Find the domain of f x = 2 x 2 - 3 x + 1 , x ∈ - 1 , 1 , where · denotes the fractional part of x .

f x = 2 x 2 - 3 x + 1 2 x 2 - 3 x + 1 ≥ 0 On solving x - 1 2 x - 1 ≥ 0 x ∈ - ∞ , 1 2 ∪ [ 1 , ∞ ) … ( i ) But range of x ∈ [ 0 , 1 ) … ( ii ) So, taking intersection in eq (i) and (ii) x ∈ 0 , 1 2 So, x ∈ - 1 , - 1 2 ∪ 0 , 1 2 ∪ 1

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.

E M B I B E

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

Amit M Agarwal Solutions for Chapter: Functions, Exercise 2: Target Exercise 3.2

Author:Amit M Agarwal

Amit M Agarwal Mathematics Solutions for Exercise - Amit M Agarwal Solutions for Chapter: Functions, Exercise 2: Target Exercise 3.2

Attempt the practice questions on Chapter 3: Functions, Exercise 2: Target Exercise 3.2 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 2: Target Exercise 3.2 with Hints & Solutions

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.2>Q 27

Find the domain of $f (x) = \sqrt{2 {\{x\}}^{2} - 3 \{x\} + 1}, x \in [- 1, 1],$ where $\{\cdot\}$ denotes the fractional part of $x$ .

HARD

JEE Main

IMPORTANT

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

Find the domain of $f (x) = \sin^{- 1} [2 x^{2} - 3],$ 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 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 $[\frac{n + 1}{2}] + [\frac{n + 2}{4}] + [\frac{n + 4}{8}] + [\frac{n + 8}{16}] + . . . = n$ .

HARD

JEE Main

IMPORTANT

Skills in Mathematics Differential Calculus for JEE Main & Advanced>Chapter 3 - Functions>Target Exercise 3.2>Q 34

Find the integral solution to the equation $[x] [y] = x + y$ . Show that all the non-integral solutions lie on exactly two lines. Determine these lines.

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