logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
0 votes

Find the value of $(a^2+\sqrt{a^2-1})^4+(a^2-\sqrt{a^2-1})^4$

$\begin{array}{1 1}(A)\;2(a^8+6a^6-5a^4-2a^2+1)\\(B)\;4\\(C)\;(a^8+6a^6-5a^4-2a^2+1)\\(D)\;a^5+6a^4-5a^3-2a+1\end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Put $a^2=x,y=\sqrt{a^2-1}$
$\therefore (x+y)^4=x^4+4C_1x^3y+4C_2x^2y^2+4C_3xy^3+4C_4y^4$------(1)
$ (x-y)^4=x^4-4C_1x^3y+4C_2x^2y^2-4C_3xy^3+4C_4y^4$------(2)
Adding (1) and (2)
$(x+y)^4+(x-y)^4=2[x^4+4C_2x^2y^2+4C_4y^4]$
$\Rightarrow 2[x^4+6x^2y^2+y^4]$
$\therefore (a^2+\sqrt{a^2-1})^4+(a^2-\sqrt{a^2-1})^4=2[(a^2)^4+6(a^2)^2(\sqrt{a^2-1})^2+(\sqrt{a^2-1})^4]$
$\Rightarrow 2[a^8+6a^4(a^2-1)+(a^2-1)^2]$
$\Rightarrow 2[a^8+6a^4(a^2-1)+a^4-2a^2+1]$
$\Rightarrow 2[a^8+6a^6-5a^4-2a^2+1]$
Hence (A) is the correct answer.
answered Jun 24, 2014 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...