DON BOSCO SCHOOL OF EXCELLENCE, EGMORE

MATHEMATICS- CLASS 12: A & B

NOTES (EX:1.3 AND CH 2 INTRODUCTION)

Note: Read once , understand and make an entry in your classwork. These will be discussed in class.

Those who didn’t collect the classwork can copy in a new classwork

Definition: Composite functions

Let f : A → B and g : B → C be two functions. Then the composition of f and g, denoted by gof,

is defined as the function gof : A -> C given by

gof(x)=g(f(x)) for all x ∈ A.

Eg:f(x) =(x+3)

g(x) =x2

Then, gof=g(f(x))=g(x+3)=(x+3)2

Ex: 1.3

1.Let f: {1,3,4} –> {1,2, 5} and g : {1, 2,5} –> {1,3} be given by f = {(1, 2), (3,5), (4,1) and g =

{(1,3), (2,3), (5,1)}. Write down g of.

Solution:

Given, f= {(1,2),(3,5),(4,1)}

and g= {(1,3),(2,3),(5,1)}

Then,

f(1) = 2, g(2) = 3 =>gof(1) = 3

f(3) = 5, g(5)= 1 =>gof(3) = 1

f(4) = 1, g(1) = 3 =>gof(4) = 3

=>gof= {(1,3), (3,1), (4,3)}

2. Let f, g and h be functions from R to R. Show that i)(f + g) oh = foh + goh,

ii)(f • g) oh = (foh) • (goh)

(Note: • denotes multiplication)

Solution:

Given, f :R –> R, g: R –> R, h: R –> R

(i) (f+g)oh(x)=(f+g)[h(x)]

= f[h(x)]+g[h(x)]

={foh} (x)+ {goh} (x)

=>(f + g) oh = foh + goh

(ii) (f • g) oh (x) = (f • g) [h (x)]

= f[h (x)] • g [h (x)]

= {foh} (x) • {goh} (x)

=> (f • g) oh = (foh) • (goh)

3. Find gof and fog, if

(i) f (x) = |x| and g (x) = |5x – 2|

(ii) f (x) = 8x³ and g (x) = .

Solution:

(i) f(x) = |x|, g(x) = |5x – 2|

∴ gof(x) = g(f(x))

= g|x|

= |5| x | – 2|

∴ fog(x) = f(g (x))

= f(|5x – 2|) (Note: ||x|| = |x|)

= ||5 x – 2|| = |5x-2|

(ii) f(x) = 8x³ and g(x) =

∴ gof(x) = g(f(x)) = g(8x³) = =.(2x)3.1

3 = 2x

∴ fog (x) = f(g (x))=f( ) = 8.( )³ = 8x

Note: Identity of Composite Function

If Let f : A → B and g : B → A be two functions.

Then if,

gof(a)=a, for all a ∈ A. We say that the composite gof=IA is the identity function on set A.

Similarly,

if fog(b)=b, for all b∈ B. We say that the composite fog=IB is the identity function on set B.

Invertible Function:

A function f:X→Y is defined to be invertible, if there exists a function

g:Y→X such that gof=IX and fog=IY. The function g is called the inverse

of f and is denoted by f-1.

- Also a function is invertible only if it is one-one and onto and

conversely, if any function is one-one and onto the n it is

invertible.

- f-1(f(x))=x

4. If f(x) =𝟒𝒙+𝟑

𝟔𝒙−𝟒 , 𝒙 ≠

𝟐

𝟑show that fof (x) = x, for all ≠

𝟐

𝟑 . What is the inverse of f?

Solution:

Given, f(x) =𝟒𝒙+𝟑

𝟔𝒙−𝟒 , 𝒙 ≠

𝟐

𝟑

Consider,fof (x) = f(f(x)) = f(𝟒𝒙+𝟑

𝟔𝒙−𝟒)

(Substituting x= 𝟒𝒙+𝟑

𝟔𝒙−𝟒 in f(x) )

∴ 𝒇𝒐𝒇(𝒙) = 𝒙 (⇒f is the inverse of itself)

Let, f(x) =y

⇒y= 𝟒𝒙+𝟑

𝟔𝒙−𝟒

x= f-1(y)

Therefore,

Thus, f is inverse of itself.

