If a given equilateral triangle $\Delta$ of side length $a$ lies in the union of five equilateral triangles of side length $b$, show that there exist four equilateral triangles of side length $b$ whose union contains $\Delta$.

Suppose $f : \mathbb { R } \rightarrow \mathbb { R }$ is differentiable and $\left| f ^ { \prime } ( x ) \right| < \frac { 1 } { 2 }$ for all $x \in \mathbb { R }$. Show that for some $x _ { 0 } \in \mathbb { R } , f \left( x _ { 0 } \right) = x _ { 0 }$.

Consider a ball that moves inside an acute-angled triangle along a straight line, until it hits the boundary, which is when it changes direction according to the mirror law, just like a ray of light (angle of incidence $=$ angle of reflection). Prove that there exists a triangular periodic path for the ball, as pictured below.

9. Prove that the equation:
$$\arctan ( x ) + x ^ { 3 } + e ^ { x } = 0$$
has one and only one real solution.

8. If $\mathrm { p } ( \mathrm { x } ) = 51 \mathrm { x } ^ { 101 } - 2323 \mathrm { x } ^ { 100 } - 45 \mathrm { x } + 1035$, using Rolle's Theorem, prove that atleast one root lies between ( $45 ^ { 1 / 100 } , 46$ ). Sol. Let $\mathrm { g } ( \mathrm { x } ) = \int \mathrm { p } ( \mathrm { x } ) \mathrm { dx } = \frac { 51 \mathrm { x } ^ { 102 } } { 102 } - \frac { 2323 \mathrm { x } ^ { 101 } } { 101 } - \frac { 45 \mathrm { x } ^ { 2 } } { 2 } + 1035 \mathrm { x } + \mathrm { c }$ $= \frac { 1 } { 2 } \mathrm { x } ^ { 102 } - 23 \mathrm { x } ^ { 101 } - \frac { 45 } { 2 } \mathrm { x } ^ { 2 } + 1035 \mathrm { x } + \mathrm { c }$. Now $\mathrm { g } \left( 45 ^ { 1 / 100 } \right) = \frac { 1 } { 2 } ( 45 ) ^ { \frac { 102 } { 100 } } - 23 ( 45 ) ^ { \frac { 101 } { 100 } } - \frac { 45 } { 2 } ( 45 ) ^ { \frac { 2 } { 100 } } + 1035 ( 45 ) ^ { \frac { 1 } { 100 } } + \mathrm { c } = \mathrm { c }$ $\mathrm { g } ( 46 ) = \frac { ( 46 ) ^ { 102 } } { 2 } - 23 ( 46 ) ^ { 101 } - \frac { 45 } { 2 } ( 46 ) ^ { 2 } + 1035 ( 46 ) + \mathrm { c } = \mathrm { c }$. So $\mathrm { g } ^ { \prime } ( \mathrm { x } ) = \mathrm { p } ( \mathrm { x } )$ will have atleast one root in given interval.

8. If $\mathrm { p } ( \mathrm { x } ) = 51 \mathrm { x } ^ { 101 } - 2323 \mathrm { x } ^ { 100 } - 45 \mathrm { x } + 1035$, using Rolle's Theorem, prove that atleast one root lies between ( $45 ^ { 1 / 100 } , 46$ ). Sol. Let $\mathrm { g } ( \mathrm { x } ) = \int \mathrm { p } ( \mathrm { x } ) \mathrm { dx } = \frac { 51 \mathrm { x } ^ { 102 } } { 102 } - \frac { 2323 \mathrm { x } ^ { 101 } } { 101 } - \frac { 45 \mathrm { x } ^ { 2 } } { 2 } + 1035 \mathrm { x } + \mathrm { c }$ $= \frac { 1 } { 2 } \mathrm { x } ^ { 102 } - 23 \mathrm { x } ^ { 101 } - \frac { 45 } { 2 } \mathrm { x } ^ { 2 } + 1035 \mathrm { x } + \mathrm { c }$. Now $\mathrm { g } \left( 45 ^ { 1 / 100 } \right) = \frac { 1 } { 2 } ( 45 ) ^ { \frac { 102 } { 100 } } - 23 ( 45 ) ^ { \frac { 101 } { 100 } } - \frac { 45 } { 2 } ( 45 ) ^ { \frac { 2 } { 100 } } + 1035 ( 45 ) ^ { \frac { 1 } { 100 } } + \mathrm { c } = \mathrm { c }$ $\mathrm { g } ( 46 ) = \frac { ( 46 ) ^ { 102 } } { 2 } - 23 ( 46 ) ^ { 101 } - \frac { 45 } { 2 } ( 46 ) ^ { 2 } + 1035 ( 46 ) + \mathrm { c } = \mathrm { c }$. So $\mathrm { g } ^ { \prime } ( \mathrm { x } ) = \mathrm { p } ( \mathrm { x } )$ will have atleast one root in given interval.

Now we consider infinite sets of whole numbers (for example, the set of all the positive whole numbers). Give examples to demonstrate that an infinitely large set of whole numbers might have zero, exactly one, or more than one target(s). Justify your answers, making it clear which is which.