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Here, you will find all the questions of ISI Entrance Paper 2011 from Indian Statistical Institute's B. Math Entrance. You will also get the solutions soon of all the previous year problems.

**Problem 1 :**

Let $a \geq 0$ be a constant such that $\sin (\sqrt{x+a})=\sin (\sqrt{x})$ for all $x \geq 0 .$ What can you say about $a$ ? Justify your answer.

**Problem 2 :**

Let $f(x)=e^{-x}$ for $x>0$. Define a function $g$ on the nonnegative real numbers as follows: for each integer $k>0$, the graph of the function $g$ on the interval $[k, k+1]$ is the straight line segment connecting the points $(k, f(k))$ and $(k+1, f (k+1) )$. Find the total area of the region which lies between the curves of $f$ and $g$.

**Problem 3 :**

For any positive integer $n$, show that $\frac{1}{2}\cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \cdot \frac{(2 n-1)}{2 n}<\frac{1}{\sqrt{2 n+1}}$.

**Problem 4 :**

If $a_{1}, \ldots, a_{7}$ are not necessarily distinct real numbers such that $1<a_{i}<13$ for all $i$, then show that we can choose three of them such that they are the lengths of the sides of a triangle.

**Problem 5 :**

For any real number $x,$ let $[x]$ denote the largest integer which is less than or equal to $x$. Let $N_{1}=2, N_{2}=3, N_{3}=5, \ldots$ be the sequence of non-square positive integers. If the $n$ th non-square positive integer satisfies $m^{2}<N_{n}<$ $(m+1)^{2},$ then show that $m=\left[\sqrt{n}+\frac{1}{2}\right]$

**Problem 6 :**

Let $R$ and $S$ be two cubes with sides of lengths $r$ and $s$ respectively, where $r$ and $s$ are positive integers. Show that the difference of their volumes equals the difference of their surface areas, if and only if $r=s$.

**Problem 7 :**

Let $A B C$ be any triangle and let $O$ be a point on the line segment $B C .$ Show that there exists a line parallel to $A O$ which divides the triangle $A B C$ into two equal parts of equal area.

**Problem 8 :**

Let $t_{1}<t_{2}<\cdots<t_{99}$ be real numbers, and consider the function $f: \mathbb{R} \rightarrow \mathbb{R}$ be given by $f(x)=\left|x-t_{1}\right|+\left|x-t_{2}\right|+\cdots+\left|x-t_{99}\right| .$ Show that $\min _{x \in R} f(x)=f\left(t_{50}\right)$

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