If $x = \left( {\frac{{1 + i}}{2}} \right),\;\;\left( {{\rm{where}}\,\;i = \sqrt { - 1} } \right)$, then the expression $2{x^4} - 2{x^2} + x + 3$ equals

$\frac{{\left( {3 + i} \right)}}{2}$

$\frac{{\left( {3 - i} \right)}}{2}$

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

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Application +4 | -1

x = \left( {\frac{{1 + i}}{2}} \right),\;\;\left( {\mathrm{where}\,\;i = \sqrt { - 1} } \right)

, then the expression

2{x^4} - 2{x^2} + x + 3

equals

View answer and explanation

Correct option: A

3 - \left( {\frac{i}{2}} \right)

Option A (

3 - \left( {\frac{i}{2}} \right)

) is correct because it is obtained by substituting the value of x and simplifying the expression using the given condition.

Practice All Chapter MCQs →

{\left| {1 - i} \right|^x} = {2^x}

|z|=1

z \neq \pm 1

z, \frac{z}{1-z^2}

\left|z_r-r\right| \leq r, \forall r=1,2,3, \ldots, n

\left|\sum\limits_{r=1}^n z_r\right|

(z+a b)^3=a^3

a \neq 0

\alpha, \beta

\gamma

|\alpha-\beta|

{z^2} + {\left| z \right|^2} = 0,

z \in C

2x = - 1 + \sqrt 3 i

{\left( {1 - {x^2} + x} \right)^6} - {\left( {1 - x + {x^2}} \right)^6}