Find the fundamental period of the following function f x = sin 12 x 1 + cos 2 6 x

f x = sin 12 x 1 - cos 2 6 x ⇒ f x = 2 sin 6 x . cos 6 x sin 2 6 x [ ∵ sin 2 A = 2 sin A cos A a n d sin 2 θ + cos 2 θ = 1 ] ⇒ f ( x ) = 2 cot 6 x period of cot 6 x is π 6 as period of f ( a x ) is equal to period of f ( x ) | a | Period of 2 cot 6 x is also π 6 as period of k f ( x ) is same as f ( x )

Find the fundamental period of the following function f x = sin 12 x 1 + cos 2 6 x

f x = sin 12 x 1 - cos 2 6 x ⇒ f x = 2 sin 6 x . cos 6 x sin 2 6 x [ ∵ sin 2 A = 2 sin A cos A a n d sin 2 θ + cos 2 θ = 1 ] ⇒ f ( x ) = 2 cot 6 x period of cot 6 x is π 6 as period of f ( a x ) is equal to period of f ( x ) | a | Period of 2 cot 6 x is also π 6 as period of k f ( x ) is same as f ( x )

Find the inverse of f x = ln x + 1 + x 2 .

Let f x = l n x + 1 + x 2 Domain : x ∈ R , Range : R y = l n x + 1 + x 2 x + 1 + x 2 = e y ... 1 x + 1 + x 2 1 + x 2 - x 1 + x 2 - x = e y 1 + x 2 - x = e - y ... 2 Equation 1 - 2 , or x = e y - e - y 2 f - 1 ( x ) = e x - e - x 2 .

Let f : [ 0 , 3 ] → [ 1 , 13 ] is defined by f ( x ) = x 2 + x + 1 , then find f - 1 ( x )

f ( x ) = x 2 + x + 1 ⇒ f ' ( x ) = 2 x + 1 Now, from [ 0 , 3 ] , f ' ( x ) > 0 Therefore, it is always increasing. Hence, it is a one-one function. Now, as it is always increasing Therefore, minimum at x = 0 ⇒ f ( 0 ) = 1 4 + 3 4 = 1 maximum at x = 3 ⇒ f 3 = 3 + 1 2 2 + 3 4 = 49 4 + 3 4 = 13 So, here co-domain is same as range Hence, function is onto. Now, since the function is bijective. Therefore, its inverse exist. f ( x ) = x 2 + x + 1 ⇒ y = x 2 + x + 1 ⇒ x 2 + x + 1 - y = 0 ⇒ x = - 1 ± 4 y - 3 2 Because x ∈ 0 , 3 , So we have to take only + sign ⇒ f - 1 x = - 1 + 4 x - 3 2

Let f : R → R be defined by f x = e 2 x - e - 2 x 2 . Is f ( x ) invertible? If yes, then find its inverse.

Given, f : R → R f x = e 2 x - e - 2 x 2 = e 2 x - 1 e 2 x 2 = e 4 x - 1 2 e 2 x A function to be invertible only if it is one-one & onto. For checking one-one Put f x 1 = f x 2 ⇒ e 4 x 1 - 1 2 e 2 x 1 = e 4 x 2 - 1 2 e 2 x 2 ⇒ e 2 x 2 + 4 x 1 - e 2 x 2 = e 4 x 2 + 2 x 1 - e 2 x 1 ⇒ e 2 x 1 - e 2 x 2 + e 2 x 2 + 4 x 1 - e 4 x 2 + 2 x 1 = 0 ⇒ e 2 x 1 - e 2 x 2 + e 2 x 1 + 2 x 2 e 2 x 1 - e 2 x 2 = 0 ⇒ e 2 x 1 - e 2 x 2 1 + e 2 x 1 + 2 x 2 = 0 ⇒ e 2 x 1 = e 2 x 2 ⇒ x 1 = x 2 Hence, one-one function. For checking onto function - lim x → ∞ f x = lim x → ∞ e 4 x - 1 2 e 2 x = lim x → ∞ e 2 x 2 - 1 2 e 2 x = ∞ Now, lim x → - ∞ f x = lim x → - ∞ e 4 x - 1 2 e 2 x = lim x → - ∞ e 2 x 2 - 1 2 e 2 x = 0 - ∞ = - ∞ Since, range ∈ R & co-domain is also R . Hence, it is an onto function. therefore, the function is invertible. For finding inverse y = e 4 x - 1 2 e 2 x Let, e 2 x = t ⇒ y = t 2 - 1 2 t ⇒ t 2 - 2 t y - 1 = 0 ⇒ t = - 2 y ± - 2 y 2 - 4 × 1 × - 1 2 ⇒ t = 2 y ± 4 y 2 + 4 2 ⇒ t = y ± y 2 + 1 ⇒ e 2 x = y ± y 2 + 1 Since, e 2 x will be always + v e , we get e 2 x = y + y 2 + 1 Taking log both sides, we get log e 2 x = log y + y 2 + 1 ⇒ 2 x = log y + y 2 + 1 ⇒ x = 1 2 log y + y 2 + 1 For inverse replace x → f - 1 x & y → x Hence, f - 1 x = 1 2 log x + x 2 + 1

