Find the domain of f x = sin - 1 2 x 2 - 3 , where · denotes the greatest integer function.

So, y = sin - 1 2 x 2 - 3 Domain of sin - 1 x is - 1 , 1 - 1 ≤ 2 x 2 - 3 ≤ 1 Case I: when 2 x 2 - 3 = - 1 - 1 ≤ 2 x 2 - 3 < 0 2 ≤ 2 x 2 < 3 1 ≤ x 2 < 3 2 So on solving, x ∈ ( - 3 2 , - 1 ) ∪ ( 1 , 3 2 ) … ( i ) Case II: when 2 x 2 - 3 = 0 0 ≤ 2 x 2 - 3 < 1 3 ≤ 2 x 2 < 4 3 2 ≤ x 2 < 2 On solving x ∈ ( - 2 , - 3 2 ) ∪ ( 3 2 , 2 ) … ( ii ) Case III: 2 x 2 - 3 = 1 1 ≤ 2 x 2 - 3 < 2 4 ≤ 2 x 2 < 5 2 ≤ x 2 < 5 2 On solving x ∈ ( - 5 2 , - 2 ) ∪ ( 2 , 5 2 ) … ( iii ) From eq. (i), (ii) and (iii) x ∈ ( - 5 2 , - 1 ) ∪ ( 1 , 5 2 )

Find the domain of f x = sin - 1 2 x 2 - 3 , where · denotes the greatest integer function.

So, y = sin - 1 2 x 2 - 3 Domain of sin - 1 x is - 1 , 1 - 1 ≤ 2 x 2 - 3 ≤ 1 Case I: when 2 x 2 - 3 = - 1 - 1 ≤ 2 x 2 - 3 < 0 2 ≤ 2 x 2 < 3 1 ≤ x 2 < 3 2 So on solving, x ∈ ( - 3 2 , - 1 ) ∪ ( 1 , 3 2 ) … ( i ) Case II: when 2 x 2 - 3 = 0 0 ≤ 2 x 2 - 3 < 1 3 ≤ 2 x 2 < 4 3 2 ≤ x 2 < 2 On solving x ∈ ( - 2 , - 3 2 ) ∪ ( 3 2 , 2 ) … ( ii ) Case III: 2 x 2 - 3 = 1 1 ≤ 2 x 2 - 3 < 2 4 ≤ 2 x 2 < 5 2 ≤ x 2 < 5 2 On solving x ∈ ( - 5 2 , - 2 ) ∪ ( 2 , 5 2 ) … ( iii ) From eq. (i), (ii) and (iii) x ∈ ( - 5 2 , - 1 ) ∪ ( 1 , 5 2 )

Find the domain of f x = sin - 1 log 2 x 2 2 , where · denotes the greatest integer function.

f x = sin - 1 log 2 x 2 2 log e x is defined when x > 0 So, log 2 x 2 2 is defined when x 2 2 > 0 x ∈ R - 0 … ( i ) Domain of sin - 1 x is - 1 , 1 So, sin - 1 log 2 x 2 2 - 1 ≤ log 2 x 2 2 ≤ 1 Case I: When log 2 x 2 2 = - 1 - 1 ≤ log 2 x 2 2 < 0 1 2 ≤ x 2 2 < 1 1 ≤ x 2 < 2 On solving x ∈ ( - 2 , - 1 ) ∪ ( 1 , 2 ) … ( i ) Case II: log 2 x 2 2 = 0 0 ≤ log 2 x 2 2 < 1 1 ≤ x 2 2 < 2 2 ≤ x 2 < 4 On solving x ∈ ( - 2 , - 2 ] ∪ [ 2 , 2 ) … ( ii ) Case III: log 2 x 2 2 = 1 1 ≤ log 2 x 2 2 < 2 log 2 2 ≤ log 2 x 2 2 < log 2 4 2 ≤ x 2 2 < 4 4 ≤ x 2 < 8 On solving x ∈ ( - 8 , - 2 ) ∪ ( 2 , 8 ) … ( iii ) From eq. (i), (ii) and (iii) x ∈ ( - 8 , - 1 ) ∪ ( 1 , 8 )

Find the domain of f x = 1 x - 2 + x - 10 - 8 , where · denotes the greatest integer function.

f x = 1 x - 2 + x - 10 - 8 It is defined when x - 2 + x - 10 - 8 ≠ 0 So, x - 2 + x - 10 - 8 = 0 when x - 2 + x - 10 = 8 So, x - 2 + x - 10 = 8 We know that x ∈ Z For x = 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 Domain is R - 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 So, x ∈ R - 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10

If a function is defined, as g x = sin x + sin x , ϕ x = sin x + cos x , 0 ≤ x ≤ π , then find the domain for f x = log ϕ x g x .

