### CSIR JUNE 2011 PART B QUESTION 26 SOLUTION

Let $\zeta$ be the primitive fifth root of unity. Define $$A = \begin{bmatrix}\zeta^{-2}&0&0&0&0\\0&\zeta^{-1}&0&0&0\\ 0&0&0&0&0\\ 0&0&0&\zeta^{1}&0\\0&0&0&0&\zeta^{2}\end{bmatrix}.$$ For a vector $v = (v_1,v_2,v_3,v_4,v_5) \in \Bbb R^5$, define $|v|_A = \sqrt{|vAv^t|}$ where $v^t$ is the transpose of $v$. If $w = (1,-1,1,1,-1)$, then $|w|_A$ equals?1)0, 2)1, 3)-1, 4)2.

Solution:
We have $wA = (\zeta^{-2},-\zeta^{-1},0, \zeta^1,-\zeta^2)$ and $wAw^t = (\zeta^{-2},-\zeta^{-1},0, \zeta^1,-\zeta^2)\begin{bmatrix}1\\-1\\1\\1\\-1\end{bmatrix}=\zeta^{-2}+\zeta^{-1}+0+\zeta^{1}+\zeta^2 = \zeta^3 + \zeta^4+\zeta^1+\zeta^2=-1$. Since for a primitive root $1+\zeta+\zeta^2+\zeta^3+\zeta^4=0$. So option 3 is correct.
### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft...