Question

Please answer the above question with explaination ✍️

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Answer 1

QUESTION :–

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

$\\ \bf (A) \: \: even \: \: function$

$\\ \bf (B) \: \: odd \: \: function$

$\\ \bf (C) \: \: neither \: \: even \: \: nor \: \: odd$

ANSWER :–

GIVEN :–

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

TO FIND :–

• Nature of function = ?

SOLUTION :–

• A function is called odd function if f(-x) = -f(x) .

• A function is called even function if f(-x) = f(x) .

• Now put x → -x in given function –

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

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

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

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

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

• Now –

$\\ \bf \implies\:f( - x) = \cos \bigg( log \bigg( \dfrac{[\sqrt{1 + {x}^{2} } - x][\sqrt{1 + {x}^{2} } + x]}{[\sqrt{1 + {x}^{2} } + x]} \bigg) \bigg)$

$\\ \bf \implies\:f( - x) = \cos \bigg( log \bigg( \dfrac{[\sqrt{1 + {x}^{2} }]^{2} - {x}^{2} }{[\sqrt{1 + {x}^{2} } + x]} \bigg) \bigg)$

$\\ \bf \implies\:f( - x) = \cos \bigg( log \bigg( \dfrac{1 + {x}^{2} - {x}^{2} }{[\sqrt{1 + {x}^{2} } + x]} \bigg) \bigg)$

$\\ \bf \implies\:f( - x) = \cos \bigg( log \bigg( \dfrac{1}{[\sqrt{1 + {x}^{2} } + x]} \bigg) \bigg)$

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

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

$\\ \bf \implies\:f( - x) = f(x)$

• Hence , The function is a even function.

• Hence , option (A) is correct.

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✯ USED IDENTITY :–

$\\ \bf (1) \:\cos( - x) =\cos (x)$

$\\ \bf (1) \: log\bigg(\dfrac{1}{x}\bigg) =-log(x)$