If f , g , h are real valued functions defined on R , then prove that f + g ∘ h = f ∘ h + g ∘ h . What can you say about f ∘ g + h ? Justify your answer.

We need to check whether ( f + g ) ∘ h = f ∘ h + g ∘ h or not We know that composite function is a function which is obtained from two given functions, where the range of one function is contained in the domain of the second function, by assigning to an element in the domain of the first function that element in the range of the second function whose inverse image is the image of the element. L.H.S : ( f + g ) ∘ h = ( f + g ) ∘ h ( x ) = f ( h ( x ) ) + g ( h ( x ) ) ⇒ ( f + g ) ∘ h = f ∘ h ( x ) + g ∘ h ( x ) = f ∘ h + g ∘ h = R.H.S ⇒ L.H.S = R.H.S, Hence proved Now f ∘ ( g + h ) = f ∘ ( g + h ) x = f ( g x + h x ≠ f ∘ h + g ∘ h Therefore f ∘ ( g + h ) ≠ f + g ∘ h

If f , g , h are real valued functions defined on R , then prove that f + g ∘ h = f ∘ h + g ∘ h . What can you say about f ∘ g + h ? Justify your answer.

We need to check whether ( f + g ) ∘ h = f ∘ h + g ∘ h or not We know that composite function is a function which is obtained from two given functions, where the range of one function is contained in the domain of the second function, by assigning to an element in the domain of the first function that element in the range of the second function whose inverse image is the image of the element. L.H.S : ( f + g ) ∘ h = ( f + g ) ∘ h ( x ) = f ( h ( x ) ) + g ( h ( x ) ) ⇒ ( f + g ) ∘ h = f ∘ h ( x ) + g ∘ h ( x ) = f ∘ h + g ∘ h = R.H.S ⇒ L.H.S = R.H.S, Hence proved Now f ∘ ( g + h ) = f ∘ ( g + h ) x = f ( g x + h x ≠ f ∘ h + g ∘ h Therefore f ∘ ( g + h ) ≠ f + g ∘ h

Find f ∘ g ( x ) and g ∘ f ( x ) , when f ( x ) = e x , g ( x ) = log x .

Given, f ( x ) = e x , g ( x ) = log x f ∘ g ( x ) = f [ g ( x ) ] = f ( log x ) = e log x = x ∵ a log a x = x log a a = x g ∘ f ( x ) = g [ f ( x ) ] = g e x = log e x = x log e = x ∵ log e e = 1 ∴ f ∘ g ( x ) = g ∘ f ( x ) = x

If f ( x ) = x + 3 4 x - 5 , g ( x ) = 3 + 5 x 4 x - 1 then show that ( f o g ) ( x ) = x .

Given that f ( x ) = x + 3 4 x - 5 , g ( x ) = 3 + 5 x 4 x - 1 To show: ( f o g ) ( x ) = x L.H.S = ( f o g ) ( x ) = f [ g ( x ) ] = f 3 + 5 x 4 x - 1 = 3 + 5 x 4 x - 1 + 3 4 3 + 5 x 4 x - 1 - 5 = ( 3 + 5 x ) + 3 ( 4 x - 1 ) 4 ( 3 + 5 x ) - 5 ( 4 x - 1 ) = 3 + 5 x + 12 x - 3 12 + 20 x - 20 x + 5 = 17 x 17 = x = R.H.S Hence proved.

Consider f x = - 2 . 5 x + 5 , for 0 ≤ x ≤ 3 . Complete the table: Function Domain Range f 0 ≤ x ≤ 3 f - 1

Given that, we have: f x = - 2 . 5 x + 5 Now, when x = 0 , then f 0 = - 2 . 5 0 + 5 = 5 [ y - intercept] When x = 1 , then f 1 = - 2 . 5 1 + 5 = 2 . 5 When x = 2 , then f 2 = - 2 . 5 2 + 5 = 0 When x = 3 , then f 3 = - 2 . 5 3 + 5 = - 2 . 5 And for, x - intercept, we have: f x = 0 , hence, - 2 . 5 x + 5 = 0 ⇒ x = 5 2 . 5 = 2 Thus, using above values, we can draw the required graph of f as, Therefore, range of the function is - 2 . 5 ≤ y ≤ 5 And also, the domain of the inverse function f - 1 is the range of the function f . The range of the inverse function f - 1 is the domain of the function f . Hence, the required table is Function Domain Range f 0 ≤ x ≤ 3 - 2 . 5 ≤ y ≤ 5 f - 1 - 2 . 5 ≤ y ≤ 5 0 ≤ y ≤ 3

Show that the function f : N → N defined by f n = n + 1 is not onto, by drawing arrow diagrams.

