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.

Solved Trigonometry Problems: 10th Grade

Here is a list of few good problems at the 10th grade. Get the full solution below and if you want
more comment or mail us at maths.programming@gmail.com!


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Circle: Geometry

Given below is a fine problem based on simple properties of circles ( tangent and chord ). Try yourself before looking at the solution!

Here is the solution!

Solved Problems : Logarithm

A collection of solved problems on logarithm meant for secondary students ( 9th and 10th grade). The collection covers almost all types of problems at the level mentioned.



Download the file here: Solved Logarithm Problems 

Solved Integration Problems for 10+2 Level - SN Dey

Hundreds of solved problems at 10+2 level or the final year at school from the topic of integration. The aim of the document to help students from different boards (ISC, CBSE and other State Boards) studying at Higher Secondary level to get access to wide variety of problems, that too solved! Most of the problems are from the books of SN Dey, which is the most sought after book in WBCHSE.

This should not be substituted for classroom teaching neither this documents aims to teach the theories of differentiation. Students are requested to go through the theory and the rules before looking at the problems. Any error in the document can be reported at: prime.maths@hotmail.com

Integration, Methods of Substitution, Integration by Parts, Partial Fractions, Special Integrations, Trigonometric Substitutions.

If you have any query, don't forget to comment below.

A  collection of completely solved special integrals for various entrance exams ( IIT and Indian Statistical Institute ) and boards exam (CBSE,ISC and other State Boards) at 10+2 level. 

Click below the links to download the files:

File 1 Introduction ( Basic Problems )

File 2 Special Integrals 

File 3 Method of Substitution- Set I

File 4 Method of Substitution- Set II

File 5 Method of Substitution- Set II

File 6 Problems based on Standard Integrals- Set I

File 7 Problems based on Standard Integrals- Set II 

File 8 Problems based on Standard Integrals- Set III

File 9 Problems based on Integration by Parts - Set I

File 10 Problems based on Integration by Parts - Set II

More solved problems to follow. Keep visiting this space.

Special Integration

Integration : A harder Problem

 Most of the studenst will fail to solve this particular integration problem. It is trickier but once you hit the right idea, you will be able to solve the integration problem easily.

A problem on inequality

Using simple formula to prove a strong inequality.

Pigeonhole Principle

The numbers 1 to 20 are placed in any order around a circle. Prove that the sum of some 3 consecutive numbers must be at least 32!

This problem uses the alternate form of pigeon hole principle which is as follows:
 If the average of n positive numbers is t, then at least one of the numbers is greater than or equal to t. Further, at least one of the numbers is less than or equal to t.

The proof is very simple, assume the contradiction and proceed!

#Solution https://youtu.be/GLQg6cSAbms

Sum of first 'n, natural numbers

A simple way to calculate the sum of first 'n' natural numbers witout the use of calculator. Infact the same procedure is used to calculate the sum of n terms of any A.P series. See it and try to obtain the formula yourself

Inequality

A challenging problem based on the inequality that square of a real number is always greater than equal to zero. Learn the trick and prepare yourself for more challenging problems based on the same ideology.

Divisibility of 2222^5555+5555^2222 by 7

For any prime p, square root of p is irrational.