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# Some more problems...

Try the followings:

• Consider a prism with a triangular base. The total area of the three faces containing a particular vertex $$A$$ is $$K$$. Show that the maximum possible volume of the prism is $$\displaystyle \sqrt{\frac{K^3}{54}}$$ and find the height of this largest prism.

• Consider an $$n^2 \times n^2$$ grid divided into $$n^2$$ subgrids of size $$n \times n$$. Find the number of ways in which you can select $$n^2$$ cells from this grid such that there is exactly one cell coming from each subgrid, one from ach row and one from each column.

• If $$a_1, \cdots, a_7 \in (1,13)$$ are not necessarily distinct reals, show that we can choose three of them such that they are lengths of the sides of a triangle.

• Show that there cannot exist a non-constant polynomial $$P(x) \in \mathbb{Z}[x]$$ such that $$P(n)$$ is prime for all positive integers $$n$$.

• (Calculus) Let $$f$$ be a twice differentiable function on the open interval $$(-1,1)$$ such that $$f(0)=1$$. Suppose that $$f$$ also satisfies $$f(x) \geq 0$$, $$f'(x) \leq 0$$ and $$f''(x) \leq f(x)$$, for all $$x \geq 0$$. Show that $$f'(0) \geq - \sqrt{2}$$.

Note by Paramjit Singh
3 years, 1 month ago

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There must be a value of $$x \in(0,1)$$ satisfying $$f'(x_{0}) = \dfrac{f(1) - 1}{1-0}$$

For this $$x$$ , $$(f'(x_{0}))^2 = (1 - f(1))^2$$

Now,

$$f''(x) \leq f(x) \Rightarrow f'(x) f''(x) \geq f'(x) f(x)$$

Hence,

$$d((f'(x))^2) \geq d((f(x))^2)$$

Integrate in limits 0 to $$x_{0}$$ to get:

$$(f'(x_{0}))^2 - (f'(0))^2 \geq (f(x_{0}))^2 - 1$$

Hence, $$(f'(0))^2 \leq 1 +(1-f(1))^2 - (f(x_{0}))^2$$

Clearly, $$(f'(0))^2 \leq 2$$ ,or $$f'(0) \geq - \sqrt{2}$$ · 3 years, 1 month ago

Good! · 3 years, 1 month ago

Problem 3 is essentially USAMO 2012 Problem 1 in disguise. Can you see why? · 3 years, 1 month ago

hey paramjeet i cant get it can u explain me in detail · 3 years, 1 month ago