If $i=\sqrt{-1}$, then $\frac{e^{x i}+e^{-x}}{2}=$

$x+\frac{x^5}{3 !}+\frac{x^3}{5 !}+\ldots x$

$i\left[x-\frac{x^3}{3 !}+\frac{x^3}{5 !}-\ldots \infty\right]$

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

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Understanding +4 | -1

i=\sqrt{-1}

, then

\frac{e^{x i}+e^{-x}}{2}=

View answer and explanation

Correct option: B

1-\frac{x^2}{2 !}+\frac{x^4}{4 !}-\ldots \infty

Option A (

1+\frac{{x^2}}{{2 !}}+\frac{{x^4}}{{4 !}}+\ldots \infty

) is incorrect. The correct answer should be

1-\frac{{x^2}}{{2 !}}+\frac{{x^4}}{{4 !}}-\ldots \infty

which represents

\cos x

. However, the provided option A is incorrect. The correct option should be B.

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}