Solution:

If `f(x) = x^n`

Then `f'(x) = nx^(n-1) AA x in R^+ , n in R`

`=> n-1 ge 0`

`=> n in [1,oo)`

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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.

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Solution:

For `x = 0`,

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

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

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

`= lim_(h -> 0) ( f(h) - f(0))/h = lim_(h -> 0) (pi cos h - pi )/2`

` = lim_(h -> 0) ( pi [cos h - 1] )/2 = lim_(h -> 0) (pi [-2 sin^2 (h//2)])/h`

`= lim_(h -> 0) pi [ - sin (h //2)/(h//2) xx sin (h //2) ] = pi [ - 1 xx 0 ] = 0`

`LHD != RHD`

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

`:.` Statement `1` is not correct.

For `x = pi/2`,

`LHD = lim_(h -> 0) ( f (pi //2 - h) - f(pi//2))/(-h)`

`= lim_(h -> 0) ( pi cos (pi/2 - h) -0)/(-h) = lim_(h - 0) ( pi sin h)/(-h) = -pi`

`RHD = lim_(h ->0) + ( f (pi//2 + h ) - f(pi//2))/h`

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

`LHD != RHD`

Hence, `f(x)` is not differentiable at `x = pi/2`.

Statement `2` is not correct.

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Solution:

Given, f" (x) = g" (x)

On integrating both sides, we get

`f'(x) = g'(x) + c`

` => f'(1) = g'(x)+ c`

` => 4 = 6 + c`

`=> c= -2`

`:. f'(x) = g'(x) - 2 `

Again, on integrating both sides, we get

`f(x) = g(x) - 2x + c_(1)`

`=> f(2) = g(2) - 2 xx 2 + c_(1)`

`=> 3 = 9 - 4 + c_(1)`

`=> c_(1) = -2`

`:. f(x)- g(x) = -2x- 2`

At `x = 4, [f(x)- g(x)] = - 8 - 2 = -10`

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Solution:

`f(x) = { tt ((2+x , x ge 0) , (2-x , x < 0))`

1. `lim_(x→1^-) f(x) = lim_(h→0) f(1-h)`

` = lim_(h→0) (2+1-h) = 3`

`lim_(x→1^+) f(x) = lim_(h→0) f(1+h)`

` = lim_(h→0) (2+1+h) = 3`

`lim_(x→1) f(x) = lim_(x→1) 2+x = 3`

mean s and exist at `x = 1`

So 1 is incorrect

2. For differentiability at `x = 0`

RHD `lim_(h→0) (f (0+h)-f(0))/h = ((2+h)-2)/h = 1`

LHD `lim_(h→0) ( f(0-h)-f(0))/h = (2+h-2)/h = 1`

Differentiable at `x= 0`

3. If function is differentiable at `x = 0` then it should be continuous too on `x = 0`

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Solution:

We have,

`f(x) = { tt ((-2, -3 <= x <= 0 ),( x -2 , 0 < x <= 3) )`

`=> f (|x | ) = { tt (( -2, -3 <= | x | <= 0),( x -2 , 0 < x <= 3 ))`

At `-3 <= | x| <= 0 , f (|x|) = -2` not possible

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

and ` | f(x) | = { tt (( |-2| , -3 <= x <= 0),( | x -2 |, 0 < x <= 3 ))`

`=> |f(x) | = { (2 , -3 <= x <= 3), ( 2 -x , 0 < x < 2 ) , ( x -2 , 2 <= x <= 3 ))` ......................(ii)

From Eqs. (i) and (ii). we get

`g(x) = f (|x|) + |f (x) | `

`g(x) = { tt (( -x , -3 <= x <= 0 ), ( 0, 0 < x < 2 ), (2x -4 , 2 <= x <= 3 ))`

`g(x)' = { tt (( -1 , -3 <= x <= 0 ), ( 0, 0 < x < 2 ), (2 , 2 <= x <= 3 ))`

The graph of g (x) is given

Since, g (x) has sharp edges at x = 0 and x = 2.

Hence, g (x) is not differential at x = 0 and x = 2.

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Solution:

From the graph of g(x), we can say that g(x) is continuous
at x = 0, x = 2 and x = -1

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Solution:

at `x =-2`

`g '(x) = -1`

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Solution:

1. For continuous,

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

`:. lim_(x -> 3^-) f(x) = lim_(x -> 3^+) f(x) = 4`

Hence, `f(x)` is continuous at `x = 4`.

2. We have, `f(x) = x^2 - 5, x <= 3`

` => f'(x)=2x => f'(0)= 0`

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

So, neither Statement `1` nor `2` is correct.

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Solution:

I. Graph of f(x) = |x| from graph we obseNe
that, the curve has sharp turn at x = 0.
So, the function f(x) = |x | is not differentiable only at x = 0.

i.e., f(x) = |x| is differentiable at x = 1

II. Given function, `f(x)=e^x`.

Now, we check the differentiability of f(x) at x = 0.

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

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

`lim_(h-> 0) (e^h- 1)/h \ \ \ \ \ \ \ \ \ text(form) 0/0`

Use L' Hospital rule,

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

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

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

Use L' Hospital rule,

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

Since, R f' (0) = L f' (0)

Hence, `f(x) =e^x` is differentiable at x = 1.

