Let f : R → R : f(x) = x² + 1.

Find f^-1 {10}

We know that, if f: X → Y such that y ∈ Y. Then f-1 (y) = {x ∈ X: f(x) = y}.

In other words, f-1 (y) is the set of pre – images of y

Let f^-1{10} = x. Then, f(x) = 10 …(i)

and it is given that f(x) =  …(ii)

So, from (i) and (ii), we get

x² + 1 = 10

⇒  x² = 10 – 1

⇒  x² = 9

⇒  x = √9

⇒  x = ± 3

∴ f ^-1{10} = {-3, 3}

Given: f(x) = x² – 3x + 4 ----- (1)

and f(x) = f(2x + 1)

Need to Find: Value of x

Replacing x by (2x + 1) in equation (1) we get,

f(2x + 1) = (2x + 1)2 – 3(2x + 1) + 4 ----- (2)

According to the given problem, f(x) = f(2x + 1)

Comparing (1) and (2) we get,

X² – 3x + 4 = (2x + 1)² – 3(2x + 1) + 4

⇒  x²– 3x + 4 = 4x² + 4x + 1 – 6x – 3 + 4

⇒  4x² + 4x + 1 – 6x – 3 + 4 – x²+ 3x – 4 = 0

⇒  3x² + x – 2 = 0

⇒  3x² + 3x – 2x – 2 = 0

⇒  3x(x + 1) – 2(x + 1) = 0

⇒  (3x – 2)(x + 1) = 0

So, either (3x – 2) = 0 or (x + 1) = 0

Values of 2/3 and -1

Given: f(x) = 1 – |x – 2|

Need to find: Where the functions are defined.

Since |x – 2| gives real no. for all values of x, the domain set can possess any real numbers.

So, the domain of the function, Df(x) = (-∞, ∞).

Now the given function is f(x) = 1 – |x – 2|, where | x – 2 | is always positive. So, the maximum value of the function is 1.

Therefore, the range of the function, Rf(x) = (-∞, 1)

f(x) = x + 3, g(x) = 3x² – 1

To find:- Set of values of x for which f(x) = g(x)

Consider,

f(x) = g(x)

x+3 = 3x² – 1

3x² – x-4=0

3x² – 4x+3x-4=0

x(3x-4) +(3x-4) = 0

(3x - 4)(x + 1) = 0

x = 4/3 or x= -1

The set values for which f(x) and g(x) have same value is { 4/3 , -1}.

Given that: f = {(0, -5), (1, -2), (3, 4), (4, 7)} be a function from Z to Z defined by linear function.

We know that, linear functions are of the form y = mx + b

Let f(x) = ax + b, for some integers a, b

Here, (0, -5) ∈ f

⇒  f(0) = -5

⇒  a(0) + b = -5

⇒  b = -5 …(i)

Similarly, (1, -2) ∈ f

⇒  f(1) = -2

⇒  a(1) + b = -2

⇒  a + b = -2

⇒  a + (-5) = -2 [from (i)]

⇒  a = -2 + 5

⇒  a = 3

∴ f(x) = ax + b = 3x + (-5)

f(x) = 3x – 5