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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
