Class 11_Chapter 2_Relations and Functions (Real Functions) Lecture 6.pdf

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Class XI
Mathematics
Chapter- 2
Relations and Functions
Lecture - 6
Dr. Pranav Sharma
Maths Learning Centre. Jalandhar.
Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar Maths Learning Centre, Jalandhar

youtube.com/@MathematicsOnlineLectures
REAL FUNCTIONS
A function 𝐟: 𝐗 → 𝐘 is called a real function, if 𝐱 ⊆ 𝐑 and 𝐲 ⊆ 𝐑.
If 𝐟(𝐱) = 𝟑𝐱𝟑
− 𝟓𝐱𝟐
+ 𝟏𝟎, find 𝐟(𝐱 − 𝟏) .
We have, 𝐟(𝐱) = 𝟑𝐱𝟑
− 𝟓𝐱𝟐
+ 𝟏𝟎.
𝐟(𝐱 − 𝟏) = 𝟑(𝐱 − 𝟏)𝟑
− 𝟓(𝐱 − 𝟏)𝟐
+ 𝟏
= 𝟑{𝐱𝟑
− 𝟏 − 𝟑𝐱(𝐱 − 𝟏)} − 𝟓(𝐱𝟐
− 𝟐𝐱 + 𝟏) + 𝟏𝟎
= 𝟑(𝐱𝟑
− 𝟑𝐱𝟐
+ 𝟑𝐱 − 𝟏) − 𝟓(𝐱𝟐
− 𝟐𝐱 + 𝟏) + 𝟏𝟎 = 𝟑𝐱𝟑
− 𝟏𝟒𝐱𝟐
+ 𝟏𝟗𝐱 + 𝟐.
If 𝐟(𝐱) = 𝐱 +
𝟏
𝐱
, show that {𝐟(𝐱)}𝟑
= 𝐟(𝐱𝟑
) + 𝟑𝐟 (
𝟏
𝐱
).
{𝐟(𝐱)}𝟑
= 𝐱𝟑
+
𝟏
𝐱𝟑
+ 𝟑 × 𝐱 ×
𝟏
𝐱
× (𝐱 +
𝟏
𝐱
) = (𝐱𝟑
+
𝟏
𝐱𝟑
) + 𝟑 (
𝟏
𝐱
+ 𝐱) = 𝐟(𝐱𝟑
) + 𝟑𝐟 (
𝟏
𝐱
)
Hence, {𝐟(𝐱)}𝟑
= 𝐟(𝐱𝟑
) + 𝟑𝐟 (
𝟏
𝐱
) .
https://youtu.be/fFEfRr2hpvw

If 𝐟(𝐱) =
𝐱−𝟏
𝐱+𝟏
′𝐱 ≠ −𝟏 then show that 𝐟{𝐟(𝐱)} =
−𝟏
𝐱
, where 𝐱 ≠ 𝟎.
𝐟{𝐟(𝐱)} = 𝐟 (
𝐱 − 𝟏
𝐱 + 𝟏
) =
{
𝐱 − 𝟏
𝐱 + 𝟏
− 𝟏}
{
𝐱 − 𝟏
𝐱 + 𝟏
+ 𝟏}
=
{(𝐱 − 𝟏) − (𝐱 + 𝟏)}
(𝐱 + 𝟏)
×
(𝐱 + 𝟏)
{(𝐱 − 𝟏) + (𝐱 + 𝟏)}
=
−𝟐
𝟐𝐱
=
−𝟏
𝐱
.
If 𝐲 = 𝐟(𝐱) =
𝐚𝐱−𝐛
𝐛𝐱−𝐚
and 𝐚𝟐
≠ 𝐛𝟐
then prove that 𝐱 = 𝐟(𝐲).
𝐟(𝐲) = 𝐟{𝐟(𝐱)} = 𝐟 (
𝐚𝐱−𝐛
𝐚𝐱−𝐚
) =
{𝐚(
𝐚𝐱−𝐛
𝐛𝐱−𝐚
)−𝐛}
{𝐛(
𝐚𝐱−𝐛
𝐛𝐱−𝐚
)−𝐚}
=
{(𝐚𝟐𝐱−𝐚𝐛)−(𝐛𝟐𝐱−𝐚𝐛)}
(𝐛𝐱−𝐚)
×
(𝐛𝐱−𝐚)
{(𝐚𝐛𝐱−𝐛𝟐)−(𝐚𝐛𝐱−𝐚𝟐)}
=
(𝐚𝟐𝐱−𝐛𝟐𝐱)
(𝐚𝟐−𝐛𝟐)
=
(𝐚𝟐−𝐛𝟐)𝐱
(𝐚𝟐−𝐛𝟐)
= 𝐱.

