text (or) x ge 1 ), ( (2x-1)^2 , text(when) 0 le x < 1 ) )`

Graphically, `f(x)` could be shown as


From given figure, it is clear that `f (x)` is continuous for `x in R`; but `f(x)` is not
differentiable at `x = 0, 1`.

`=> f(x)` is continuous for all `x in R`

`=> f(x)` is differentiable for all `x in R- {0, 1}`.

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Solution:

Given, `f(x) = |x- 1 |e^x`

Since, |x - 1| is not differentiable at x = 1

So, `f(x) = |x- 1| e^x` is not differentiable at `x = 1`

Hence, the required set is R - { 1 } .

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Solution:

`lim_(x-> 0) (2f(x)-3f(2x)+f(4x))/x^2` `[0/0 text(form)]`


`lim_(x-> 0) (2f'(x)-6f'(2x)+4f'(4x))/(2x)` [ using L hospital s rule]


`lim_(x-> 0) (2f''(x)-12f''(2x)+16f''(4x))/2` [using L hospital s rule]


` = (2f''(0)-12f''(0)+16f''(0))/2`


` = 3f''(0) = 3xx2 = 6`

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Solution:

`f(x) = [x]sin ( pi x)`

If `x` is just less than `k , [x] = k- 1`

`:. f(x) = ( k -1) sin ( pi x)`, when `x < k, AA k in I`

Now, `LHD` at `x = k`,

`= lim_(x->k) ( (k-1) sin (pi x) - k sin (pi k) )/(x-k)`

`= lim_(x->k) ( (k-1) sin (pi x) )/(x -k )` [ As `sin (pi k) = 0, k in` integer ]

`= lim_(h->0) ( (k-1) sin (pi (k-h) ) )/(-h )` [Let `x = (k -h)` ]

`= lim_(h->0) ( (k-1) (-1)^(k-1) sin (h pi ) )/(-h)`

`= lim_(h->0) (k-1) (-1)^(k-1) (sin h pi )/(h pi ) xx (- pi)`

`= (k-1 ) (-1)^k pi`

Therefore, (a) is the correct answer.

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Solution:

We have, `f(x) = | 2x -1 | sinx`

At `x =1/2, f(x)` is not differentiable.

Hence, `f(x)` is differentiable in `R- (1/2)`

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

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

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

and `Lf' (1/2) = lim_(h->0) (f (1/2 -h) - f (1/2))/( -h)`

`=lim_(h->0) ( |2 (1/2 -h)^(-1) | -sin(1/2 - h) -0) /(-h)`

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

`:. Rf' (1/2) ne Lf' (1/2)`

So. `f(x)` is not differentiable at `x=1/2`

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Solution:

`f (|x|) = { ( ( |x| -3 , |x| < 0), ( |x|^2 -3 |x| +2 , |x| ge 0) )`

Where,` |x | < 0` is not possible, thus neglecting, we get

`f(|x|) = { tt ( ( | x|^2 -3|x| +2 , |x| ge 0) )`

`f (|x|) = { tt ( (x^2 +3x +2 , x < 0), ( x^2 -3x +2 , x ge 0) )` ...........(i)

Again, `| f (x) | = { tt ( ( |x-3| , x < 0), ( | x^2 -3x +2 | , x < 0) )`

`| f (x) | = { tt ( ( (3-x) , x < 0), ( (x^2 -3x +2) , 0 le x < 1 ), ( - (x^2 -3x +2) , 1 le x < 2) , ( (x^2 -3x +2) , 2 le x) )` .........(ii)

Now, from Eqs. (i) and (ii), `g(x) = f ( | x |)+ | f(x) |`

`g(x) = { tt ( (x^2 +2x + 5, x< 0), ( 2x^2 -6x +4, 0 le x < 1) ,( 0, 1 le x < 2), ( 2x^2 -6x +4, x ge 2) )`


and `g'(x) = { tt ( (4x -6, 0 < x < 1), ( 0, 1 < x < 2), (4x -6 , x > 2) )`

Therefore, `g (x)` is continuous in `R - { 0}` and `g(x)` is differentiable in
`R-{0, 1, 2}`.

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Solution:

(a) To check continuity at `x = 0`

Here, `f (0)=0`

Also, `lim_(x->0) f(x) = lim_(x->0) xsin (1/x)`

`= 0 xx ` (`A` finite quantity that lies between `-1` to `+1`)

`=0 `

`:. lim_(x->0) f(x) = f(0)` and hence, `f (x)` is continuous.

Now,

(b) To check differentiability at `x = 0`

( `LHD` at `x =0`)

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

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

(A number which oscillates between `-1` and `+1`)

(`LHD` at `x = 0`) doesn't exist.

Similarly, it could be shown that `RHD` at `x = 0` doesn't exist.

Hence, `f(x)` is continuous but not differentiable.

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Solution:

Since, `f(x)` is continuous at `x = 0`, therefore

`lim_(x->0) f(x) = f(0) => lim_(x -> 0) x^n sin 1/x = 0, AA n > 0`

`f(x)` is differentiable at `x = 0`, if `lim_(x -> 0) ( f(x) - f(0))/(x - 0)`

exists finitely.

