Find the period of function f x = cos 3 x + sin 3 π x .

Given, f x = cos 3 x + sin 3 π x Let g x = cos 3 x & h x = sin ( 3 π x ) So it is clear from the graph fundamental period of sin x = 2 π and cos x = 2 π . We know that if the period of f ( x ) is T , then the period of f ( a x + b ) will be T a . Using the above concept, period of g x = 2 π 3 Period of h x = 2 π 3 π = 2 3 Period of f x = LCM of 2 π 3 , 2 3 = does not exist Hence, it does not exist (due to different irrationality).

Find the periods of function f x = sin π x n ! - cos π x n + 1 ! .

Given, f x = sin π x n ! - cos π x n + 1 ! Let g x = sin π x n ! & h x = cos π x n + 1 ! It is clear from graph fundamental period of sin x = 2 π and cos x = 2 π We know that if the period of f ( x ) is T , then the period of f ( a x + b ) will be T a . Using the above concepts, we can write, Period of g x = 2 π π n ! = 2 n ! Period of h x = 2 π π n + 1 ! = 2 n + 1 ! Period of f x = LCM of 2 n ! , 2 n + 1 ! Hence, period of f ( x ) = 2 n + 1 !

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

E M B I B E

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

Amit M Agarwal Solutions for Chapter: Functions, Exercise 5: Target Exercise 3.5

Author:Amit M Agarwal

Amit M Agarwal Mathematics Solutions for Exercise - Amit M Agarwal Solutions for Chapter: Functions, Exercise 5: Target Exercise 3.5

Attempt the practice questions on Chapter 3: Functions, Exercise 5: Target Exercise 3.5 with hints and solutions to strengthen your understanding. Skills in Mathematics Differential Calculus for JEE Main & Advanced solutions are prepared by Experienced Embibe Experts.

Questions from Amit M Agarwal Solutions for Chapter: Functions, Exercise 5: Target Exercise 3.5 with Hints & Solutions

MEDIUM

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.5>Q 1

Find the period of function $f (x) = \cos 3 x + \sin \sqrt{3} π x$ .

MEDIUM

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.5>Q 1

Find the periods of function $f (x) = \sin \frac{π x}{n!} - \cos \frac{π x}{(n + 1)!}$ .

EASY

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.5>Q 1

Find the periods of function $f (x) = x - [x - b]$ , where, $[.]$ denotes G.I.F. and $b \in I$ .

MEDIUM

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.5>Q 1

Find the periods of function $f (x) = e^{\ln (\sin x)} + \tan^{3} x - cosec (3 x - 5)$ .

HARD

JEE Main

IMPORTANT

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

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

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

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