Solution:

`f(x)` is defined , if `3x^2 - 4x + 5 >= 0`

`=> 3 [ x^2 - 4/3 x + 5/3 ] >= 0`

` => 3 [ ( x - 2/3 )^2 + (11)/9 ] >= 0`

Which is true for all real `x`.

`:.` Domain `(f) = (- oo , oo) = R`

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Solution:

`f(x) = sqrt(cos^(-1) ((1-|x|)/2))`

`[because ` The value of cos lies between `-1` to `+1]`


`-1 le (1-|x|)/2 le 1`


`=> -2-1 le -|x| le 2-1`


`=> -3 le -|x| le 1`


`=> -1 le |x| le 3`


`=> x in [-3 , 3]`

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Solution:

For `f(x)` to be defined,

`x + 2 >= 0 => x >= -2`

and ` 1 - x > 0 ` and ` 1 - x != 1`

`=> 1 - x > 0` and `1 - x != 1`

`=> x < 1` and `x != 0`

`:. x in [- 2, 0 [ cup ] 0, 1 [`

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Solution:

`fog (x) = f [ g(x) ] = f ( | 3x + 4| )`

Since, the domain of `f` is `[ - 3, 5]`.

`:. - 3 <= | 3x + 4| <= 5`

`=> | 3x + 4 | <= 5`

`=> -5 <= 3x + 4 <= 5`

`=> - 9 <= 3x <= 1 => - 3 <= x <= 1/3`

`:. ` Domain of ` fog` is ` [ -3 , 1/3 ]`

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Solution:

`f (x ) = log sqrt ( -( cos x + 1/2 ) )`

`- ( cos x + 1/2 ) > 0`

`=> cos x < - 1/2`

` x in ( 2 n pi + ( 2 pi ) /3 , 2n pi + ( 4 pi)/3)`

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Solution:

`f(x) = sqrt ( log_(0.3) | (x -2 )/x| )`

for f(x) to be defined

`0 < | ( x -2)/x | <= 1`

`=> -1 <= (x-2)/x <= 1` and `(x -2)/x != 0` ............(i)

Solving LHS

`(x -2 )/x +1 >= 0`

`=> x < 0 , x >= 1` ..............(ii)

Solving RHS

`(x -2 ) /x - 1 <= 1`

`x > 0 ` ..........(iii)

hence from (i) , (ii) and (iii)

`x in [ 1, oo ) - { 2}`

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Solution:

`f(x) = sqrt((x^2 -3x -10) ln^2 (x-3))`

`f(x) = sqrt((x-5) (x+2) ln^2 (x-3))`

`(x - 5)(x + 2) ln^2(x - 3) ge 0` and `x > 3`

`[5, oo) cup ln (x-3) =0 => [5, oo) cup {4}`

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Solution:

`f(x) = sqrt(x^2 -|x|) +1/(sqrt(9-x^2))`

Let `|x| =t`

`f(x) = sqrt(t^2-t) + 1/(sqrt(9-t^2))`

`t(t - 1) ge 0` and `9 - t^2 > 0`

`|x| ge 1 \ \ \ \ x ge 1 ` or `x le -1`

or `|x| le 0=> x=0`

`(-3,-1) cup [1,3) cup {0}`

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Solution:

`=> log_2 log_(1//2) (x) ge 0 => log_(1//2) (x) > (2^0)`

`=> log_(1//2) x >1 => x < (1/2)^1`

Also, `log_(1//2) x >0`

`=> x in (0,1/2)`

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Solution:

`f(x) = sqrt(log_(1/2) ((5x-x^2)/4))`

`(5x-x^2)/4 > 0`

`=> x (5-x) > 0 => x(x-5) < 0`

`x in (0,5)`..................(i)

Also, `log_(1/2) ((5x-x^2)/4) ge 0`

`=> (5x-x^2)/4 le (1/2)^n => 5x-x^2 le 4`

`=> x^2 -5x +4 ge 0`

`x in (-oo ,1] cup [4,oo)`................(ii)

Using (i) and (ii)

`x in (0,1] cup [4,5)`

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Solution:

`f(x) = a sin x + b, a > 0 , b in R`

`f()x = a sin x + b`

` - 1 le sin x + b`

` - a + b + f(x) le a + b`

Range `in [b - a, b + a]`

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Solution:

`y = log_e (3x^2 - 4x + 5)`

`y` is defined if `3x^2 - 4x + 5 > 0`

`D < 0` and coefficeint of `x^2 > 0`

hence domain is `R` and log is increaising function.

