The range of the function f x = sin - 1 x 2 + 1 2 + cos - 1 x 2 - 1 2 , where . , is the greatest integer function, is

Given, f x = sin - 1 x 2 + 1 2 + cos - 1 x 2 - 1 2 Let first find the domain of f x We know that, the domain of sin - 1 x and cos - 1 x is x ∈ - 1 , 1 . Therefore, here - 1 ≤ x 2 + 1 2 ≤ 1 and - 1 ≤ x 2 - 1 2 ≤ 1 As, x 2 + 1 2 and x 2 - 1 2 , is an integer value So, x 2 + 1 2 and x 2 - 1 2 , can be equal to - 1 , 0 , 1 . But x 2 + 1 2 ≠ - 1 as x 2 + 1 2 ≥ 1 2 , for all x ∈ R x 2 - 1 2 = - 1 ⇒ - 1 ≤ x 2 - 1 2 < 0 ⇒ - 1 2 ≤ x 2 < 1 2 ⇒ 0 ≤ x 2 < 1 2 ⇒ x 2 - 1 2 < 0 ⇒ x ∈ - 1 2 , 1 2 . So, when - 1 2 < x < 1 2 , then x 2 + 1 2 = 0 and x 2 - 1 2 = - 1 When, x 2 - 1 2 = 1 ⇒ x 2 = 3 2 ⇒ x = 3 2 We can see that, for 1 2 < x < 3 2 1 < x 2 + 1 2 < 2 and 0 < x 2 - 1 2 < 1 ⇒ x 2 + 1 2 = 1 and x 2 - 1 2 = 0 For x > 3 2 x 2 + 1 2 = 2 and x 2 - 1 2 = 1 But x 2 + 1 2 = 2 , does not satisfy the domain of sin - 1 x , hence x > 3 2 is not in the domain of f x Hence, domain of f x is - 1 2 , 1 2 ∪ 1 2 , 3 2 . Now for a range of f x , When, - 1 2 < x , 1 2 Then, x 2 + 1 2 = 0 and x 2 - 1 2 = - 1 So, sin - 1 x 2 + 1 2 = 0 and cos - 1 x 2 - 1 2 = π f x = π for - 1 2 < x < 1 2 When 1 2 < x < 3 2 x 2 + 1 2 = 1 and x 2 - 1 2 = 0 So, sin - 1 x 2 + 1 2 = sin - 1 ( 1 ) = π 2 and cos - 1 x 2 - 1 2 = c o s - 1 ( 0 ) = π 2 f x = π 2 + π 2 = π Hence, the range f x is π .

tan - 1 ( tan θ ) = π + θ , - 3 π 2 < θ < - π 2 θ , - π 2 < θ < π 2 - π + θ , π 2 < θ < 3 π 2 , sin - 1 ( sin θ ) = - π - θ , - 3 π 2 < θ < - π 2 θ , - π 2 ≤ θ ≤ π 2 π - θ , π 2 < θ ≤ 3 π 2 cos - 1 ( cos θ ) = - θ , - π ≤ θ < 0 θ , 0 ≤ θ ≤ π 2 π - θ , π < θ ≤ 2 π Based on the above results, prove each of the following: (i) cos - 1 x = sin - 1 1 - x 2 if 0 < x < 1 (ii) sin - 1 x = cos - 1 1 - x 2 if 0 < x < 1 (iii) cos - 1 x = π + tan - 1 1 - x 2 x if - 1 < x < 0

(i) Let cos - 1 ( x ) = θ , where θ ∈ [ 0 , π ] ⇒ cos ( cos - 1 ( x ) ) = cos θ ⇒ cos θ = x , x ∈ [ - 1 , 1 ] Now, sin - 1 1 - x 2 = sin - 1 sin θ = sin - 1 ( sin θ ) ∵ sin θ ≥ 0 , for θ ∈ [ 0 , π ] So, sin - 1 1 - x 2 = sin - 1 sin θ = θ , if 0 ≤ θ ≤ π 2 π - θ , if π 2 < θ ≤ π = cos - 1 x , if x ∈ [ 0 , 1 ] π - cos - 1 x , if x ∈ [ - 1 , 0 ) ⇒ sin - 1 1 - x 2 = cos - 1 x , if x ∈ ( 0 , 1 ) (ii) Let sin - 1 x = θ , where θ ∈ - π 2 , π 2 ⇒ sin ( sin - 1 x ) = sin θ ⇒ x = sin θ , where x ∈ [ - 1 , 1 ] Now, cos - 1 1 - x 2 = cos - 1 cos θ = cos - 1 ( cos θ ) ∵ cos θ ≥ 0 , for θ ∈ [ - π 2 , π 2 ] So, cos - 1 1 - x 2 = cos - 1 ( cos θ ) = - θ , if θ ∈ [ - π 2 , 0 ) θ , if θ ∈ 0 , π 2 = - sin - 1 ( x ) , if x ∈ [ - 1 , 0 ) sin - 1 x , if x ∈ [ 0 , 1 ) Hence, sin - 1 x = cos - 1 1 - x 2 , when x ∈ ( 0 , 1 ) (iii) Let cos - 1 x = θ , where θ ∈ [ 0 , π ] ⇒ cos ( cos - 1 x ) = cos θ ⇒ x = cos θ , where θ ∈ [ 0 , π ] Now, tan - 1 1 - x 2 x = tan - 1 sin θ cos θ = tan - 1 tan θ , ∵ tan θ ≥ 0 , where θ ∈ [ 0 , π 2 ) So, tan - 1 1 - x 2 x = tan - 1 tan θ = θ , if θ ∈ [ 0 , π 2 ) θ - π , if θ ∈ ( π 2 , π ] = cos - 1 x , when x ∈ [ 0 , 1 ) - π + cos - 1 x , when x ∈ [ - 1 , 0 ) ⇒ - π + cos - 1 x = tan - 1 1 - x 2 x , when x ∈ [ - 1 , 0 ) ⇒ cos - 1 ( x ) = π + tan - 1 1 - x 2 x

