Question

Consider the function $g\left( x \right)$ defined as $g\left( x \right).\left( {{x^{{2^{2008}} - 1}} - 1} \right) = \left( {x + 1} \right)\left( {{x^2} + 1} \right)\left( {{x^4} + 1} \right).......\left( {{x^{{2^{2007}}}} + 1} \right) - 1$. The value of $g\left( 2 \right)$ equals to…

Answer 1

We have the equation and to find the value we can proceed with the right-hand side by multiplying and dividing with $\left( {x - 1} \right)$ and then solve the right-hand side equation for the value of $x$. In this way, we can find the value for the given function.

Formula used:
We have $\left( {{x^{{2^1}}} - 1} \right)\left( {{x^{{2^1}}} + 1} \right)$
And on expanding and solving more we can write it as
$\Rightarrow \left( {{x^{{2^2}}} - 1} \right)$

Complete step by step solution:
We have the equation given as-
$g\left( x \right).\left( {{x^{{2^{2008}} - 1}} - 1} \right) = \left( {x + 1} \right)\left( {{x^2} + 1} \right)\left( {{x^4} + 1} \right).......\left( {{x^{{2^{2007}}}} + 1} \right) - 1$
Taking the RHS, we get
$\Rightarrow \left( {x + 1} \right)\left( {{x^2} + 1} \right)\left( {{x^4} + 1} \right).......\left( {{x^{{2^{2007}}}} + 1} \right) - 1$
Now multiplying dividing the above equation with$\left( {x - 1} \right)$, we get
$\Rightarrow \dfrac{{\left( {x - 1} \right)}}{{\left( {x - 1} \right)}} \times \left( {x + 1} \right)\left( {{x^2} + 1} \right)\left( {{x^4} + 1} \right).......\left( {{x^{{2^{2007}}}} + 1} \right) - 1$
Now on expanding the equation, we get
$\Rightarrow \dfrac{{\left( {{x^{{2^1}}} - 1} \right)\left( {{x^{{2^1}}} + 1} \right)\left( {{x^{{2^2}}} + 1} \right)\left( {{x^{{2^3}}} + 1} \right).....\left( {{x^{{2^{2007}}}} + 1} \right)}}{{\left( {x - 1} \right)}} - 1$, let suppose it equation $1$
As we know from the formula that
$\Rightarrow \left( {{x^{{2^1}}} - 1} \right)\left( {{x^{{2^1}}} + 1} \right)$, can be written as
$\Rightarrow {\left( {{x^{{2^1}}}} \right)^2} - 1$
And on solving the equation, we get
$\Rightarrow {x^{{2^1} \times 2}} - 1$
And it can also be written as
$\Rightarrow \left( {{x^{{2^2}}} - 1} \right)$
So on further expanding the equation$1$, we get
$\Rightarrow \dfrac{{\left( {{x^{{2^2}}} - 1} \right)\left( {{x^{{2^2}}} + 1} \right).....\left( {{x^{{2^{2007}}}} + 1} \right)}}{{\left( {x - 1} \right)}} - 1$
And on solving, it can be written as
$\Rightarrow g\left( x \right).\left( {{x^{{2^{2008}} - 1}} - 1} \right) = \dfrac{{x\left( {{x^{{2^{2008}} - 1}} - 1} \right)}}{{\left( {x - 1} \right)}}$
And from here, taking the $g\left( x \right)$one side then we will get
$\Rightarrow g\left( x \right) = \dfrac{x}{{x - 1}}$
Now since, we have to calculate for the values of$g\left( 2 \right)$, so on putting the$x = 2$, we get
$\Rightarrow g\left( 2 \right) = \dfrac{2}{{2 - 1}}$
And on solving it will be equal to
$\Rightarrow g\left( 2 \right) = 2$

Hence, $g\left( 2 \right) = 2$ will be the correct answer.

Note:
For solving this type of question, we have to do a lot of complicated solutions so we should always try to do the step-by-step process as it will help to eliminate the errors and will get through the solution easily. And also we have to remember some formulas which help in expanding the equation and make the calculation lot easier. So in this way we solve this type of problem easily.