Minimum value of `3x^2 - 4x + 5` is `-D/(4a)`

`=> (-(-44))/(4(3)) = 11/3 => y ge log_e (11/3)`

Range `in [ log_e (11/3) , oo)`

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Solution:

`y= 3- 2^x`

Domain is `x in R`

`0 le 2^x le oo`

Range `in (-oo , 3)`

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Solution:

Clearly, `f( x) = text()^(7- x)P_(x - 3)` is defined for positive

integer values of `x` satisfying

`7 - x > 0; x - 3 >= 0` and `x - 3 <= 7 - x`

i.e. `x < 7 , x >= 3` and `x <= 5`

i.e. `x = 3, 4, 5`

`:. ` Domain of `f = {3, 4, 5}`

Hence, range of `f = {f(3), f(4), f(5)}`

`= {text()^4P_0 , text()^3P_1 , text()^2P_2 } = {1,2, 3}`

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Solution:

For any `x, y in R`., we have

`f(x)=f(y)`

`=> (x-m)/(x-n) =(y-m)/(y-n)`

`=> x=y`

So, `f` is one-one.

Let `a in R`. such that `f(x) =alpha`

`=> (x-m)/(x-n) =alpha`

`=> x= (m- n alpha)/(1- alpha)`

Clearly, ` x notin R`. for `alpha = 1`, So, `f` is not onto.

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Solution:

(i) Total number of functions
Every element in A has 5 options for image, hence
Total number of functions = 54 = 625.

(ii) Number of injective (one-one) mapping.
4 elements in A needs four images hence number of one one functions `= text()^5C_4 xx 4! = 120`.

(iii) Number of many-one functions
Number of many-one mapping
= Total number of mapping - number of one-one mapping
`= 5^4 - text()^5C_4 xx 4! = 505`

(iv) Number of onto function = 0

(v) Number of into functions `= 5^4 = 625`

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Solution:

`f(x) = x^2 - 4x + 5`

Minima at x = 2

at x = 2, y = 4 - 8 + 5 = 1

For function to be one-one it should be monotonic.

Hence, for `x in [2, oo)`, f(x) is increasing

at `x = 2, y = 1` Hence `y in [1, oo)`

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Solution:

`f(x) = e^x + e^-x`

Domain `in R`

`y = e^x + 1/e^x`

`=> y = e^x + 1/e^x >= 2`

Range `[ 2, oo)`

also `f (x) = f (-x)`

hence function is many one into

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Solution:

I `f(x) = x `and `g(x) =- x` or `f(x) = x` and `g(x) =- x^3`

II `f(x) = x` and `g (x) = x^3`

llI `f(x) = sin x` which is odd but not one-one or `f(x) = x^2 sin x` which is odd but many one]

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Solution:

(i) Here, `f: R -> R` is defined by `f(x) = 3 - 4x`

Let `x_1, x_2 in R` such that `f(x_1) = f(x_2)`

`=> 3 -4x_1 = 3 -4x_2 => -4x_1 = -4x_2 => x_1 = x_ 2`

Therefore, `f` is one-one.

For any real number `y` in `R`, there exists `(3-y)/4` in `R` such that



`f( (3-y)/4) =3 -4 ((3-y)/4) =y`

Therefore, `f` is onto. Hence, `f` is bijective.




(ii) Here `f : R -> R` is defined as

`f(x) = 1 + x^2`

Let `x_1, x_2 in R` such that `f(x_1) = f(x_2)`

`=> 1 + x_(1)^(2) = 1 + x_(2)^(2) => x_(1)^(2) = x_(2)^(2) => x_1 = pm x_2`

For instance, `f(1) = f(- 1) = 2`

Therefore, `f(x_1) = f(x_2)` does not imply that `x_1 = x_2`

Therefore, `f` is not one-one.

Consider an element `- 2` in co-domain `R`.

It is seen that `f(x) = 1 + x^2` is positive for all `x in R`.

Thus, there does not exist any `x` in domain `R` such that `f(x) = - 2`.

Therefore, `f` is not onto. Hence, `f` is neither one-one nor onto.

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Solution:

Given that `A= {1, 2, 3}` and `B = {4, 5, 6, 7}`

Now, `f: A-> B` is defined as `f = { (1, 4), (2, 5), (3, 6)}`

Therefore, `f(1) = 4, f(2) = 5, f(3) = 6`

It is seen that the images of distinct elements of `A` under fare distinct.

Hence, function `f` is one-one.

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Solution:

Given that, `A=[-1,1]`

(i) `f(x) = x/2`

Let `f(x_1)=f(x_2)`

`=>(x_1)/2=(x_2)/2=> x_1=x_2`

So, `f(x)` is one-one.

