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 \(f'(0) \geq - \sqrt{2}\).

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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 \(f'(0) \geq - \sqrt{2}\) – Jatin Yadav · 3 years, 1 month ago

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– Paramjit Singh · 3 years, 1 month ago

Good!Log in to reply

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

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

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it. :) – Paramjit Singh · 3 years, 1 month agoLog in to reply