Mathematics Olympiad ~ Vinod Sing, Kolkata $Problem$ #1  If $A$ and $B$ are different matrices satisfying $A^3 = B^3$ and $A^2B = B^2A$, find $det(A^2+B^2)$  Since $A$ and $B$ are different matrices $A-B \neq O$, Now $(A^2+B^2)(A-B) = A^3-A^2B+B^2A-B^3$  =$O$ since $A^3 = B^3 and A^2B = B^2A$  This shows that $(A^2+B^2)$ has a zero divisor, so it is not invertible hence $det(A^2+B^2) = 0$  $Problem$ #2  Prove that $\mathbf{\prod_{n=2}^{\infty} \left( 1 - \frac{1}{n^2} \right) = \frac{1}{2}}$  $\prod_{n=2}^{N} \left( 1 - \frac{1}{n^2} \right) = \prod_{n=2}^{N} \left( 1 - \frac{1}{n} \right) \prod_{n=2}^{N} \left( 1 + \frac{1}{n} \right)$  $= \prod_{n=2}^{N} \left( \frac{n-1}{n} \right) \prod_{n=2}^{N} \left( \frac{n+1}{n} \right) = \frac{1}{2}\frac{2}{3}\dots\frac{N-1}{N} \frac{3}{2}\frac{4}{3}\dots\frac{N+1}{N} = \frac{N+1}{2N}$  Letting $N \to$ to $\infty$, we see that $\mathbf{\prod_{n=2}^{\infty} \left( 1 - \frac{1}{n^2} \right) = \frac{1+\frac{1}{N}}{2} = \frac{1}{2}}$  $Problem$ #3  Find all real numbers $x$ for which $\sqrt{3-x}-\sqrt{x+1} > \frac{1}{2}$  Let $f(x)$ $=\sqrt{3-x}-\sqrt{x+1}$. First note that $f(x)$ is defined for $-1 \leq x \leq 3$  $f'(x) = \frac{-1}{2\sqrt{3-x}} - \frac{1}{\sqrt{x+1}} = - \big(\frac{1}{2\sqrt{3-x}} + \frac{1}{\sqrt{x+1}}\big) < 0 \Rightarrow f(x)$ is strictly decreasing  Now $f(-1) = 2 > \frac{1}{2}$ and $f(3) = -2 < \frac{1}{2}$ Since $f(x)$ is continuous, $\exists$ at least one x $\in$ ${(-1,3)}$ suct that $f(x) = \frac{1}{2}$  $f(x) = \frac{1}{2} \Rightarrow \sqrt{3-x}-\sqrt{x+1} = \frac{1}{2} \Rightarrow 64x^2-128x+33 = 0 \Rightarrow x = 1 \pm \frac{\sqrt{31}}{8}$ but $x = 1 + \frac{\sqrt{31}}{8}$ does not satisfy $\sqrt{3-x}-\sqrt{x+1} = \frac{1}{2}$ Check yourself! So the only solution is $x = 1 - \frac{\sqrt{31}}{8}$  Since $f(x)$ is strictly decreasing, the given inequality is true for $x \in {[-1,1 - \frac{\sqrt{31}}{8}\big)}$ 

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