Determine whether the function is even or odd f x + y + f x - y = 2 f x · f y ; where f 0 ≠ 0 and x , y ∈ R .

f ( x + y ) + f ( x - y ) = 2 f ( x ) f ( y ) Let x = 0 , y = 0 f ( 0 + 0 ) + f ( 0 - 0 ) = 2 f ( 0 ) f ( 0 ) ⇒ 2 f ( 0 ) [ f ( 0 ) - 1 ] = 0 ∵ f ( 0 ) ≠ 0 ⇒ f ( 0 ) = 1 Now, let x = 0 ⇒ f ( 0 + y ) + f ( 0 - y ) = 2 f ( 0 ) f ( y ) ⇒ f ( y ) + f ( - y ) = 2 f ( 0 ) f ( y ) ⇒ f ( - y ) = 2 × 1 × f ( y ) - f ( y ) ⇒ f ( - y ) = f ( y ) ⇒ Given function is even.

A function, defined for all real numbers is defined for x ≥ 0 as follows: f x = x x , 0 ≤ x < 1 2 x , x ≥ 1 How, is f defined for x ≤ 0 , if f is even?

As for given function Given, f x = x 2 0 ≤ x < 1 2 x x ≥ 1 (because x = x when x > 0 ) If f ( x ) is an even function on R then f ( - x ) = f ( x ) Therefore, for x ≤ 0 , by putting - x in place of x in f x so f - x = f x = x 2 - 1 < x < 0 - 2 x x ≤ - 1

A function defined for all real numbers is defined for x ≥ 0 as follows: f x = x x , 0 ≤ x < 1 2 x , x ≥ 1 . How is f defined for x ≤ 0 , if i f is odd ? ii f is even ?

Given, f x = x 2 0 ≤ x < 1 2 x x ≥ 1 i If f x is an odd function on R , then f - x = - f x Therefore, for x ≤ 0 f x = - x 2 - 1 < x < 0 2 x x ≤ - 1 . ii If f x is an even function an R , then f - x = f x Therefore, for x ≤ 0 f x = x 2 , - 1 < x < 0 - 2 x , x ≤ - 1

Show that the function f x = 2 x sin x + tan x 2 x + 21 π π - 41 , is symmetric about the origin.

As for given function f x = 2 x ( sin x + tan x ) 2 x + 21 π π - 41 f x = 2 x ( sin x + tan x ) 2 x π + 2 × 21 - 41 = 2 x ( sin x + tan x ) 2 x π + 1 Replacing x by ( - x ) ⇒ f - x = 2 - x ( - sin x - tan x ) 2 - x π - 1 + 1 = 2 x ( sin x + tan x ) 2 - x π - 1 + 1 f - x = 2 x ( sin x + tan x ) - 2 x π - 2 + 1 = - f x ⇒ Odd Function ⇒ symmetric about origin.

Find the periods of function f x = sin 3 x + cos 6 x , where [ · ] denotes the greatest integer function.

Given, f x = sin 3 x + cos 6 x ; where · denotes G.I.F. Let, g x = sin 3 x ; where · denotes G.I.F. & h x = cos 6 x We know that if the period of f ( x ) is ' T ' then the period of f ( a x + b ) will be T a . For f ( x ) = g ( h ( x ) ) , if period of h ( x ) is ' T ' then period of g ( h ( x ) ) will be ' T ' . Using this Period of g x = 2 π 3 Period of h x = π 6 Period of f x = L C M o f 2 π 3 , π 6 LCM of l m , r s , p q = L C M o f l , r , p H C F o f m , s , q Hence, period of f ( x ) = 2 π 3

Find the period of function: f x = 1 2 sin x cos x + cos x sin x

For the period, firstly we have to write an equivalent definition of f ( x ) . Case I: When x ∈ 0 , π 2 ⇒ f x = 1 2 sin x cos x + cos x sin x ⇒ f x = 1 2 tan x + 1 tan x ⇒ f x = 1 + tan 2 x 2 tan x ⇒ f x = 1 sin 2 x ⇒ f x = cosec 2 x Case II: When x ∈ π 2 , π ⇒ f x = 1 2 sin x cos x - cos x sin x ⇒ f x = 1 2 tan x - 1 tan x ⇒ f x = - 1 - tan 2 x 2 tan x ⇒ f x = - 1 tan 2 x ⇒ f x = - cot 2 x Case III: When x ∈ π , 3 π 2 f x = 1 2 - sin x cos x - cos x sin x ⇒ f x = - 1 2 tan x + 1 tan x ⇒ f x = - 1 + tan 2 x 2 tan x ⇒ f x = - 1 sin 2 x ⇒ f x = - cosec 2 x Case IV: When x ∈ 3 π 2 , 2 π f x = 1 2 - sin x cos x + cos x sin x ⇒ f x = 1 2 - tan x + 1 tan x ⇒ f x = 1 2 1 - tan 2 x tan x ⇒ f x = 1 tan 2 x ⇒ f x = cot 2 x Hence we can write, f x = cosec 2 x ; x ∈ 0 , π 2 - cot 2 x ; x ∈ π 2 , π - cosec 2 x ; x ∈ π , 3 π 2 cot 2 x ; x ∈ e 3 π 2 , 2 π cosec 2 x ; x ∈ 2 π , 5 π 2 - cot 2 x ; x ∈ 5 π 2 , 3 π - cosec 2 x ; x ∈ 3 π , 7 π 2 cot 2 x ; x ∈ 7 π 2 , 4 π and so on As we can see in the equivalent definition of f x it repeats its values after 2 π . Hence, f x is a periodic function, whose fundamental period is 2 π .

E M B I B E

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.4>Q 1

MEDIUM

JEE Main

IMPORTANT

Earn 100

Determine whether the function is even or odd: $f (x) = sgn (x) + x^{2}$

Important Questions on Functions

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.4>Q 1

Determine whether the function is even or odd

f (x + y) + f (x - y) = 2 f (x) \cdot f (y);

where

f (0) \neq 0

and

x, y \in R

.

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.4>Q 2

Determine whether function;

f (x) = {(- 1)}^{[x]}

is even, odd or neither of two (where

[.]

denotes the greatest integer functions).

MEDIUM

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.4>Q 3

A function, defined for all real numbers is defined for $x \geq 0$ as follows:

$f (x) = \{\begin{cases} x |x|, 0 \leq x < 1 \\ 2 x, x \geq 1 \end{cases}$

How, is $f$ defined for $x \leq 0$ , if $f$ is even?

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.4>Q 3

A function defined for all real numbers is defined for $x \geq 0$ as follows: $f (x) = \{\begin{cases} x |x|, 0 \leq x < 1 \\ 2 x, x \geq 1 \end{cases}$ . How is $f$ defined for $x \leq 0$ , if $(i) f$ is odd ? $(ii) f$ is even ?

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.4>Q 4

Show that the function

f (x) = \frac{2 x (\sin x + \tan x)}{2 [\frac{x + 21 π}{π}] - 41}

, is symmetric about the origin.

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.4>Q 5

If $f : [- 20, 20] \to R$ defined by $f (x) = [\frac{x^{2}}{a}] \sin x + \cos x$ , is an even function, then find the set of values of $' a'$ . (Where $[\cdot]$ denotes the greatest integral function).