The domain of cosh-1 5x is

Question

The domain of cosh-1 5x is (A) R (B) [0, ∞) (C) ( (1/5) , ∞ ) (D) [ (1/5) , ∞ )

Tardigrade Admin · Accepted Answer

beginarrayl cosh -1 x= log [x+√x2-1] cosh -1 5 x=(1/5) log [x+√x2-1] endarray Range of log is from 1 to ∞ as x>0 for cosh -1 x to be defined ∴ text range is =[(1/5), ∞]

Q. The domain of $cos h^{- 1} 5 x$ is

R

$[0, \infty)$

$(\frac{1}{5}, \infty)$

$[\frac{1}{5}, \infty)$

$< b r / > < b r / > cosh^{- 1} x = lo g [x + x^{2} - 1] < b r / > cosh^{- 1} 5 x = \frac{1}{5} lo g [x + x^{2} - 1] < b r / > < b r / >$
Range of $lo g$ is from 1 to $\infty$ as $x > 0$ for $cosh^{- 1} x$ to be defined
$< b r / > ∴ range is = [\frac{1}{5}, \infty] < b r / >$

Q. The domain of cosh−15x is

R

[0,∞)

(51​,∞)

[51​,∞)

<br/><br/>cosh−1x=log[x+x2−1​]<br/>cosh−15x=51​log[x+x2−1​]<br/>​<br/> Range of log is from 1 to ∞ as x>0 for cosh−1x to be defined <br/>∴ range is =[51​,∞]<br/>

Q. The domain of $cos h^{- 1} 5 x$ is

$[0, \infty)$

$(\frac{1}{5}, \infty)$

$[\frac{1}{5}, \infty)$

$< b r / > < b r / > cosh^{- 1} x = lo g [x + x^{2} - 1] < b r / > cosh^{- 1} 5 x = \frac{1}{5} lo g [x + x^{2} - 1] < b r / > < b r / >$
Range of $lo g$ is from 1 to $\infty$ as $x > 0$ for $cosh^{- 1} x$ to be defined
$< b r / > ∴ range is = [\frac{1}{5}, \infty] < b r / >$