If 2cos-1-1-x-2- -2- then x-10-1-21-2

The correct option is B 0Given equation is 2cos^ 1√( ( 1+x/2 ))=π/2 ⇒ cos^ 1√( ( 1+x/2 ))=π/4⇒ cosπ/4=√(1+x)/√(2) ⇒1/√(2)=√(1+x)/√(2)⇒ 1=√(1+x)⇒ x=0. ...

if 2cos^-1√(1...

Question

if $2 c o s^{- 1} \sqrt{\frac{1 + x}{2}} = \frac{π}{2}$ , then x=

1

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

0

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

\frac{- 1}{2}

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

\frac{1}{2}

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

Open in App

Solution

The correct option is B $0$
Given equation is $2 c o s^{- 1} \sqrt{(\frac{1 + x}{2})} = \frac{π}{2}$
$\Rightarrow c o s^{- 1} \sqrt{(\frac{1 + x}{2})} = \frac{π}{4} \Rightarrow c o s \frac{π}{4} = \frac{\sqrt{1 + x}}{\sqrt{2}}$
$\Rightarrow \frac{1}{\sqrt{2}} = \frac{\sqrt{1 + x}}{\sqrt{2}} \Rightarrow 1 = \sqrt{1 + x} \Rightarrow x = 0$ .

Suggest Corrections

Similar questions

Q. Prove:

{tan}^{- 1} (\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}}) = \frac{π}{4} - \frac{1}{2} {cos}^{- 1} x, - \frac{1}{\sqrt{2}} \leq x \leq 1

[Hint:

p u t x = cos 2 θ]

Q. If

2 {cos}^{- 1} \sqrt{\frac{1 + x}{2}} = \frac{π}{2}

, then

x =

Q. if

2 c o s^{- 1} \sqrt{\frac{1 + x}{2}} = \frac{π}{2}

, then x=

Q. For

0 \leq {cos}^{- 1} x \leq π

and

- \frac{π}{2} \leq {sin}^{- 1} x \leq \frac{π}{2}

, the value of

cos ({sin}^{- 1} x + 2 {cos}^{- 1} x)

x = \frac{1}{5}

is:

Q. Show that

{cos}^{- 1} (2 x \sqrt{1 - x^{2}}) = (\frac{π}{2}) - 2 {cos}^{- 1} x, \frac{1}{\sqrt{2}} \leq x \leq 1

if 2cos−1√1+x2=π2, then x=

The correct option is B 0Given equation is 2cos−1√(1+x2)=π2 ⇒cos−1√(1+x2)=π4⇒cosπ4=√1+x√2 ⇒1√2=√1+x√2⇒1=√1+x⇒x=0.

if $2 c o s^{- 1} \sqrt{\frac{1 + x}{2}} = \frac{π}{2}$ , then x=