5. State with reason whether following functions have inverse (i) f: {1,2,3,4}–>{10} with f = {(1,10), (2,10), (3,10), (4,10)} (ii) g: {5,6,7,8}–>{1,2,3,4} with g = {(5,4), (6,3), (7,4), (8,2)} (iii) h: {1,2,3,4,5}–>{7,9,11,13} with h = {(2,7), (3,9), (4,11), (5,13)} Solution:

(i) f: {1,2,3,4} –> {10} with f = {(1,10), (2,10), (3,10), (4,10)} we can observe that 1,2,3,4 have the same image 10 f is not one-one. Since

=> f has no inverse (or not invertible).

(ii) g: {5,6,7,8} –> {1,2,3,4} with g = {(5,4), (6,3) , (7,4), (8,2)}

we can observe that, 5 and 7 have the same image 4. => g is not one-one.

∴g is not invertible. (iii) Given, h = {(2,7), (3,9), (4,11), (5,13)}

we can observe that, h is one-one and onto.

∴ h is invertible.

6. Show that f: [-1,1] –> R, given by f(x) = is one-one. Find the inverse of the function f: [-1,1] –> Range f.

(Hint – For y ∈ Range f, y = f (x) = for some x in [- 1,1], i.e., x = ) Solution:

Let f(x) = y =𝒙

𝒙+𝟐

Taking inverse of f;

f-1(f(x)) = f-1(y)

x =f-1(y)

=>y(x+2) =x

=>yx +2y = x

=>x(y-1) = -2y

∴f-1(y) =x= 𝟐𝒚

𝟏−𝒚

7. Consider f: R –> R given by f (x) = 4x + 3. Show that f is invertible. Find the inverse of f. Solution:

To prove: f is invertible.

i.e) To prove that f is one – one and onto

f: R—>R given by f(x) = 4x + 3

Let x1 ,x2 ∈R Assume, f(x1) = f(x2),

then 4x1 + 3 = 4x2 + 3

4x1 = 4x2

x1 = x2

∴f is one-one Also let y = 4x + 3

4x = y – 3

∴ Thus, for each value of y ∈ R and belonging to co-domain of y has a pre-image in its domain.

∴f is onto i.e. f is one-one and onto.

∴f is invertible and f-1 (y) = x =

8. Consider f: R+ –> [4, ∞] given by f (x) = x² + 4. Show that f is invertible with the

inverse f-1 of fgiven by f-1 (y) = √𝒚 − 𝟒 , where R+ is the set of all non-negative real

numbers.

Solution:

To prove: f is invertible.

i.e) To prove that f is one – one and onto

Let x1 ,x2∈R+

Assume, f(x1) = f(x2)

=>x12 + 4 = x2

2 + 4

=>x12 = x2

2

=>x1 = x2 As x ∈R+ (Since only positive values are considered)

=>f is one-one

Let y = x² + 4

=>x² = y – 4

=> x = ±√𝒚 − 𝟒

x being > 0, -ve sign not to be taken

x = √𝒚 − 𝟒

∴ f-1 (y) = x = √𝒚 − 𝟒 , y ≥ 4

For every y ≥ 4, g (y) has real positive value.

∴The inverse of f is f-1 (y) = √𝒚 − 𝟒

9. Consider f: R+ –> [- 5, ∞) given by f (x) = 9x² + 6x – 5. Show that f is invertible with

Solution:

To prove: f is invertible.

i.e) To prove that f is one – one and onto.

Assume, f (x1)=f (x2)

9x1² + 6x1 – 5= 9x2² + 6x2 – 5

9(x12-x2

2) +6(x1-x2)=0

3(x1-x2)(3(x1+x2) + 2)=0

x1= x2 (Since Domain is only positive real numbers)

f is one – one

Let y= f(x) = 9x² + 6x – 5

i.e) 9x² + 6x – 5-y=0

=9x² + 6x –( 5+y)=0

Solving the above equation using quadratic formula, we get;

x= −𝟔±√𝟑𝟔+𝟑𝟔(𝟓+𝒚)

𝟏𝟖

= −𝟔±𝟔√𝟏+(𝟓+𝒚)

𝟏𝟖

=−𝟏±√𝟔+𝒚

𝟑

X=√𝟔+𝒚−𝟏

𝟑 (Since domain is positive)

Consider, f(x) =f(√𝟔+𝒚−𝟏

𝟑)

=9(√𝟔+𝒚−𝟏

𝟑)² + 6(

√𝟔+𝒚−𝟏

𝟑) – 5

= (6+y) -2√𝟔 + 𝒚 +1 +2√𝟔 + 𝒚 -2 -5

= y

∴ f is onto.

