Tan (pi/2)

We most of the time think tan(pi/2) = infinite , 

and our reasoning is 

tan (pi/2) = sin(pi/2) / cos(pi/2) = 1/0 = infinite , this expression is true 

but when we find value of tan(pi/2) its exactly not defined, since it has two values (+/- infinite)[check image] which cannot be of any function [Functions have only one image].

And having two values at (pi/2) we treat this as not-defined.

SO WHEN EVER SOMEONE ASK TO YOU (generally in interviews) :

What is TAN , a function or relation: Answer is Relation

And at odd multiple of (pi/2) "tan" is not defined.

Use of basic concepts to solve questions in seconds

Q. With the speed of 50 Km/hr Ram reaches his office from home at 11:30 AM, and with speed of 90 Km/hr he reaches at 10:30 AM, what should be his speed if he wants to reach at 10:59 AM (in all cases he starts at same time and and reaches at same day)

a] 70 KM/hr
b] 70.5 Km/hr
c] 71.5 KM/hr
d] 72.5 Km/hr
e] None of these
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Sol: We know that we should use speed average, as shown in video http://www.youtube.com/watch?v=q0Y1YjwFol8 (watch this on non mobile device for text and animations and in full mode)


Now come to solution part:

we know speed average is HM. and we also know HM is less than AM (mean) 

and given speeds are 50 and 90, so AM (50,90) = 70 its AM but we are looking at HM

so, answer would be less than 70

check option less than 70 and eliminate the options, final answer is None of these
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NOTE : THESE TYPE OF QUESTIONS, FIND AM THEN ELIMINATE OPTIONS LESS THAN AM (since AM is easy to find)




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Concept : AM >= GM

An approach to solve a higher degree equation.

Its a trend : Past years easy & conceptual questions of JEE, SCRAs & AIEEE are today's higher level questions of MBA Test Prep & SSC .

Aim of sharing this question is, to let you know, an approach to solve a higher degree equation.

Learn the basic graphs of different equations.

Factorian

A factorion is a natural number that equals the sum of the factorials of its decimal digits. For example, 145 is a factorion because 1! + 4! + 5! = 1 + 24 + 120 = 145.

There are just four factorions (in base 10) and they are 1, 2, 145 and 40585 

~Text is copied from Wiki~

What they can ask : 

There are how many natural number(s) which is(are) equal to summation of factorials of all its digits.

So remember answer is 4 and numbers are 1, 2, 145 and 40585.

Easiest way to solve a Right Angled Triangle "Challenger"

Challenger Concept :

How to proceed: Just plot the question on coordinate geometry plane, right angle at origin, name the coordinates of all points, then use basic concepts and get the answer.

The line AC and DE are intercept and M, N and are midpoints so I gave a name to this theorem IMT (my alma-mater), here IMT stands for Intercept-Midpoint-Theorem.

Comparison between a^b & b^a

One of the most important concept of Geometry : Integral Sided Triangle by Alcuin's sequence

One of the most important concept of Geometry : Integral Sided Triangle by Alcuin's sequence

Courtesy : Counting integral triangle by Krier and Manvel

Tangential Circles

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Did you know

► Last two digits of 76^n are 76 (where n is a natural number).

► If last two digits of ab^n = ab then ab = 00, 01, 25 or 76.

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In your examination, you might get a question :

Find the last two digits of 2345678765676^76876545434^5467876^876567 .

Ans is 76 (since last two digits of base is 76, hardly 2 sec.)

General concept of Last Digit

How to find last digit of A^B

Step I : Find B mod 4* (or remainder by 4*) [say R=0**,1,2,3]

Step II : Write the last digit of base [say L]

Step III: Solve L^R 

[**Note: If R=0, put R=4 instead of 0, or For Odd base direct answer is 1, and for even it's 6]

Step IV : The last digit of L^R is your required last digit.

e.g : Find the last digit of 1234567^55555555

Step I : 55555555 mod 4 = 3 [easy just check for last two digits]

Step II: 7

Step III : 7^3 =343

Step IV : Required Answer = 3

*Last digit is nothing but remainder by 10. The cyclicity of 10 is 4

Bimedian

Important Concept : Bimedian

► The two bimedians in a quadrilateral and the line segment joining the midpoints of the diagonals in that quadrilateral are concurrent and are all bisected by their point of intersection

► In a convex quadrilateral, the two bimedians have equal length if and only if the two diagonals are perpendicular

► In a convex quadrilateral, the two bimedians are perpendicular if and only if the two diagonals have equal length

Question on Cyclicity

Find remainder (13^402) by 8000

By Cyclicity ( C ); one of the easiest method.

C(p^n * q^n) = LCM( C(p^n ) , C (q^n)) [Where p and q are primes]

Geometry

Level CAT-2013

An Important Output

Last Term Left

Q.  If 135791357913579......... is a 1000 terms series. If A student of 3E Learning removes odd placed term and creates a new series and repeats the same process again and again and removed all the terms by this process  . What would be the last term removed by him.

Geometry

Concept : Triangle, Parallel lines, Similarity, Areas

Probability In Real Life, Expected in CAT 2013

Q. In 3E Learning centre 3 out of 7 students like probability. Assume total of 7^10 researcher took data of 10 students to check either they like probability or not. Then how many researcher would you expect to report only 2 students like probability.





Sol: