Check whether the function defined by f x + λ = 1 + 2 f x - f 2 x , ∀ x ∈ R , λ > 0 is periodic. If yes, then find its period is.

We have, f ( x + λ ) = 1 + 2 f ( x ) - f 2 ( x ) . . . ( i ) f ( x + λ ) - 1 = 2 f ( x ) - f 2 ( x ) Squaring on both sides, we get f x + λ - 1 2 = 2 f ( x ) - f 2 ( x ) Adding and subtracting 1 on the right side, we get f x + λ - 1 2 = - f 2 ( x ) - 2 f ( x ) + 1 + 1 ⇒ f x + λ - 1 2 = 1 - f ( x ) - 1 2 . . . ( ii ) Replacing x by x + λ , Then, f ( x + 2 λ ) - 1 2 = 1 - ( f ( x + λ ) - 1 ) 2 Using equation ii , we get ⇒ f ( x + 2 λ ) - 1 2 = 1 - ( 1 - ( f ( x ) - 1 ) 2 ) ⇒ f ( x + 2 λ ) - 1 2 = ( f ( x ) - 1 ) 2 Taking square root on both sides, we get: f ( x + 2 λ ) - 1 = f ( x ) - 1 . . . ( iii ) Now, from equation i i.e., f ( x + λ ) = 1 + 2 f ( x ) - f 2 ( x ) 2 f ( x ) - f 2 ( x ) ≥ 0 f ( x ) 2 - f ( x ) ≥ 0 f ( x ) f ( x ) - 2 ≤ 0 f x ∈ 0 , 2 Also, f ( x + λ ) ≥ 1 f ( x ) ≥ 1 f ( x ) ∈ [ 1 , ∞ ) Therefore, f ( x ) ∈ [ 1 , 2 ] From iii i . e . , f ( x + 2 λ ) - 1 = f ( x ) - 1 , we get f ( x + 2 λ ) - 1 = f ( x ) - 1 f ( x + 2 λ ) = f ( x ) Hence, period is 2 λ .

Let f x + p = 1 + 2 - 3 f x + 3 f x 2 - f x 3 1 3 , ∀ x ∈ R where p > 0 , then prove that f x is periodic.

f x + p = 1 + 2 - 3 f x + 3 f x 2 - f x 3 1 3 f x + p = 1 + 1 + 1 - 3 f x + 3 f x 2 - f x 3 1 3 f x + p = 1 + 1 + ( 1 - f ( x ) ) 3 1 3 f x + p - 1 = 1 - ( f ( x ) - 1 ) 3 1 3 Let, f ( x ) - 1 = g ( x ) then f x + p - 1 = 1 - ( f ( x ) - 1 ) 3 1 3 becomes g x + p = 1 - g ( x ) 3 1 3 Now replace , x → x + p g x + 2 p = 1 - g ( x + p ) 3 1 3 g x + 2 p = 1 - 1 - g ( x ) 3 1 3 g x + 2 p = g ( x ) 3 1 3 g x + 2 p = g ( x ) g x + 2 p + 1 = g ( x ) + 1 f ( x + 2 p ) = f ( x ) Hence, f ( x ) is a periodic function with period 2 p .

Let f x be a function such that, f x - 1 + f x + 1 = 3 f x , ∀ x ∈ R . If f 5 = 100 , then find ∑ r = 0 99 f 5 + 12 r .

