Question

verify my answers :)

Answer 1

q.1)
\[I=\int\limits_{-1}^{1}x-[x]dx\]
[x] is the greatest integer function
Im getting answer 1

q.2)
\[\Delta=\left[\begin{matrix}loga_{n} & loga_{n+1} & loga_{n+2} \\ loga_{n+3} & loga_{n+4} & loga_{n+5} \\ loga_{n+6} & loga_{n+7} & loga_{n+8} \end{matrix}\right]\]
This is a determinant not matrix ok and a1 a2 a3 etc. are in G.P.
I'm getting answer as 0

Answer 2

\[\begin{align}
I&=\int\limits_{-1}^{1}x-[x]dx\\~\\
&=\int\limits_{-1}^{1}x \,-\int\limits_{-1}^{1}[x]\,dx\\~\\
&=0-\int\limits_{-1}^{1}[x]\,dx

\end{align}\]

In interval [-1, 0), the value of [x] is -1 and in interval [0,1), the value of [x] is 0
therefore,
\[\begin{align}
-\int\limits_{-1}^{1}[x]\,dx&=-\int\limits_{-1}^{0}[x]\,dx-\int\limits_{0}^{1}[x]\,dx\\~\\
&=-(-1)-0\\~\\
&=1

\end{align}\]

Answer 3

this is what I did:
\[I=\int\limits_{-1}^{0} (x+1) dx + \int\limits_{0}^{1}dx\]
from -1 to 0 [x]=-1 so x-(-1)=x+1
from 0 to 1 [x]=0
and got answer 1

Answer 4

0->1 it should be x right

Answer 5

yeah sorry i entered wrong it is \[I=\int\limits_{-1}^{0} (x+1)dx+\int\limits_{0}^{1}xdx\]
I got answer 1

Answer 6

looks neat

Answer 7

Let \(r\) be the common ratio of the GP, then \(\log a_{n+1}-\log a_{n}=\log \frac{a_{n+1}}{a_n}=\log r\)


C3-C2 and C2-C1 gives
\[\begin{align}
&\begin{vmatrix}\log a_{n} & \log r & \log r\\ \log a_{n+3} & \log r & \log r \\ \log a_{n+6} & \log r & \log r\end{vmatrix}\\~\\
\end{align}\]

since the last two columns are same, the determinant is 0

Answer 8

that's exactly what I did yippie