If x + √ {({x^2} + 1)} = a, then x =

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Basic of Logarithms

easy

If $x + \sqrt {({x^2} + 1)} = a,$ then $x =$

A

${1 \over 2}(a + 1/a)$

B

${1 \over 2}(a - 1/a)$

C

$(a + {a^{ - 1}})$

D

None of these

Solution

(b) $x + \sqrt {{x^2} + 1} = a \Rightarrow \sqrt {{x^2} + 1} = a – x$

$ \Rightarrow $${x^2} + 1 = {(a – x)^2} = {x^2} – 2ax + {a^2}$

$ \Rightarrow $$x = {{1 – {a^2}} \over { – 2a}} = {{{a^2} – 1} \over {2a}} = {1 \over 2}\left( {a – {1 \over a}} \right)$.

Standard 11

Mathematics

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