Given, the function f : N → N defined by f n = n + 1 . N = 1 , 2 , 3 , 4 , 5 , 6 , . . . . . . . f 1 = 1 + 1 = 2 f 2 = 2 + 1 = 3 f 3 = 3 + 1 = 4 f 4 = 4 + 1 = 5 f 5 = 5 + 1 = 6 Thus, f = 1 , 2 , 2 , 3 , 3 , 4 , 4 , 5 , 5 , 6 , . . . . . . . . . . . . Different elements have different images. Here, the range is not equal to the codomain. Therefore, the given function is not an onto function.

The function f : N → R is defined by f n = 2 n . The range of the function is

a subset of set of all even positive integers

the set of all even positive integers

EASY

10th Tamil Nadu Board

IMPORTANT

Earn 100

If $f, g, h$ are real valued functions defined on $R,$ then prove that $(f + g) \circ h = f \circ h + g \circ h .$ What can you say about $f \circ (g + h)$ ? Justify your answer.

Important Questions on Relations and Functions

EASY

10th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Relations and Functions>Exercise>Q 42

Find $(f \circ g) (x) and (g \circ f) (x)$ , when $f (x) = e^{x}, g (x) = \log x$ .

MEDIUM

10th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Relations and Functions>Exercise>Q 43

f (x) = \frac{x + 3}{4 x - 5}, g (x) = \frac{3 + 5 x}{4 x - 1}

then show that

(f o g) (x) = x

MEDIUM

10th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Relations and Functions>Exercise>Q 44

Consider $f (x) = - 2.5 x + 5, for 0 \leq x \leq 3$ . Draw the graph of $f$ on a pair of axes and draw the graph of $f^{- 1}$ on the same set of axes.

MEDIUM

10th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Relations and Functions>Exercise>Q 45

Consider $f (x) = - 2.5 x + 5, for 0 \leq x \leq 3$ . Complete the table:

Function	Domain	Range
$f$	$0 \leq x \leq 3$
$f^{- 1}$

EASY

10th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Relations and Functions>Exercise>Q 46

The graph shows the equivalence between Pounds Sterling $(UK £)$ and US Dollars $(US $)$ .

Write down the coordinates of a point on the graph that gives you some accurate information about the relationship and find the relationship between $x and u$

Question Image

EASY

10th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Relations and Functions>Exercise>Q 47

Consider the functions

f (x) = 2 x + 1

and

g (x) = m x + 3

, where

m

is a constant. The graphs of these functions intersect at the point

(- 5, - 9)

. Find the point of intersection of the graph of

g

with the

x -

axis.

MEDIUM

10th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Relations and Functions>Exercise>Q 48

The managers of a factory are trying to determine what to charge for an item they make. The demand for the item (a measure of how many are likely to sell) can be modelled by

D (p) = 800 - 2 p

, where

p

is the selling price in dollars. The supply for the item (the number that can be made) can be modelled by

S (p) = 3 p

. Sketch both functions on the same set of axes.

EASY

10th Tamil Nadu Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 1 - Relations and Functions>Exercise>Q 49

Find the intersection point for each of the following pair of functions.

$f (x) = 2 x + 3, g (x) = 1 - 4 x$

If f,g,h are real valued functions defined on R, then prove that f+g∘h=f∘h+g∘h. What can you say about f∘g+h? Justify your answer.

Important Questions on Relations and Functions

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If $f, g, h$ are real valued functions defined on $R,$ then prove that $(f + g) \circ h = f \circ h + g \circ h .$ What can you say about $f \circ (g + h)$ ? Justify your answer.