Trigonometry: Elimination Problem

Solving a system of in-equations: A problem form High School

In this video, I have solve a system of in-equations, using elementary ideas of high school mathematics. Using the fact that, square of any real number is always greater than equal to zero, we can at the same time frame and solve unique problems. Watch the full video, to learn.

Less calculations, more thought: Problem form Definite Integral

A very interesting problem from, definite integrals. This problem is slightly different form the usual problems of definite integral, as the function given here is not the usual polynomial or trigonometric functions. Watch the full video for the solution and learn something new and different.

Algebra Problem: Can you Solve it?

Using elementary identity from middle school, his problem can be solved. But it's really challenging and tricky if you don't hit the right idea. Check out the video to learn something new and interesting.

Let x,y,z be distinct real
numbers.

Prove that

∛(x-y)+∛(y-z)+∛(z-x)  ≠0

You should be able to solve this! Minimum value Problem

A problem of minimisation. Given an expression, for which we have to find the minimum value. A very interesting problem, with simple properties of logarithmic functions to be used. Watch the full video to see the solution and learn something new and different.

A very interesting problem from Indian Statistical Institute: Geometry

This problem, requires some knowledge of a regular polygon. Only the elementary properties is required. If you don't know anything about regular polygons, just google it! Lot of thinking is required to solve this problem. A very good problem where the calculations are minimal but the logical part is required throughout. Check the full video for the solution and share your views about the problem.

Problem form American Invitational Mathematics Examination: Can you solve it?

A very beautiful problem with elegant solution. Suitable for students preparing for IIT and other entrance exams. Even for school students this problem is suited who wants to learn something different. As the problem doesn't require more than knowledge of multiplication of variables and solving simultaneous linear equations, a 10th grade students can also understand it!

An interesting property of cyclic quadrilateral.

In a cyclic quadrilateral the product of its diagonals is equal to the sum of the products of it's opposite sides. Watch the full video for the proof. 

Definite Integral : Problem with a twist!

Sum of squares of 5 consecutive natural numbers!

Here's a difficult problem of proving that a given expression is not a perfect square. A direct approach will be a nightmare ( even not sure it can be proved) but use of a simple property of perfect squares will ease the problem. We know that any perfect squares leaves a remainder either 1 or 0 when being divided by 3 or 4. This simple result would be used to solve the problem.

Number of solutions of the given equation

Here the equation to be solved is very simple |2x-[x]|=4. But unfortunately the traditional methods of solving will not apply! You have solved equations by methods like factorisation, vanishing method, hit and trial.... But in many equations these methods don't work or requires huge computation and guess work to solve it. Here is an equation which can be solved from the simple fact that if one side of an equation is an integer the we look to the other side only and see for what values its also an integer. This simple but powerful method can be used in many problems

A very hard integration problem! Can you solve it? ( Practice for IIT-JEE)

A very hard integration problem based on the method of substitution. It took almost an hour to figure this out! Can you solve it on your own?! This problem is well suited for students preparing for IIT-JEE.

Inequalities

Show that $|x|+|y| \leq |x+y|+|x-y| \quad and \quad \frac{|x+y|}{1+|x+y|} \leq \frac{|x|}{1+|x|} + \frac{|y|}{1+|y|} \quad for \quad x,y \in \mathbb{R} $ $$$$ We have $\quad|x+y+x-y| \leq |x+y|+|x-y| $ $$$$ $ \implies 2|x| \leq |x+y|+|x-y| \dots A$ $$$$ Now, $|x+y+y-x| \leq |x+y|+|y-x|$ $$$$ $ \implies 2|y| \leq |x+y|+|-1||x-y| $ $$$$ $\implies 2|y| \leq |x+y|+|x-y| \dots B$ $$$$ Adding $A$ and $B$ we get $ |x|+|y| \leq |x+y|+|x-y|$ The second inequality holds $$$$ $iff$ $\frac{(1+|x|)(1+|y|)|x+y|}{1+|x+y|} \leq |x|(1+|y|)+|y|(1+|x|)$ $$$$ $ iff \quad (1+|x|)(1+|y|)|x+y| \leq (|x|+|y|+2|xy|)(1+|x+y|)$ $$$$ (multiply and cancel out terms) $$$$ $iff \quad |x+y| \leq |x|+|y| +|xy|(2+|x+y|)$ $$$$ Now for any real $x$ and $y$, $|xy|(2+|x+y|) \geq 0$ and we know that $|x+y| \leq |x|+|y|$ $$$$ Thus the last inequality holds, which in turn proves the original ineqality.