A1. Exactly Rs. \(5000\).

A2. Exactly Rs. \(500\).

Consider the person who donated Rs. 500. Suppose the neighbor to the left donates \(500+x\). Then the one on the right donates \(500-x\). But continuing leftward, the amounts donated are \(500+2 x, 500+3 x, \ldots,\) forcing \(x\) to be \(0,\) since you come around to the neighbor to the right.

There are 201 integers.

The function is strictly decreasing as \(x\) goes from \(-\infty\) to \(\infty\). The range of the function is \((0,201.8)\). By continuity, all the integers in the range appear as values of \(f(x)\).

Notation.Let \(G\) and \(P\) denote the GDP and the population at the start of the period, respectively. Let \(x\) be the percentage change of the population.

The new per-capita GDP is then given by \(\frac{1.078 G}{(1+x) P}\). The percent increase in per-capita GDP is \(10 \%=0.1\). So we have \(\frac{1.078}{1+x}=1.1 .\) Solving for \(x\) we get:

Per-capita GDP is \(\frac{\text { GDP }}{\text { population }}\).

\[1+x=\frac{1.078}{1.1}=0.98\] So population decreased by \(2 \%\)

Put \(t=\sqrt[3]{x}\) to get: \begin{align} \left(t^{3}+4\right)&=(1+t)^{3} \\ 4&=t^{3}+3t^2+3t+1 \\ t^{2}+t-1&=0\\ \end{align} Solving for \(t\) and then \(x\) we get: \begin{align*} x &= \left( \frac{-1\pm\sqrt{5}}{2} \right)^{3} \\ &= -2\pm \sqrt{5} \end{align*}

The equation is of the form \(a^b=1\). This can happen only in three cases:
(i) Either \(a=1\), which means \(x^2-2x=1\). So \(x=1\pm\sqrt{2}\).
(ii) or \(a=-1\) and \(b\) is an even integer. So \(x=1\) works.
(iii) or \(b=0\) and \(a\neq 0\). Which happens when \(x=-3\) or \(1\). The value \(x=2\) makes \(a=b=0\)
So \(x=-3,1,1 \pm \sqrt{2}\) are the only solutions.

Let \(a=\sqrt[3]{6\sqrt{3}+10}\) and \(b=\sqrt[3]{6\sqrt{3}-10}\). We want to find the value of \(a-b\). We have the following: \begin{align} a^3 - b^3 &= 20 \\ ab &= \sqrt[3]{ (6\sqrt{3})^2-10^2 } = \sqrt[3]{108-100} = 2 \end{align} Substituting the values above in \( (a-b)^3 \) gives a cubic in \( (a-b) \). \begin{align} (a-b)^3 &= a^3 - b^3 - 3ab(a-b) \\ (a-b)^{3}&=20-6(a-b) \end{align} This cubic has one real root \(a-b=2\) and two complex roots. Hence, the value of the given expression is 2.

We will make use of the identity: \(\log_{x} a+\log_{x} b=\log_{x} a b\).

\begin{align} \frac{1}{1+\log_{a^{2} b}\left(\frac{c}{a}\right)}&=\frac{1}{\log_{a^{2} b}\left(a^{2} b\right)+\log_{a^{2} b}\left(\frac{c}{a}\right)} \\ &=\frac{1}{\log_{a^{2} b}\left(\frac{c}{a} \cdot a^{2} b\right)} \\ &=\frac{1}{\log_{a^{2} b}(a b c)}\\ &\\ &=\log_{a b c}\left(a^{2} b\right) \end{align}

Hence the given expression is:

\(\log_{a b c}\left(a^{2} b\right)+\log_{a b c}\left(b^{2} c\right)+\log_{a b c}\left(c^{2} a\right)\) \(=\log_{a b c}(a b c)^{3}\) \(=3\)

No-No-Yes-No.
Suppose \(A\) has nonzero determinant. Then for any \(\vec{b},\) we see that \(A \vec{x}=\vec{b}\) has the unique solution \(\vec{x}=A^{-1} \vec{b}\). If determinant of \(A\) is zero then we can make two cases:
(i) If \(A\) is the zero matrix, then \(A \vec{x}=\vec{b}\) has infinitely many solutions for \(\vec{b}=\overrightarrow{0}\) and no solutions otherwise.

(ii) If \(A\) is nonzero then it is easy to see that we are solving two equations in two variables whose left hand sides are proportional. So if the two right hand constants that make up \(\vec{b}\) are in the same proportion, then we will have infinitely many solutions (because one of the variables can be arbitrary). If the constants are not in the same proportion, then the two equations will be inconsistent and we will have no solutions.

\[\frac{\left(2^{x}\right)^{3}+\left(3^{x}\right)^{3}}{\left(2^{x}\right)^{2} \cdot 3^{x}+\left(3^{x}\right)^{2} \cdot 2^{x}}=\frac{7}{6}\]

Putting \(a=2^x\) and \(b=3^x\) in the above equation, we get:

\[ \frac{a^3 + b^3}{a^2b+ab^2} = \frac{7}{6} \]

\[ \frac{ (a+b)(a^2+b^2-ab) }{ab(a+b)} = \frac{7}{6} \]

\[ \frac{ (a+b)(a^2+b^2-ab) }{ab(a+b)} = \frac{7}{6} \]

\[ 6a^2 - 13ab + 6b^2 = 0 \]

\[ 6t^{2}-13t+6=0 \text{ where } t=a/b \]

The above quadratic has \(t = 3/2\) and \(t=2/3\) as the roots. So \(x=\pm 1\) satisfies the original equation.