Test Index

AP EAMCET Engineering 20 Apr 2019 Shift 2 Paper

Section: Mathematics

© examsnet.com

Question : 10 of 160

Marks: +1, -0

If

(1+x)^{n}

a_0+a_1x+a_2x^2

a_nx^n

a_0-a_2+a_4-a_6

\frac{n\pi}{4}

, then k =

$2^{n}$
$2^{2n}$
$\frac{2^{n}}{2}$
$2^{\frac{n}{2}}$

Solution:

Consider the expression

(1+x)^{n}=a_{0}+a_{1}x+a_{2}x^{2}+\dots+a_{n}x^{n}

Substitute

x

for

i

in the above expression.

(1+i)^{n}=\left[\begin{array}{l}(a_{0}-a_{2}+a_{4}-a_{6}+\dots)+i(a_{1}-a_{3}+a_{5}-a_{7}+\dots)\end{array}\right]

\left[\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)\right]^{n}=\left[\begin{array}{l}(a_{0}-a_{2}+a_{4}-a_{6}+\dots)+i(a_{1}-a_{3}+a_{5}-a_{7}+\dots)\end{array}\right]

Compare the real and imaginary parts

2^{\frac{n}{2}}\cos\frac{n\pi}{4}=(a_{0}-a_{2}+a_{4}-a_{6}+\dots)

2^{\frac{n}{2}}\cos\frac{n\pi}{4}=k\cos\frac{n\pi}{4}

k=2^{\frac{n}{2}}

© examsnet.com

Go to Question:

Prev Question

Next Question