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 14 - Inverse Trigonometric Functions>Exercise-2>Q 22

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Earn 100

$x \in (\frac{- π}{2}, \frac{π}{2}), {cosec}^{- 1} (cosec x)$ is

100% studentsanswered this correctly

Important Questions on Inverse Trigonometric Functions

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Alpha Question Bank for Engineering: Mathematics>Chapter 14 - Inverse Trigonometric Functions>Exercise-3>Q 2

The value of

\cot (\sum_{n = 1}^{23} \cot^{- 1} (1 + \sum_{k = 1}^{n} 2 k))

is

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Alpha Question Bank for Engineering: Mathematics>Chapter 14 - Inverse Trigonometric Functions>Exercise-3>Q 5

If

α = 3 \sin^{- 1} (\frac{6}{11})

and

β = 3 \cos^{- 1} (\frac{4}{9})

where the inverse trigonometric functions take only the principal values, then the correct option(s) is(are)

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Alpha Question Bank for Engineering: Mathematics>Chapter 14 - Inverse Trigonometric Functions>Exercise-3>Q 8

Let

\tan^{- 1} y = \tan^{- 1} x + \tan^{- 1} (\frac{2 x}{1 - x^{2}})

, where

|x| < \frac{1}{\sqrt{3}}

. Then a value of

y

is,

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Alpha Question Bank for Engineering: Mathematics>Chapter 14 - Inverse Trigonometric Functions>Exercise-3>Q 9

If

\cos^{- 1} (\frac{2}{3 x}) + \cos^{- 1} (\frac{3}{4 x}) = \frac{π}{2} (x > \frac{3}{4})

, then

x

, is equal to.

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Alpha Question Bank for Engineering: Mathematics>Chapter 14 - Inverse Trigonometric Functions>Exercise-3>Q 10

The value of

\cot (\sum_{n = 1}^{19} \cot^{- 1} (1 + \sum_{p = 1}^{n} 2 p))

is.

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Alpha Question Bank for Engineering: Mathematics>Chapter 14 - Inverse Trigonometric Functions>Exercise-3>Q 11

For any positive integer

n

, define

f_{n} : (0, \infty) \to R

as

f_{n} (x) = \sum_{j = 1}^{n} \tan^{- 1} (\frac{1}{1 + (x + j) (x + j - 1)})

for all

x \in (0, \infty)

(Here, the inverse trigonometric function

\tan^{- 1} x

assumes values in

(- \frac{π}{2}, \frac{π}{2})

)
Then, which of the following statement(s) is (are) TRUE?

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Alpha Question Bank for Engineering: Mathematics>Chapter 14 - Inverse Trigonometric Functions>Exercise-4>Q 1

The range of the function

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

where

[.]

, is the greatest integer function, is

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Alpha Question Bank for Engineering: Mathematics>Chapter 14 - Inverse Trigonometric Functions>Exercise-4>Q 2

$\tan^{- 1} (\tan θ) = \{\begin{array}{l} π + θ, - \frac{3 π}{2} < θ < - \frac{π}{2} \\ θ, - \frac{π}{2} < θ < \frac{π}{2} - π + θ, \frac{π}{2} < θ < \frac{3 π}{2} \end{array}, \sin^{- 1} (\sin θ) = \{\begin{cases} - π - θ, - \frac{3 π}{2} < θ < - \frac{π}{2} \\ θ, - \frac{π}{2} \leq θ \leq \frac{π}{2} π - θ, \frac{π}{2} < θ \leq \frac{3 π}{2} \end{cases}$

$\cos^{- 1} (\cos θ) = \{\begin{aligned} - θ, & - π \leq θ < 0 \\ θ, & 0 \leq θ \leq π \\ 2 π - θ, & π < θ \leq 2 π \end{aligned}$

Based on the above results, prove each of the following:

(i) $\cos^{- 1} x = \sin^{- 1} \sqrt{1 - x^{2}}$ if $0 < x < 1$
(ii) $\sin^{- 1} x = \cos^{- 1} \sqrt{1 - x^{2}}$ if $0 < x < 1$
(iii) $\cos^{- 1} x = π + \tan^{- 1} \frac{\sqrt{1 - x^{2}}}{x}$ if $- 1 < x < 0$