If 𝐟(𝐱) =
𝐱−𝟏
𝐱+𝟏
then prove that 𝐟(𝟐𝐱) =
𝟑𝐟(𝐱)+𝟏
𝐟(𝐱)+𝟑
.
𝟑𝐟(𝐱)+𝟏
𝐟(𝐱)+𝟑
=
𝟑(
𝐱−𝟏
𝐱+𝟏
)+𝟏
(𝐱−𝟏)
(𝐱+𝟏)
+𝟑
=
(𝟑𝐱−𝟑)+(𝐱+𝟏)
(𝐱+𝟏)
×
(𝐱+𝟏)
(𝐱−𝟏)+(𝟑𝐱+𝟑)
=
𝟐𝐱−𝟏
𝟐𝐱+𝟏
= 𝐟(𝟐𝐱).
Let the functions 𝐟 and 𝐠 be defined by
𝐟(𝐱) = (𝐱 − 𝟑) and 𝐠(𝐱) = {
𝐱𝟐−𝟗
𝐱+𝟑
, 𝐰𝐡𝐞𝐧 𝐱 ≠ −𝟑
𝐤, 𝐰𝐡𝐞𝐧 𝐱 = −𝟑.
Find the value of 𝐤 such that 𝐟(𝐱) = 𝐠(𝐱) for all 𝐱 ∈ 𝐑.
We have, 𝐟(𝐱) = 𝐠(𝐱) for all 𝐱 ∈ 𝐑. 𝐟(−𝟑) = 𝐠(−𝟑) ⇒ (−𝟑 − 𝟑) = 𝐤 ⇒ 𝐤 = −𝟔.

If 𝐟(𝐱) =
𝟏
(𝟏−𝐱)
then show that 𝐟[𝐟{𝐟(𝐱)}] = 𝐱.
𝐟{𝐟(𝐱)} =
𝟏
{𝟏−
𝟏
(𝟏−𝐱)
}
=
𝟏−𝐱
−𝐱
=
𝐱−𝟏
𝐱
⇒ 𝐟[𝐟{𝐟(𝐱)}] =
{
𝟏
(𝟏−𝐱)
−𝟏}
𝟏
(𝟏−𝐱)
=
𝐱
(𝟏−𝐱)
×
(𝟏−𝐱)
𝟏
= 𝐱.
If 𝐟(𝐱) =
𝟐𝐱
(𝟏+𝐱𝟐)
then show that 𝐟( 𝐭𝐚𝐧 𝛉) = 𝐬𝐢𝐧 𝟐𝛉.
𝐟( 𝐭𝐚𝐧 𝛉) =
𝟐 𝐭𝐚𝐧 𝛉
(𝟏 + 𝐭𝐚𝐧𝟐𝛉)
=
𝟐 𝐭𝐚𝐧 𝛉
𝐬𝐞𝐜𝟐𝛉
=
𝟐 𝐬𝐢𝐧 𝛉
𝐜𝐨𝐬 𝛉
× 𝐜𝐨𝐬𝟐
𝛉 = 𝟐 𝐬𝐢𝐧 𝛉 ⋅ 𝐜𝐨𝐬 𝛉 = 𝐬𝐢𝐧 𝟐𝛉.
If 𝐲 = 𝐟(𝐱) =
𝟑𝐱+𝟏
𝟓𝐱−𝟑
, prove that 𝐱 = 𝐟(𝐲) .
𝐟(𝐲) =
𝟑𝐲 + 𝟏
𝟓𝐲 − 𝟑
=
𝟑 (
𝟑𝐱 + 𝟏
𝟓𝐱 − 𝟑
) + 𝟏
𝟓 (
𝟑𝐱 + 𝟏
𝟓𝐱 − 𝟑
) − 𝟑
=
𝟗𝐱 + 𝟑 + 𝟓𝐱 − 𝟑
𝟏𝟓𝐱 + 𝟓 − 𝟏𝟓𝐱 + 𝟗
=
𝟏𝟒𝐱
𝟏𝟒
= 𝐱.

