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Quantitative Ability – POINTS TO REMEMBER(Algebra)

Quantitative Ability – POINTS TO REMEMBER

1.If an equation (i.e. f(x) = 0) contains all positive co-efficients of any powers of x, it has no positive roots.
Eg: x^3+3x^2+2x+6=0 has no positive roots

2.For an equation, if all the even powers of x have same sign coefficients and all the odd powers of x have the opposite sign coefficients, then it has no negative roots.

3.For an equation f(x)=0 , the maximum number of positive roots it can have is the number of sign changes in f(x) ; and the maximum number of negative roots it can have is the number of sign changes in f(-x)

4.Complex roots occur in pairs, hence if one of the roots of an equation is 2+3i, another has to be
2-3i and if there are three possible roots of the equation, we can conclude that the last root is real. This real root could be found out by finding the sum of the roots of the equation and subtracting (2+3i)+(2-3i)=4 from that sum.

5.For a cubic equation ax^3+bx^2+cx+d=o
Sum of the roots = - b/a
Sum of the product of the roots taken two at a time = c/a
Product of the roots = -d/a

6.For a bi-quadratic equation ax^4+bx^3+cx^2+dx+e = 0
Sum of the roots = - b/a
Sum of the product of the roots taken two at a time = c/a
Sum of the product of the roots taken three at a time = -d/a
Product of the roots = e/a

7.If an equation f(x)= 0 has only odd powers of x and all these have the same sign coefficients or if f(x) = 0 has only odd powers of x and all these have the same sign coefficients, then the equation has no real roots in each case (except for x=0 in the second case)

8.Consider the two equations
a1x+b1y=c1
a2x+b2y=c2
Then,
If a1/a2 = b1/b2 = c1/c2, then we have infinite solutions for these equations.
If a1/a2 = b1/b2 <> c1/c2, then we have no solution.
If a1/a2 <> b1/b2, then we have a unique solution.

9.a + b = a + b if a*b>=0
else, a + b >= a + b

10.The equation ax^2+bx+c=0 will have max. value when a<0>0. The max. or min. value is given by (4ac-b^2)/4a and will occur at x = -b/2a

11.If for two numbers x + y=k (a constant), then their PRODUCT is MAXIMUM if x=y (=k/2). The maximum product is then (k2)/4.

12. If for two numbers x*y=k (a constant), then their SUM is MINIMUM if x=y (=root(k)). The minimum sum is then 2*root (k).

13.Product of any two numbers = Product of their HCF and LCM. Hence product of two numbers = LCM of the numbers if they are prime to each other.

14. For any 2 numbers a, b where a>b
a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic means respectively)
(GM)^2 = AM * HM

15.For three positive numbers a, b, c
(a + b + c) * (1/a + 1/b + 1/c)>=9

16.For any positive integer n
2<= (1 + 1/n)^n <=3

17. a^2 + b^2 + c^2 >= ab + bc + ca
If a=b=c, then the case of equality holds good.

18.a^4 + b^4 + c^4 + d^4 >= 4abcd (Equality arises when a=b=c=d=1)

19.(n!)^2 > n^n

20.If a + b + c + d=constant, then the product a^p * b^q * c^r * d^s will be maximum if a/p = b/q = c/r = d/s

21.x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + .......+ a^(n-1) ) ......Very useful for finding multiples. For example (17-14=3 will be a multiple of 17^3 - 14^3)

22.e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ........to infinity
Note: 2 < e < 3

23.log(1+x) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 .........to infinity [Note the alternating sign . .Also note that the logarithm is with respect to base e]

24.(m + n)! is divisible by m! * n!

25.When a three digit number is reversed and the difference of these two numbers is taken, the middle number is always 9 and the sum of the other two numbers is always 9.

26.Any function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y)

The sum of first n natural numbers = n(n+1)/2

The sum of squares of first n natural numbers is n(n+1)(2n+1)/6

The sum of cubes of first n natural numbers is (n(n+1)/2)2/4

The sum of first n even numbers= n (n+1)

The sum of first n odd numbers= n^2

3 Responses to "Quantitative Ability – POINTS TO REMEMBER(Algebra)"

Unknown (visit their site)

Sir in property 24 you said

24.(m + n)! is divisible by m! * n!

How this can be possible, considering the fact that if m>n then
(m+n)! = (m+n)*(m+n-1)*(m+n-2)*.... (m+n-n+1)*(m+n-n)!

here last term is (m)!
so the whole term is divisible by (m)! but not by (n)!
how it can be divisible by (m)!*(n)!
also since m>n so they individually are divisible but as a product how they can?
please elaborate.

Anonymous (visit their site)

Sir, could you check the point 7.
It has repeatition of a phrase.

Amol (visit their site)

Hi Nikhil,
just take few examples
let m=2,n=3
so m+n=5, (m+n)! is divisible by
m!*n!
5! is divisible by 2!*3! i.e. 120 is divisible by 2*6= 12.
you can check for higher values also. m is greter or less doesent matter...because we are taking m+n as a complete value....
still any doubts call me...

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