If f x = ( α - 1 ) x , x ∈ Q c - α 2 x + α + 3 x - 1 , x ∈ Q and g x = x , x ∈ Q c 1 - x , x ∈ Q are inverse to each other. Then find all possible values of α .

Given: f x = α - 1 x ; x ∈ Q c - α 2 x + α + 3 x - 1 ; x ∈ Q g x = x ; x ∈ Q c 1 - x ; x ∈ Q Also, f x = g - 1 x and g x = f - 1 x Now, finding g - 1 x :- g x = y ⇒ y = x ; x ∈ Q c ⇒ x = g - 1 x ⇒ g - 1 x = x ( replacing x → g - 1 x and y → x ) And, g x = 1 - x ; x ∈ Q ⇒ y = 1 - x ⇒ x = 1 - y ⇒ g - 1 x = 1 - x ; x ∈ Q Now putting f x = g - 1 x α - 1 x = x and - α 2 x + α + 3 x - 1 = 1 - x a - 1 = 1 and 3 - α 2 x + α - 1 = 1 - x ⇒ a = 2 and 3 - α 2 x + α - 1 = - x + 1 On comparing, we get 3 - α 2 = - 1 and α - 1 = 1 a = ± 2 and a = 2 So, α = 2 .

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-1>Q 169

MEDIUM

JEE Main/Advance

IMPORTANT

Earn 100

Find the fundamental period of the following function $f (x) = \frac{\sin 12 x}{1 + \cos^{2} 6 x} \begin{array}{l} \end{array}$

Important Questions on Functions

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-1>Q 170

Find the fundamental period of the following function $f (x) = \sec^{3} x + {cosec}^{3} x$

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-1>Q 171

Let $f : D \to R,$ where $D$ is the domain of $f$ . Find the inverse of $f$ , if it exists, $f (x) = 1 - 2^{- x}$ .

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-1>Q 172

Find the inverse of

f (x) = \ln (x + \sqrt{1 + x^{2}})

.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-1>Q 173

Let

f : [0, 3] \to [1, 13]

is defined by

f (x) = x^{2} + x + 1,

then find

f^{- 1} (x)

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-1>Q 174

Let

f : R \to R

be defined by

f (x) = \frac{e^{2 x} - e^{- 2 x}}{2} .

Is

f (x)

invertible? If yes, then find its inverse.

EASY

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-1>Q 175

If $f (x) = - x | x |,$ then find $f^{- 1} (x)$ and hence find the number of solutions of $f (x) = f^{- 1} (x)$

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 24 - Functions>Exercise-1>Q 176

If $f (x) = \{\begin{aligned} (α - 1) x, & x \in Q^{c} \\ - α^{2} x + α + 3 x - 1, & x \in Q \end{aligned}$ and $g (x) = \{\begin{cases} x, & x \in Q^{c} \\ 1 - x, & x \in Q \end{cases}$ are inverse to each other. Then find all possible values of $α .$