Operations On Functions
(i) Sum Of Two Real Functions
Let 𝐟 : 𝐃𝟏 → 𝐑 and 𝐠: 𝐃𝟐 → 𝐑, where 𝐃𝟏 ⊆ 𝐑 and 𝐃𝟐 ⊆ 𝐑. Then,
(𝐟 + 𝐠): (𝐃𝟏 ∩ 𝐃𝟐) → 𝐑: (𝐟 + 𝐠)(𝐱) = 𝐟(𝐱) + 𝐠(𝐱) for all 𝐱 ∈ (𝐃𝟏 ∩ 𝐃𝟐) .
(ii) Difference Of Two Real Functions
(𝐟 − 𝐠): (𝐃𝟏 ∩ 𝐃𝟐) → 𝐑: (𝐟 − 𝐠)(𝐱) = 𝐟(𝐱) − 𝐠(𝐱) for all 𝐱 ∈ (𝐃𝟏 ∩ 𝐃𝟐) .
(iii) Multiplication Of Two Real Functions
(𝒇𝒈): (𝐃𝟏 ∩ 𝐃𝟐) → 𝐑: (𝒇𝒈) (𝐱) = 𝐟(𝐱) ⋅ 𝐠(𝐱) for all 𝐱 ∈ (𝐃𝟏 ∩ 𝐃𝟐).

(iv) Quotient Of Two Real Functions
Let 𝐟 : 𝐃𝟏 → 𝐑 and 𝐠: 𝐃𝟐 → 𝐑, where 𝐃𝟏 ⊆ 𝐑 and 𝐃𝟐 ⊆ 𝐑.
Let 𝐃 = (𝐃𝟏 ∩ 𝐃𝟐) − {𝐱: 𝐠(𝐱) = 𝟎}. Then,
(
𝐟
𝐠
) : 𝐃 → 𝐑: (
𝐟
𝐠
) (𝐱) =
𝐟(𝐱)
𝐠(𝐱)
for all 𝐱 ∈ 𝐃.
(V) Reciprocal of a Function
Let 𝐟 : 𝐃𝟏 → 𝐑 and let 𝐃 = 𝐃𝟏 − {𝐱: 𝐟(𝐱) = 𝟎}.Then,
𝟏
𝐟
: 𝐃 → 𝐑: (
𝟏
𝐟
) (𝐱) =
𝟏
𝐟(𝐱)
for all 𝐱 ∈ 𝐃.
(vi) Scalar Multiple of a Function
Let 𝐟 : 𝐃 → 𝐑 and let 𝛂 be a scalar (a real number). Then,
(𝛂𝐟): 𝐃 → 𝐑: (𝛂𝐟)(𝐱) = 𝛂 ⋅ 𝐟(𝐱) for all 𝐱 ∈ 𝐃.

Let 𝐟: 𝐑 → 𝐑: 𝐟(𝐱) = 𝐱𝟐
and 𝐠: 𝐑 → 𝐑: 𝐠(𝐱) = 𝟐𝐱 + 𝟏.
Find (i) (𝐟 + 𝐠)(𝐱) (ii) (𝐟 − 𝐠)(𝐱) (iii) (𝒇𝒈)(𝒙) (iv) (
𝒇
𝒈
) (𝒙)
dom (𝐟) = 𝐑 and dom (𝐠) = 𝐑 dom (𝐟) ∩ dom (𝐠) = (𝐑 ∩ 𝐑) = 𝐑.
(i) (𝐟 + 𝐠): 𝐑 → 𝐑 is given by
(𝐟 + 𝐠)(𝐱) = 𝐟(𝐱) + 𝐠(𝐱) = 𝐱𝟐
+ (𝟐𝐱 + 𝟏) = (𝐱 + 𝟏)𝟐
.
(ii) (𝐟 − 𝐠): 𝐑 → 𝐑 is given by
(𝐟 − 𝐠)(𝐱) = 𝐟(𝐱) − 𝐠(𝐱) = 𝐱𝟐
− (𝟐𝐱 + 𝟏) = (𝐱𝟐
− 𝟐𝐱 − 𝟏) .
(iii) (𝒇𝒈): 𝐑 → 𝐑 is given by
(𝐟𝐠)(𝐱) = 𝐟(𝐱) ⋅ 𝐠(𝐱) = 𝐱𝟐
⋅ (𝟐𝐱 + 𝟏) = (𝟐𝐱𝟑
+ 𝐱𝟐
) .
(iv) {𝐱: 𝐠(𝐱) = 𝟎} = {𝐱: 𝟐𝐱 + 𝟏 = 𝟎} = {
−𝟏
𝟐
}.
𝐝𝐨𝐦 (
𝐟
𝐠
) = 𝐑 ∩ 𝐑 − {
−𝟏
𝟐
} = 𝐑 − {
−𝟏
𝟐
}.
The function
𝐟
𝐠
: 𝐑 − {
−𝟏
𝟐
} → 𝐑 is given by (
𝐟
𝐠
) (𝐱) =
𝐟(𝐱)
𝐠(𝐱)
=
𝐱𝟐
𝟐𝐱+𝟏
, 𝐱 ≠
−𝟏
𝟐
.

