The set for which $2 {cos}^{- 1} x = {cos}^{- 1} (2 x^{2} - 1)$ is valid is

x \in [0, 1]

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

x \in (0, 1)

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

x \in [0, 1)

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

x \in (0, 1]

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

Open in App

Solution

The correct option is B $x \in [0, 1]$
$2 {cos}^{- 1} x = {cos}^{- 1} (2 x^{2} - 1)$
Range of ${cos}^{- 1} f (x)$ is $[0, π]$
$0 \leq {cos}^{- 1} f (x) \leq π$
$0 \leq {cos}^{- 1} (2 x^{2} - 1) \leq π$
$0 \leq 2 {cos}^{- 1} x \leq π$
then $0 \leq {cos}^{- 1} x < \frac{π}{2}$
$x \in [0, 1]$
$D$ is correct

Suggest Corrections

Similar questions

$t a n^{- 1} \sqrt{x} = \frac{1}{2} c o s^{- 1} (\frac{1 - x}{1 + x}), x ϵ [0, 1] .$

Q. Prove:

{tan}^{- 1} \sqrt{x} = \frac{1}{2} {cos}^{- 1} (\frac{1 - x}{1 + x}), x \in [0, 1]

Q. Show that

{tan}^{- 1} \sqrt{x}

\frac{1}{2} {cos}^{- 1} (\frac{1 - x}{1 + x}), x \in [0, 1]

$L e t f (x) = \frac{x - [x]}{1 + x - [x]}, x ϵ R, [] d e n t o e s t h e g r e a t e s t i n t e g e r f u n c t i o n . T h e n, t h e r a n g e o f f i s$

Q. Let

f (x) = \frac{x - [x]}{1 + x - [x]}, x ϵ R

, where [ x] denotes the greatest integer function. Then, the range of f is

Join BYJU'S Learning Program

The set for which 2cos−1x=cos−1(2x2−1) is valid is

The correct option is B x∈[0,1]2cos−1x=cos−1(2x2−1)Range of cos−1f(x) is [0,π]0≤cos−1f(x)≤π0≤cos−1(2x2−1)≤π0≤2cos−1x≤πthen 0≤cos−1x<π2x∈[0,1]D is correct

The set for which $2 {cos}^{- 1} x = {cos}^{- 1} (2 x^{2} - 1)$ is valid is