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Solution:

Given function, `f(x) = {tt {(2-x, text(for) 1<= x <= 2),( 3x - x^2, text(for) x > 2))`

and whole function defined in `1 <= x < oo`.

I. Since, the function is polynomial, so it is continuous

as well as differentiable in its domain `[1, oo) - {2}`.

Now, we check the continuity of the function at `x = 2`.

`LHL = f(2 - 0)= lim_(h -> 0) f(2- h)`

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

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

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

` = lim_(h -> 0) 3(2 + h) - (2 + h)^2`

` = 3(2 + 0) - (2 + 0)^2`

` = 6 - 4 = 2`

and `f(2) = 2 - 2 = 0`

`∵ f(2) = LHL != RHL`

So, the function is not continuous at every point in the interval

`[1, oo)` i.e., not continuous at `x = 2`.

II. We also check the differentiability of the function at

`x=1.5`.

`Rf'(1.5) = lim_(h -> 0) (f(1.5 + h) - 0 f(1.5))/h`

` lim_(h -> 0)( 2- (1.5+ h)- (2 - 1.5))h`

` = lim_(h -> 0)( 0.5- h - 0.5)/h`

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

`Lf' (1.5) = lim_(h -> 0) (f(1. 5- h) f(1.5))/(-h)`

` = lim_(h -> 0) (2-(1.5- h)-(2 - 1.5))/(-h)`

` = lim_(h -> 0) ( 0.5 + h - 0.5)/(-h)`

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

`∵ Lf' (1.5) = Rf' (1.5)`

So, the function is differentiable at `x = 1.5`

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Solution:

Given function, `f(x) = tt ({(2-x, text(for) 1<= x <= 2),( 3x - x^2, text(for) x > 2))`

and whole function defined in `1 <= x < oo`.

`∵ f(x) = tt ({(2-x, text(for) 1<= x <= 2),( 3x - x^2, text(for) x > 2))`

` => f'(x) = tt ({(-1, text(for) 1<= x <= 2),( 3 - 2x, text(for) x > 2))`...........(1)

So, the differentiable coefficient of `f(x)` at `x = 3` is

`f ' (3) = 3 - 2(3) = 3- 6 = - 3 quad [∵ f' (x) = 3 - 2x` for `x > 2]`

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Solution:

Given function,

`f:R -> R, f(x) = { tt ( (x^2 , x ge 0), (-x , x < 0) )`

`text (For continuity at x = 0)`

`LHL = f (0^-) = lim_(h->0) f(0-h)`

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

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

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

`:. f(0) = LHL= RHL`

Hence, `f(x)` is continuous at `x = 0`.

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

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

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

`:. L f'(0) ne R f'(0)`

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

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Solution:

I. The derivative, where the function attains maxima or
minima must be zero.

II. If a function is differentiable at a point, then it must be
continuous at that point but if a function is continuous at a
point, then it is not necessarily that function is differentiable at
that point.

So, both statements are correct.

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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 .`

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Solution:

Given function, `f (x) = | x - 3 |`

I. `LHL` at `x= 3, f(3-0) = lim_(x->3^-) f(3)`

`= lim_(h->0) f(3-h) = lim_(x->0) h =0`

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

RHL at `x = 3, f (3 + 0)`

`lim_(x->3^+) = lim_(h->0) f(3+h)`

`= lim_(h->0) |3+h -3 | = lim_(h->0) | h | =0`

and `f(3) = |3-3 | =0`

Here, `f(x)` is continuous at `x = 3`.


II. `Lf' (0) = lim_(x->0) (f(0-h) -f (0) )/(-h) = lim_(h->0) (|-h -3 | - |-3 |)/(-h)`

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


`Rf' (0) = lim_(h->0) (f (0+h) -f (0) )/h = lim_(h->0) ( | h-3| - |-3 | )/h`

`= lim_(h->0) (-h +3 -3)/h = lim_(h->0) (-h)/h`

`= -1`

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

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

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Solution:

`sin x` is periodic, dontinuous at every point on `( -oo, oo )`,
differentiable at every point on `( -oo, oo )`, has a period `2pi , sin x`
increases on `( 0,pi/2)` and decreases on `(pi/2,pi)`

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Solution:

`cos x` is periodid, continuous and differentiable at
every point on `(-oo, oo)` and has a period `2pi,cos x` decreases on
`( 0,pi/2)` and increases on `(pi/2,pi)`

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Solution:

tan x is a periodic function with period `pi` and is
discontinuous at `x = (m pi)/2` . Also, `tan x` is not differentiable at every
point on `(- oo, oo)` and increases on `( 0,pi/2)` and increases on
`(pi/2, pi)`

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Solution:

Redefined the function,

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

`:. LHL = lim_(x->0^-) f(x) = lim_(h->0) f(0-h) = lim_(h->0) (3 +h) = 3`

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

`= lim _(h->0) (3-h) = 3`

and `f(0) = 3-0 =3`

`=> LHL =RHL = f(0)`

Hence, `f(x)` is continuous at `x = 0`.

Now, `RHD = f'(0^-) = lim_(h->0) (f(0) -f (0-h) )/h`

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

and `RHD= f'(0^+) = lim_(h->0) (f (0+h) -f(0) )/h`

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

`=> LHD= RHD`

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

Hence, both Statements I and II are correct.

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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)`.

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Solution:

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