Consider the function ^@ f(x) = \dfrac{ 1 }{ 2^x + \sqrt{ 2 } }. | Math Question and Answer

Answer:

^@ 6 ^@

Step by Step Explanation:

Given, ^@ f(x) = \dfrac{ 1 }{ 2^x + \sqrt{ 2 } } ^@
Here, ^@ f(-5) ^@ can be written as^@ f(1 -6). ^@
Similarly, ^@ f(-4) = f(1-5), f(-3) = f(1-4), \space\space\space\space\space\space\space\space \ldots f(0) = f(1-1) ^@
We have
^@ \begin{align} & f(x) = \dfrac{ 1 }{ 2^x + \sqrt{ 2 } } \\ \implies & f(1-x) = \dfrac{ 1 }{ 2^{ 1-x } + \sqrt{ 2 } } \\ \implies & f(x) + f(1 - x) = \dfrac{ 1 } { 2^{ x } + \sqrt{ 2 } } + \dfrac{ 1 }{ 2^{ 1-x } + \sqrt{ 2 } } \\ \implies & f(x) + f(1 - x) = \dfrac{ \sqrt{ 2 } } { \sqrt{ 2 } \left( 2^{ x } + \sqrt{ 2 } \right) } + \dfrac{ 2^x }{ 2^x \left( 2^{ 1-x } + \sqrt{ 2 } \right) } \\ \implies & f(x) + f(1 - x) = \dfrac{ \sqrt{ 2 } } { 2^{ x } \sqrt{ 2 } + 2 } + \dfrac{ 2^x }{ 2 + 2^x \sqrt{ 2 } } \\ \implies & f(x) + f(1 - x) = \dfrac{ \sqrt{ 2 } + 2^x } { 2 + 2^{ x } \sqrt{ 2 } } \\ \implies & f(x) + f(1 - x) = \dfrac{ \sqrt{ 2 } + 2^x } { \sqrt{ 2 } \left( \sqrt{ 2 } + 2^{ x } \right) } \\ \implies & f(x) + f(1 - x) = \dfrac{ 1 } { \sqrt{ 2 } } \\ \end{align} ^@
Now,
^@ \begin{align} & \sqrt{ 2 } \left[ f(-5) + f(-4) + f(-3) + f(-2) + f(-1) + f(0) + f(1) + f(2) + f(3) + f(4) + f(5) + f(6) \right] \\ \implies & \sqrt{ 2 } \left[ \{ f(6) + f(1-6) \} + \{ f(5) + f(1-5) \} + \ldots + \{ f(2) + f(1-2) \} + \{ f(1) + f(1 -1) \} \right] \\ \implies & \sqrt{ 2 } \left[ \dfrac{ 6 }{ \sqrt{ 2 } } \right] \\ \implies & 6 \\ \end{align} ^@
Hence, the value of the given expression is ^@ 6. ^@

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