Now, let `y=x/2`

`=> x=2y notin A, forall y in A`

As for `y=1 in A, x=2 in A`

So, `f(x)` is not onto.

(ii) `g(x) =| x|`

Let `g(x_1)=g(x_2)`

`=> |x_1|=|x_2|=>x_1= pm x_2`

So, `g(x)` is not one-one.

Now,

`g(x_1) = g(x_2 )`

`|x_1|=|x_2| => x _1 =x_2`

So, `g(x)` is not onto, also, `g(x)` is not bijective.

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Solution:

The mapping `f:C-> R`

Given `f(z)=|z|, forall z in C`

`f(1)=|1|=1`

`f(-1)=|-1|=1`

`f(1)=f(-1)`

But `1 ne -1`

So, `f(z)` is not one-one. Also, `f(z)` is not onto as there is no pre-image for any negative element of `R` under the mapping `f(z)`.

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Solution:

Given that `A = R − {3} , B = R − {1}`

Consider the function

`f : A -> B` defined by `f (x) = ((x - 2)/(x - 3))`

Let `x, y in A` such that `f (x) = f (y)`

` => (x - 2)/(x - 3) = (y - 2)/( y - 3)`

`=> (x - 2) (y - 3) = (y - 2)( x- 3)`

`=> xy - 3x - 2y + 6 = xy - 3y - 2x + 6`

` => - 3x - 2y = - 3y - 2x`

` => 3x - 2x = 3y - 2y`

`=> x = y`

`:. f` is one-one.

To check onto replace f(x) with y and find `x` in terms of `y`

`=> x = (2 - 3y)/(1 - y)` ............(1)

Here unique value of `x` exist for every value of `y ` except `y = 1`

So function is onto in `{R} -1`

Therefore, `f^(-1)` exists.

To find inverse

Replace `y` by `x` and `x` by `f^(-1) (x)` in the above equation (1),

we have,

` f^(-1) (x) = ( 2- 3x)/(1 - x) , x != 1`

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Solution:

(a) It is easy to see that f is one-one and onto, so that `f` is invertible with the inverse `f ^(–1)` of f given by `f ^(–1) = {(1, 1), (2, 2), (3, 3)} = f`.

(b) Since `f (2) = f (3) = 1, f` is not one-one, so that `f` is not invertible.

(c) It is easy to see that `f` is one-one and onto, so that `f` is invertible with `f ^(–1) = {(3, 1), (2, 3), (1, 2)}`.

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Solution:

`f : R -> (0, 1) , f(x) = (2^x)/(1+ 2^x)`

`y = (2^x)/(1+ 2^x) => y + 2^x y =2^x`

`=> 2^x = y/(1-y) => x = log_2 (y/(1-y))`

`=> f^(-1) = y = log_2 (x/ (1-x))`

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Solution:

Since `sqrt (x^2 +1 ) > sqrt (x^2) AA x in R`

Hence` x + sqrt (x^2 +1) > 0 AA x in R`

` f(x)` is one-one onto hence invertible

`y = log_a (x + sqrt (x^2 +1) )`

`a^y = x + sqrt (x^2 +1)` ..........(i)

`a^(-y) = 1/(x+ sqrt (x^2 +1) ) = -x + sqrt (x^2 +1)` .........(ii)

(i) - (ii)

`a^y - a^(-y) = 2x => x =1/2 (a^y - a^(-y) )`

Hence `f^(-1) (x) = 1/2 (a^x - a^(-x) )`

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Solution:

`y=(3x-2)/(x+4)`

`=> xy + 4y = 3x - 2 => x(y-3) = 4y -2`

`=> x= (4y+2)/(3-y) => f^(-1) (y)= (4y+2)/(3-y)`

`= f^(-1) (x) = (4x+2)/(3-x) =(x+1/2)/((-x)/4 +3/4)`

`b= 1/2 , c=(-1)/4 , d= 3/4`

`b+c+d =1`

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Solution:

Since `sqrt (x^2 +1 ) > sqrt (x^2) AA x in R`

Hence` x + sqrt (x^2 +1) > 0 AA x in R`

` f(x)` is one-one onto hence invertible

`y = log_a (x + sqrt (x^2 +1) )`

`a^y = x + sqrt (x^2 +1)` ..........(i)

`a^(-y) = 1/(x+ sqrt (x^2 +1) ) = -x + sqrt (x^2 +1)` .........(ii)

(i) - (ii)

`a^y - a^(-y) = 2x => x =1/2 (a^y - a^(-y) )`

Hence `f^(-1) (x) = 1/2 (a^x - a^(-x) )`

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