`=> lim_(x -> 0) (f(x)- f(0))/(x - 0)` exists finitely

` => lim_(x->0) x^(n-1) sin 1/x` exists finitely

` => n-1 > 0 => n > 1`

Hence, `f(x)` is continuous but not differentiable at

`x = 0`, if `n in (0,1)`.

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Solution:

For the domain of `f(x)`

(i) `2 xx - 4 >= 0 => x >= 2`

(ii) `x + 2 sqrt(2x - 4) >= 0 => 2 sqrt(2x - 4) >= -x`

which holds if `x` is positive.

(iii) `x -2 sqrt(2x -4) >= 0 => 2 sqrt(2x - 4) <= x`

`=> x >= 0` and `4(2x - 4) <= x^2`

`=> x^2 - 8x + 16 >= 0 => x in R`

`:. f (x)` is defined if `x >= 2`

Further `f (x)` is not differentiable if the expressions

inside the radical sign vanish, i.e., if `2x - 4 = 0`

or `x + 2 sqrt( 2x - 4) = 0` or `x - 2 sqrt(2x - 4) = 0`

or if `x = 2` or `4`

`:. f(x)` is differentiable in `(2, oo) - {4}`

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Solution:

`[d/(dx) (f(x) * g(x)) ]_(x=a) = f' (a) g (a) + lim_(h->0) (g (a+h) - g(a) )/h * f(a) `

If `f(a) ne 0 => g' (a)` must exist.

Also, if `g(a)` is discontinuous, `f( a)` must be `0` for `f(x) * g(x)` to be
differentiable.

Hence, (b) and (c) are the correct answers.

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Solution:

`lim_(x->y) | (f(x) -f(y))/(x-y) | le lim_(x->y) | x - y |`

or `| f'(x) | le 0 `

`=> f'(x) = 0 => f(x)` is constant.

As `f(0) = 0`

`:. f(1) = 0`

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Solution:

We have,

`f(x) = { tt( (x^2 +3x+p , text(if) x le 1) ,(qx +2, text(if) x >1) )` is differentiable at `x=1`.

`:. f' (1) =lim_(x->1) ( f(x) -f(1))/(x-1)`

LHD `= lim_(x-> 1^-) ( (x^2 +3x + p) -(1+3 +p))/(x-1)`

`= lim_(h-> 0)( [ (1-h)^2 + 3 (1-h) + p] - [1+3+p])/( (1-h)-1)`

`= lim_(h->0) ([1+h^2 -2h +3 -3h +p] -[4+p ])/(-h)`

`= lim_(h->0) ( [h^2 -5h + p+4 -4 -p])/(-h) = lim_(h->0) (h [h-5]) /(-h)`

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

RHD ` f'(1) =lim_(x->1^+) ( f(x) - f(1))/(x-1) = lim_(x-> 1^+) ( (qx+2) - (1+3 +p) )/(x-1)`

`= lim_(h-> 0) ( [q(1+h) +2 ] - (4+p)) /(1+ h-1)`

`= lim_(h->0) ( [ q +qh +2 -4 -p])/h = lim_(h->0) ( qh + (q-2-p)) /h`

`=> q-2 -p =0 => p -q =-2` ......................(i)

`=> lim_(h-> 0) (qh +0)/h =q` [for existing the limit]

If `L H D f'(1) = R H D f'(1) `, then `5 =q`

`=> p-5 = -2 => p =3`

`:. p =3` and `q =5`

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Solution:

We have, `f(x) = { tt ((x^2, x ge 0),(-x^2, x < 0))`

Clearly, `f(x)` is differentiable for all `x > 0` and
for all `x < 0`. So, we check the differentiability
at `x = 0`.

Now, (RHD at `x =a`)

`= (d/(dx)(x^2))_(x=0) =(2x)_(x=0)=0`

`:.` (LHD at `x=0`) `=(d/(dx)(-x^2))_(x=0)= (-2x)_(x=0)=0`

(LHD at `x = 0`) = (RHD at `x = 0`)

So, `f (x)` is differentiable for all `x` ie, the set of all
points where `f(x)` is differentiable is `(- oo, oo)` ie,
R.

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Solution:

We have, `lim_(x -> 1^-) f(x) = lim_(x -> 1^-) (1- x) = 0`

`lim_(x -> 1^+) f(x) = lim_(x -> 1^+) (1 - x) (2 -x) = 0`

and `f(1) = 0`

`:. lim_(x -> 1^-) f(x) = lim_(x -> 1^+) f(x) = f(1)`

So, `f(x)` is continuous at `x = 1`.

Now, (LHD at `x =1`) `= [ d/(dx) (1 -x) ]_(x = 1) = -1`

(`RHD` at `x = 1`) `= [d/(dx) (1 - x) (2 - x) ]_(x = 1) = - 1`

`:.` (`LHD` at `x =1`) = (`RHD` at `x = 1`)

So, `f( x)` is differentiable able at `x = 1`.

Now, `lim_(x -> 2^-) f(x) = lim_(x -> 2^-) (1 -x)(2 -x) = 0`

and `lim_(x -> 2^+) f(x) = lim_(x -> 2^+) (3 -x) = 1` .

`:. lim_(x -> 2^-) f(x) = lim_(x -> 2^+) f(x)`

Thus, `f( x)` is discontinuous and not differentiable

at `x = 2`.

Hence, the only point of discontinuity and

non-differentiability of `f( x)` is `x = 2` .

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