If cos h-1 x = 2 loge ( √2 + 1 ), then x =

Question

If cos h-1 x = 2 loge ( √2 + 1 ), then x = (A) 1 (B) 2 (C) 4 (D) 3

Tardigrade Admin · Accepted Answer

We have, cosh -1 x=2 log θ(√2+1) ⇒ log e(x+√x2-1)= log e(√2+1)2 ⇒ x+√x2-1=(√2+1)2 ⇒ x+√x2-1=3+2 √2=3+√8 On comparing rational and irrational part, we get x=3

Q. If $cos h^{- 1} x = 2 lo g_{e} (2 + 1)$ , then x =

1

2

4

3

We have, $cosh^{- 1} x = 2 lo g_{θ} (2 + 1)$
$\Rightarrow lo g_{e} (x + x^{2} - 1) = lo g_{e} (2 + 1)^{2}$
$\Rightarrow x + x^{2} - 1 = (2 + 1)^{2}$
$\Rightarrow x + x^{2} - 1 = 3 + 22 = 3 + 8$
On comparing rational and irrational part, we get
$x = 3$

Q. If cosh−1x=2loge​(2​+1), then x =

1

2

4

3

We have, cosh−1x=2logθ​(2​+1) ⇒loge​(x+x2−1​)=loge​(2​+1)2 ⇒x+x2−1​=(2​+1)2 ⇒x+x2−1​=3+22​=3+8​ On comparing rational and irrational part, we get x=3

Q. If $cos h^{- 1} x = 2 lo g_{e} (2 + 1)$ , then x =

We have, $cosh^{- 1} x = 2 lo g_{θ} (2 + 1)$
$\Rightarrow lo g_{e} (x + x^{2} - 1) = lo g_{e} (2 + 1)^{2}$
$\Rightarrow x + x^{2} - 1 = (2 + 1)^{2}$
$\Rightarrow x + x^{2} - 1 = 3 + 22 = 3 + 8$
On comparing rational and irrational part, we get
$x = 3$