Inverse trigonometric functions have restricted domain and range- - and - denotes greatest integer function and modulus function respectively- thenInverse Trigonometric-Domain-Rangell Function-llthlineend array The solution set for - -1 x-2-6- -1 x-9 - 0- isA- - 3- 2B- - 2- -C- - 3-D- - 3- -

The correct option is C ( ∞, cot3][cot^ 1x]^2 6[cot^ 1x]+9≤ 0 ⇒ ([cot^ 1x] 3)^2 ≤ 0 ⇒ [cot^ 1x]=3 ⇒ 3 ≤ cot^ 1x lt; π ⇒ cot3 ≥ x gt; ∞ ⇒ x ϵ ( ∞, cot3] ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Finding Inverse Using Elementary Transformations

Inverse trigo...

Question

Inverse trigonometric functions have restricted domain and range, [.] and |.| denotes greatest integer function and modulus function respectively, then
$\begin{matrix} I n v e r s e T r i g o n o m e t r i c & D o m a i n & R a n g e F u n c t i o n y = s i n^{- 1} x & x ϵ [- 1, 1] & y ϵ [\frac{- π}{2}, \frac{π}{2}] y = c o s^{- 1} x & x ϵ [- 1, 1] & y ϵ [0, π] y = t a n^{- 1} x & x ϵ R & y ϵ (\frac{- π}{2}, \frac{π}{2}) y = c o t^{- 1} x & x ϵ R & y ϵ (0, π) y = s e c^{- 1} x & x ϵ R - (- 1, 1) & y ϵ [0, π] - {\frac{π}{2}} y = c o s e c^{- 1} x & x ϵ R - (- 1, 1) & y ϵ [\frac{- π}{2}, \frac{π}{2}] \end{matrix}$
The solution set for $[c o t^{- 1} x]^{2} - 6 [c o t^{- 1} x] + 9 \leq 0$ , is

(- \infty, c o t 3]

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

[c o t 3, c o t 2)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[c o t 3, \infty]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[c o t 2, \infty]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $(- \infty, c o t 3]$
$[c o t^{- 1} x]^{2} - 6 [c o t^{- 1} x] + 9 \leq 0 \Rightarrow ([c o t^{- 1} x] - 3)^{2} \leq 0 \Rightarrow [c o t^{- 1} x] = 3 \Rightarrow 3 \leq c o t^{- 1} x < π \Rightarrow c o t 3 \geq x > - \infty \Rightarrow x ϵ (- \infty, c o t 3]$

Suggest Corrections

Similar questions

Q. Inverse trigonometric functions have restricted domain and range, [.] and |.| denotes greatest integer function and modulus function respectively, then

\begin{matrix} I n v e r s e T r i g o n o m e t r i c & D o m a i n & R a n g e F u n c t i o n y = s i n^{- 1} x & x ϵ [- 1, 1] & y ϵ [\frac{- π}{2}, \frac{π}{2}] y = c o s^{- 1} x & x ϵ [- 1, 1] & y ϵ [0, π] y = t a n^{- 1} x & x ϵ R & y ϵ (\frac{- π}{2}, \frac{π}{2}) y = c o t^{- 1} x & x ϵ R & y ϵ (0, π) y = s e c^{- 1} x & x ϵ R - (- 1, 1) & y ϵ [0, π] - {\frac{π}{2}} y = c o s e c^{- 1} x & x ϵ R - (- 1, 1) & y ϵ [\frac{- π}{2}, \frac{π}{2}] \end{matrix}

The equation

s i n^{- 1} x = | x - a |

will have at least one solution, if

The complete solution set of the inequality $[c o t^{- 1} x]^{2} - 6 [c o t^{- 1} x] + 9 \leq 0,$ where $[.]$ denotes greatest integer function is

Q. Complete solution set

[{cot}^{- 1} x] + 2 [{tan}^{- 1} x] = 0

, where

[.]

denotes the greatest integer function, is

The complete solution set of the inequality $[c o t^{- 1} x]^{2} - 6 [c o t^{- 1} x] + 9 \leq 0,$ where $[.]$ denotes greatest integer function is

Q. Domain and range of trigonometric functions.

Join BYJU'S Learning Program

The correct option is A (−∞,cot3][cot−1x]2−6[cot−1x]+9≤0⇒([cot−1x]−3)2≤0⇒[cot−1x]=3⇒3≤cot−1x<π⇒cot3≥x>−∞⇒xϵ(−∞,cot3]

The correct option is A $(- \infty, c o t 3]$
$[c o t^{- 1} x]^{2} - 6 [c o t^{- 1} x] + 9 \leq 0 \Rightarrow ([c o t^{- 1} x] - 3)^{2} \leq 0 \Rightarrow [c o t^{- 1} x] = 3 \Rightarrow 3 \leq c o t^{- 1} x < π \Rightarrow c o t 3 \geq x > - \infty \Rightarrow x ϵ (- \infty, c o t 3]$