f ( x + 1 ) + f ( x - 1 ) = 3 f ( x ) . . . ( i ) Replace x → x + 1 , then f ( x ) + f ( x + 2 ) = 3 f ( x + 1 ) . . . ( i i ) Replace x → x - 1 , then from equation ( i ) f ( x - 2 ) + f ( x ) = 3 f ( x - 1 ) . . . ( i i i ) Adding equation i i and i i i , and from i f ( x + 2 ) + f ( x - 2 ) + 2 f ( x ) = 3 3 f ( x ) f ( x + 2 ) + f ( x - 2 ) + 2 f ( x ) = 3 f ( x ) f ( x + 2 ) + f ( x - 2 ) = f ( x ) . . . ( i v ) Replace, x → x + 2 f ( x + 4 ) + f ( x ) = f ( x + 2 ) . . . ( v ) Replace , x → x - 2 f ( x ) + f ( x - 4 ) = f ( x - 2 ) . . . ( v i ) Adding equation v & v i and from i v , we get f ( x + 4 ) + f ( x - 4 ) = - f ( x ) . . . ( v i i ) Replace , x → x + 4 f ( x + 8 ) + f ( x ) = - f ( x + 4 ) . . . v i i i Replace, x → x - 4 f ( x ) + f ( x - 8 ) = - f ( x - 4 ) . . . ( i x ) Adding v i i i & i x and from v i i i.e., f ( x + 4 ) + f ( x - 4 ) = - f ( x ) we get, f ( x + 8 ) + f ( x - 8 ) = - f ( x ) Replace, x → x + 4 f ( x + 12 ) + f ( x - 4 ) = - f ( x + 4 ) . . . ( x i ) ∵ f ( x + 4 ) + f ( x - 4 ) = - f ( x ) we can write f ( x + 12 ) - f ( x + 4 ) - f ( x ) = - f ( x + 4 ) f ( x + 12 ) = f ( x ) Hence, period of f ( x ) is 12 . ∑ r = 0 99 f ( 5 + 12 r ) = f ( 5 ) + f ( 5 + 12 ) + f ( 5 + 24 ) + . . . + f ( 5 + 99 × 12 ) Since, the function is periodic we can write f 5 = f 5 + 12 = f 5 + 24 = . . . = f 5 + 12 · 99 = f ( 5 ) + f ( 5 ) + f ( 5 ) + f ( 5 ) + f ( 5 ) + . . . . . . . . u p t o 100 t e r m s = 100 f ( 5 ) = 100 × 100 = 10000

f x = x - 1 , - 1 ≤ x ≤ 0 x 2 , 0 < x ≤ 1 and g x = sin x , then find h x = f g x + f g x .

Given functions f x = x - 1 , - 1 ≤ x ≤ 0 x 2 , 0 < x ≤ 1 and g x = sin x f g x = g x - 1 , - 1 ≤ g x ≤ 0 g x 2 , 0 < g x ≤ 1 = sin x - 1 , - 1 ≤ sin x ≤ 0 sin 2 x , 0 < sin x ≤ 1 = - 1 , sin x = 0 sin 2 x , sin x ≠ 0 = - 1 , sin x = 0 sin 2 x , - 1 ≤ sin x < 0 sin 2 x , 0 < sin x ≤ 1 f g x = g x - 1 , - 1 ≤ g x ≤ 0 g x 2 , 0 < g x ≤ 1 = sin x - 1 , - 1 ≤ sin x ≤ 0 sin 2 x , 0 < sin x ≤ 1 Then, f g x = 1 - sin x , - 1 ≤ sin x ≤ 0 sin 2 x , 0 < sin x ≤ 1 = 1 - 1 , sin x = 0 sin x , - 1 ≤ sin x < 0 sin 2 x , 0 < sin x ≤ 1 Now, h x = f g x + f g x = sin 2 x - 0 , sin x = 0 sin x + 1 , - 1 ≤ sin x < 0 2 sin 2 x , 0 < sin x ≤ 1 = sin 2 x - sin x + 1 , - π 2 ≤ x < 0 2 sin 2 x , 0 ≤ x ≤ π 2

Let f x be defined on - 2 , 2 and is given by f x = - 1 , - 2 ≤ x ≤ 0 x - 1 , 0 ≤ x ≤ 2 and g x = f x + f x , then find g x .

