Question

Check whether the following function is odd or even or neither
$f(x) = log(x + \sqrt{1 + {x}^{2} } )$
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Answer 1

GIVEN :–

• A function $\bf f(x) = \log(x + \sqrt{1 + {x}^{2} } )$

TO FIND :–

• Type of function odd or even ?

SOLUTION :–

$\\ \implies \bf f(x) = \log(x + \sqrt{1 + {x}^{2} } )$

• Now replace x → -x ,

$\\ \implies \bf f( - x) = \log \left( - x + \sqrt{1 + { (- x)}^{2} } \right)$

$\\ \implies \bf f( - x) = \log \left( - x + \sqrt{1 + {x}^{2} } \right)$

• Now add f(x) & f(-x) –

$\\ \implies \bf f(x) + f( - x) =\log \left(x + \sqrt{1 + {x}^{2} } \right) + \log \left( - x + \sqrt{1 + {x}^{2} } \right)$

$\\ \implies \bf f(x) + f( - x) =\log \left[ \left(x + \sqrt{1 + {x}^{2} } \right) \left( - x + \sqrt{1 + {x}^{2} } \right) \right]$

$\\ \implies \bf f(x) + f( - x) =\log \left[ \left(\sqrt{1 + {x}^{2}} + x\right) \left(\sqrt{1 + {x}^{2} } - x \right)\right]$

$\\ \implies \bf f(x) + f( - x) =\log \left[ \left(\sqrt{1 + {x}^{2}}\right)^{2} - \left(x\right)^{2} \right]$

$\\ \implies \bf f(x) + f( - x) =\log(1 + {x}^{2} - x^{2})$

$\\ \implies \bf f(x) + f(-x) =\log(1)$

$\\ \implies \bf f(x) + f(-x) =0$

• We know that –

☆ f(x) + f(-x) = 0 , it mean f(x) is an odd function.

☆ f(x) - f(-x) = 0 , it mean f(x) is an even function.

• Hence , Given function f(x) is an odd function.