Let 𝐟(𝐱) = √𝐱 and 𝐠(𝐱) = 𝐱 be two functions defined over the set of nonnegative real
numbers. Find: (i) (𝐟 + 𝐠)(𝐱) (ii) (𝐟 − 𝐠)(𝐱) (iii) (𝐟𝐠)(𝐱) (iv)
𝐟
𝐠
(𝐱)
𝐟 : [𝟎, ∞) → 𝐑: 𝐟(𝐱) = √𝐱 𝐠: [𝟎, ∞) → 𝐑: 𝐠(𝐱) = 𝐱.
dom (𝐟) = [𝟎, ∞) dom (𝐠) = [𝟎, ∞).
dom (𝐟) ∩ dom (𝐠) = [𝟎, ∞) ∩ [𝟎, ∞) = [𝟎, ∞).
(i) (𝐟 + 𝐠): [𝟎, ∞) → 𝐑 is given by (𝒇 + 𝒈)(𝒙) = 𝒇(𝒙) + 𝒈(𝒙) = (√𝒙 + 𝒙) .
(ii) (𝐟 − 𝐠): [𝟎, ∞) → 𝐑 is given by (𝒇 − 𝒈)(𝒙) = 𝒇(𝒙) − 𝒈(𝒙) = (√𝒙 − 𝒙) .
(iii) (𝒇𝒈): [𝟎, ∞) → 𝐑 is given by (𝒇𝒈)(𝒙) = 𝒇(𝒙) ⋅ 𝒈(𝒙) = (√𝒙 × 𝒙) = 𝒙𝟑/𝟐
.
(iv) {𝐱: 𝐠(𝐱) = 𝟎} = {𝟎}.
𝒅𝒐𝒎 (
𝒇
𝒈
) = 𝒅𝒐𝒎(𝒇) ∩ 𝒅𝒐𝒎 (𝒈) − {𝒙: 𝒈(𝒙) = 𝟎} = [𝟎, ∞) − {𝟎} = (𝟎, ∞)
So,
𝒇
𝒈
∶ (𝟎, ∞) → 𝑹 is given by (
𝐟
𝐠
) (𝐱) =
𝐟(𝐱)
𝐠(𝐱)
=
√𝐱
𝐱
=
𝟏
√𝐱
, 𝐱 ≠ 𝟎.