Given that f x is defined on - 2 , 2 and f x = - 1 , - 2 ≤ x ≤ 0 x - 1 , 0 ≤ x ≤ 2 f x = - 1 , - 2 ≤ x ≤ 0 x - 1 , 0 ≤ x ≤ 2 = x - 1 , x ∈ - 2 , 2 = - x - 1 , - 2 ≤ x < 0 x - 1 , 0 ≤ x ≤ 1 x - 1 , 1 ≤ x ≤ 2 Then f x = - 1 , - 2 ≤ x ≤ 0 x - 1 , 0 ≤ x ≤ 2 ⇒ f x = 1 , - 2 ≤ x ≤ 0 1 - x , 0 ≤ x ≤ 1 x - 1 , 1 ≤ x ≤ 2 Now, g x = f x + f x = - x - 1 + 1 , - 2 ≤ x < 0 x - 1 + 1 - x , 0 ≤ x < 1 2 x - 1 , 1 ≤ x ≤ 2 = - x , - 2 ≤ x < 0 0 , 0 ≤ x < 1 2 x - 1 , 1 ≤ x ≤ 2

Let two functions are defined as g x = x 2 , - 1 ≤ x < 2 x + 2 , 2 ≤ x ≤ 3 and f x = x + 1 , x ≤ 1 2 x + 1 , 1 < x ≤ 2 , then find g o f .

Given functions: g x = x 2 , - 1 ≤ x < 2 x + 2 , 2 ≤ x ≤ 3 and f x = x + 1 , x ≤ 1 2 x + 1 , 1 < x ≤ 2 Then g o f x = g f x = f x 2 , - 1 ≤ f x < 2 f x + 2 , 2 ≤ f x ≤ 3 = x + 1 2 , - 1 ≤ x + 1 < 2 and x ≤ 1 x + 1 + 2 , 2 ≤ x + 1 ≤ 3 and x ≤ 1 2 x + 1 2 , - 1 ≤ 2 x + 1 < 2 and 1 < x ≤ 2 2 x + 1 + 2 , 2 ≤ 2 x + 1 ≤ 3 and 1 < x ≤ 2 = x + 1 2 , - 2 ≤ x < 1 ∩ x ≤ 1 x + 3 , 1 ≤ x ≤ 2 and x ≤ 1 2 x + 1 2 , - 1 ≤ x < 1 2 and 1 < x ≤ 2 2 x + 3 , 1 2 ≤ x ≤ 1 and 1 < x ≤ 2 = x + 1 2 , - 2 ≤ x < 1 x + 3 , x = 1 2 x + 1 2 , x ∈ ϕ 2 x + 3 , x ∈ ϕ = x + 1 2 , ∀ x ∈ - 2 , 1

E M B I B E

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.5>Q 2

HARD

JEE Main

IMPORTANT

Earn 100

Find the period of the real-valued function satisfying, $f (x) + f (x + 4) = f (x + 2) + f (x + 6)$ .

Important Questions on Functions

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.5>Q 3

Check whether the function defined by

f (x + λ) = 1 + \sqrt{2 f (x) - f^{2} (x)}, \forall x \in R, λ > 0

is periodic. If yes, then find its period is.

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.5>Q 4

Let

f (x + p) = 1 + {\{2 - 3 f (x) + 3 {(f (x))}^{2} - {(f (x))}^{3}\}}^{\frac{1}{3}}, \forall x \in R

where

p > 0,

then prove that

f (x)

is periodic.

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.5>Q 5

Let

f (x)

be a function such that,

f (x - 1) + f (x + 1) = \sqrt{3} f (x), \forall x \in R

. If

f (5) = 100

, then find

\sum_{r = 0}^{99} f (5 + 12 r)

.

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.6>Q 1

Consider the real-valued function satisfying

2 f (\sin x) + f (\cos x) = x .

Find the domain and range of

f (x)

, using the Principal domain and range.

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.6>Q 2

If $f (x)$ is defined in $[- 3, 2]$ , then find the domain of definition of $f ([|x|])$ and $f ([2 x + 3])$

where $[.]$ denotes the greatest integer function.

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.6>Q 3

$f (x) = \{\begin{cases} x - 1, - 1 \leq x \leq 0 \\ x^{2}, 0 < x \leq 1 \end{cases}$ and $g (x) = \sin x$ , then find $h (x) = f (|g (x)|) + |f (g (x))|$ .