Thus, f is invertible and f-1(y) = x=√𝟔+𝒚−𝟏

𝟑

Hence proved.

10.

Solution:

11.

Solution:

12.

Solution:

(Mark the answers of 13 and 14 in text book itself)

Note:

1) Binary Operation: A binary operation * on a set A is a function,

*: AXA → A

We denote *(a,b) by a*b.

(just that f is replaced by the symbol *. Also its domain is a set of cartesian product)

Eg: + : 𝑹 𝑿 𝑹 → 𝑹 is given by (a,b) → a+b

(Refer Eg 29 in Ncert Book)

Thus * takes the form of the operation or the expression as defined in the question.

2) ⋁ : 𝑹 𝑿 𝑹 → 𝑹 is given by (a,b) → max{a,b}

3) ∧ : 𝑹 𝑿 𝑹 → 𝑹 is given by (a,b) → min{a,b}

4) A binary operation * on a set X is called Commutative if a*b=b*a, for every a, b ∈ 𝑿

5) A binary operation * on a set X is called Associative if (a*b)*c=a*(b*c), for every a,

b,c ∈ 𝑿

6) In a binary operation * on a set X, if there exists an element 𝒆 ∈ 𝑿, it is called as

identity element for the operation * if a*e=e*a=a for every a∈ 𝑿.

(Note: Binary Operations is not in the curriculum as given by cbse 2019-2020, so we are not doing Ex:1.4. Just be sure of the definitions)

***********End of Ch 1**********************

Ch – 2: Inverse Trigonometric Functions

Date: 25/03/19

Refer your text book and draw neatly the following images:

1) y=sinx (fig 2.1 (i))

2) y=sin-1 x (fig 2.1 (ii))

3) y=cosx (fig 2.2(i))

4) y=cos-1 x (fig 2.2 (ii))

5) y=tanx (fig 2.5 (i))

6) y=tan-1 x (fig 2.5 (ii))

7) y=cosecx (fig 2.3 (i))

8) y=cosec-1 x (fig 2.3 (ii))

9) y=secx (fig 2.4 (i))

10) y=sec-1 x (fig 2.4 (ii))

11) y=cotx (fig 2.6 (i))

12) y=cot-1 x (fig 2.6 (ii))

Domain and Range comparision of trigonometric functions and their inverses:( Referring the above diagrams)

S.No.

Trigonometric Functions Inverse Trigonometric Functions

1 Sinx: R → [−𝟏, 𝟏] sin-1 x : [-1,1] →[- 𝝅

𝟐, 𝒄]

2 cosx: R → [−𝟏, 𝟏] cos-1 x : [-1,1] →[0,𝝅]

3 tanx:

R – {x:x=(2n+1) 𝝅

𝟐, 𝒏 ∈ 𝒁} →

R

tan-1 x : R → (- 𝝅

𝟐,

𝝅

𝟐)

4 cosecx:

R – {x:x=n𝝅, 𝒏 ∈ 𝒁}

→ R-(-1,1)

cosec-1 x :

R-(-1,1) →[- 𝝅

𝟐,

𝝅

𝟐] − {𝟎}

5 secx:

R – {x:x=(2n+1) 𝝅

𝟐, 𝒏 ∈ 𝒁}

→ R – (-1,1)

sec-1 x : R-(-1,1) →[0,𝝅] -

{ 𝝅

𝟐}

6 cotx:

R – {x:x=n𝝅, 𝒏 ∈ 𝒁} → R

cot-1 x : R → (- 𝝅

𝟐,

𝝅

𝟐)