g x = sin x + sin x , ϕ x = sin x + cos x , 0 ≤ x ≤ π We have to find domain of f x = log ϕ x g x It is defined only when g x > 0 ϕ x > 0 ϕ x ≠ 1 And log ϕ x g x ≥ 0 So, g x > 0 ⇒ sin x + sin x > 0 Given, x ∈ [ 0 , π ] Hence, 2 sin x > 0 sin x > 0 x ∈ 0 , π ....... 1 Now, ϕ x > 0 ⇒ sin x + cos x > 0 ⇒ 2 1 2 cos x + 1 2 sin x > 0 ⇒ sin π 4 + x > 0 ⇒ x ∈ 0 , 3 π 4 ...... 2 Now, ϕ x ≠ 1 ⇒ sin x + cos x ≠ 1 sin π 4 + x ≠ 1 2 ⇒ x ≠ 0 , π 2 ..... 3 log ϕ x g x ≥ 0 ⇒ Case I: g x ≥ 1 if ϕ x > 1 ⇒ sin x + sin x ≥ 1 & sin x + cos x > 1 Given that, x ∈ 0 , π ⇒ 2 sin x ≥ 1 & sin x + π 4 > 1 2 ⇒ sin x ≥ 1 2 & sin x + π 4 > 1 2 x ∈ π 6 , 5 π 6 & ∈ 0 , π 2 ⇒ x ∈ π 6 , π 2 .......... 4 Case II: g x ≤ 1 if 0 < ϕ x < 1 ⇒ sin x + sin x ≤ 1 & 0 < sin x + cos x < 1 ∵ x ∈ 0 , π ⇒ 0 ≤ 2 sin x ≤ 1 & 0 < 2 sin x + π 4 < 1 ⇒ 0 ≤ sin x ≤ 1 2 & 0 < sin x + π 4 < 1 2 ⇒ x ∈ 0 , π 6 ∪ 5 π 6 , π & x ∈ ϕ ⇒ x ∈ ϕ ..... 5 Using equation 1 , 2 , 3 , 4 & 5 The final domain of f x is x ∈ π 6 , π 2

Solve the equations, y = 1 3 sin x + sin x + sin x and y + y = 2 cos x , where · denotes the greatest integer function.

Given, y = 1 3 sin x + sin x + sin x and y + y = 2 cos x Where · denotes G.I.R. Using x + n = x + n , n ∈ I , we can write y = 1 3 sin x + sin x + sin x ⇒ y = sin x … ( i ) and y + y = 2 cos x ⇒ y = cos x … ( ii ) Using y = sin x and y = cos x we can write, sin x = cos x As shown in figure, there is 0 intersection points between both curves. Hence, no solution.

Find the range of f x = 9 - x 2 .

Given function: y = 9 - x 2 It is defined only when 9 - x 2 ≥ 0 ⇒ x 2 - 9 ≤ 0 ⇒ x + 3 x - 3 ≤ 0 ⇒ x ∈ - 3 , 3 For range 9 - x 2 ∈ ( - ∞ , 9 ] ⇒ 9 - x 2 ∈ [ 0 , 3 ] ; - ∞ , 0 will be rejected because for x , x ≥ 0 .

Find the range of f x = x 1 + x 2 , x ∈ R

Let y = x 1 + x 2 ; x ∈ R ⇒ y 1 + x 2 = x ⇒ y + y x 2 - x = 0 ⇒ y x 2 - x + y = 0 ; x ∈ R If a quadratic equation has real roots then, D ≥ 0 i.e., b 2 - 4 a c ≥ 0 for y x 2 - x + y = 0 , we can write, - 1 2 - 4 y y ≥ 0 ⇒ 1 - 4 y 2 ≥ 0 ⇒ 4 y 2 - 1 ≤ 0 ⇒ y - 1 2 y + 1 2 ≤ 0 ⇒ y ∈ - 1 2 , 1 2

E M B I B E

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.2>Q 28

HARD

JEE Main

IMPORTANT

Earn 100

Find the domain of $f (x) = \sin^{- 1} [2 x^{2} - 3],$ where $[\cdot]$ denotes the greatest integer function.

Important Questions on Functions

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.2>Q 29

Find the domain of

f (x) = \sin^{- 1} [\log_{2} (\frac{x^{2}}{2})],

where

[\cdot]

denotes the greatest integer function.

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.2>Q 30

Find the domain of

f (x) = \frac{1}{[|x - 2|] + [|x - 10|] - 8},

where

[\cdot]

denotes the greatest integer function.

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.2>Q 31

If a function is defined, as

g (x) = |\sin x| + \sin x, ϕ (x) = \sin x + \cos x, 0 \leq x \leq π

, then find the domain for

f (x) = \sqrt{\log_{ϕ (x)} g (x)}

.

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.2>Q 32

Solve the equations,

y = \frac{1}{3} [\sin x + [\sin x + [\sin x]]]

and

[y + [y]] = 2 \cos x

, where

[\cdot]

denotes the greatest integer function.

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.2>Q 33

If $[x] = [\frac{x}{2}] + [\frac{x + 1}{2}],$ where $[\cdot]$ denotes the greatest integer function, and $n$ is a positive integer, then show that $[\frac{n + 1}{2}] + [\frac{n + 2}{4}] + [\frac{n + 4}{8}] + [\frac{n + 8}{16}] + . . . = n$ .