Let 𝐟 and 𝐠 be real functions defined by 𝐟(𝐱) = √𝐱 − 𝟏 and 𝐠(𝐱) = √𝐱 + 𝟏.
Find (i) (𝐟 + 𝐠)(𝐱) (ii) (𝐟 − 𝐠)(𝐱) (iii) (𝒇𝒈)(𝒙) (iv) (
𝐟
𝐠
) (𝐱) .
𝐟(𝐱) = √𝐱 − 𝟏 is defined for all real values of 𝐱 for which 𝐱 − 𝟏 ≥ 𝟎, dom (𝐟) = [𝟏, ∞).
𝐠(𝐱) = √𝐱 + 𝟏 is defined for all real values of 𝐱 for which 𝐱 + 𝟏 ≥ 𝟎, dom (𝐠) = [−𝟏, ∞)
dom (𝐟) ∩ dom (𝐠) = [𝟏, ∞) ∩ [−𝟏, ∞) = [𝟏, ∞).
(i) (𝐟 + 𝐠): [𝟏, ∞) → 𝐑 is given by (𝒇 + 𝒈)(𝒙) = 𝒇(𝒙) + 𝒈(𝒙) = (√𝒙 − 𝟏 +
√𝒙 + 𝟏) .
(ii) (𝐟 − 𝐠): [𝟏, ∞) → 𝐑 is given by (𝒇 − 𝒈)(𝒙) = 𝒇(𝒙) − 𝒈(𝒙) = (√𝒙 − 𝟏 −
√𝒙 + 𝟏) .
(iii) (𝒇𝒈): [𝟏, ∞) → 𝐑 is given by (𝐟𝐠)(𝐱) = 𝐟(𝐱) ⋅ 𝐠(𝐱) = √𝐱 − 𝟏 × √𝐱 + 𝟏 = √𝐱𝟐 − 𝟏.
(iv) {𝐱: 𝐠(𝐱) = 𝟎} = {𝐱: √𝐱 + 𝟏 = 𝟎} = {𝐱: 𝐱 + 𝟏 = 𝟎} = {−𝟏}.
dom (𝐟) ∩ dom (𝐠) − {𝐱: 𝐠(𝐱) = 𝟎} = [𝟏, ∞) ∩ [−𝟏, ∞)−{𝟏} = [𝟏, ∞)
𝐟
𝐠
→ [𝟏, ∞) → 𝐑 is given by (
𝐟
𝐠
) (𝐱) =
𝐟(𝐱)
𝐠(𝐱)
=
√𝐱−𝟏
√𝐱+𝟏
.

Let 𝐑+
be the set of all positive real numbers. Let 𝐟 : 𝐑+
→ 𝐑: 𝐟(𝐱) = 𝐥𝐨𝐠𝐞𝐱.
Find (i) range (f) (ii) {𝐱: 𝐱 ∈ 𝐑+
and 𝐟(𝐱) = −𝟐}.
(iii) Find out whether 𝐟(𝐱𝐲) = 𝐟(𝐱) + 𝐟(𝐲) for all 𝐱, 𝐲 ∈ 𝐑+
(i) For every 𝐱 ∈ 𝐑+
, we have 𝐥𝐨𝐠𝐞𝐱 = 𝐑. So, range (𝐟) = 𝐑.
(ii) 𝐟(𝐱) = −𝟐 ⇒ 𝐥𝐨𝐠𝐞𝐱 = −𝟐 ⇒ 𝐱 = 𝐞−𝟐
. So, {𝐱: 𝐱 ∈ 𝐑+
and 𝐟(𝐱) = −𝟐} = {𝐞−𝟐}.
(iii) 𝐟(𝐱𝐲) = 𝐥𝐨𝐠𝐞(𝐱𝐲) = 𝐥𝐨𝐠𝐞𝐱 + 𝐥𝐨𝐠𝐞𝐲 = 𝐟(𝐱) + 𝐟(𝐲) .
Let 𝐟 : 𝐑 → 𝐑: 𝐟(𝐱) = 𝟐𝐱
. Find (i) range (f) (ii) {𝐱: 𝐟(𝐱) = 𝟏}.
(iii) Find out whether 𝐟(𝐱 + 𝐲) = 𝐟(𝐱) ⋅ 𝐟(𝐲) for all 𝐱, 𝐲 ∈ 𝐑.
(i) 𝐟(𝐱) = 𝟐𝐱
> 𝟎 for every 𝐱 ∈ 𝐑. If 𝐱 ∈ 𝐑+
, there exists 𝐥𝐨𝐠𝟐𝐱 such that
𝐟(𝐥𝐨𝐠𝟐𝐱) = 𝟐𝐥𝐨𝐠𝟐𝐱
= 𝐱. range (𝐟) = 𝐑+
.
(ii) 𝐟(𝐱) = 𝟏 ⇒ 𝟐𝐱
= 𝟏 = 𝟐𝟎
⇒ 𝐱 = 𝟎. So, {𝐱 ∶ 𝐟(𝐱) = 𝟏} = {𝟎}.
(iii) 𝐟(𝐱 + 𝐲) = 𝟐𝐱+𝐲
= 𝟐𝐱
× 𝟐𝐲
= 𝐟(𝐱) ⋅ 𝐟(𝐲) .

Class 11_Chapter 2_Relations and Functions (Real Functions) Lecture 6.pdf

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Class 11_Chapter 2_Relations and Functions (Real Functions) Lecture 6.pdf