45 - Algebra Questions Answers

Left hand derivative of given function at x = k (k is an integer) is

   lim_{h -> 0} {f(k-h) - f(k)}/ {k-h-k}  

= lim_{h -> 0} {[k-h] sin π(k-h) - [k] sin πk}/{-h}                  now 2 cases arise for even and odd k

first case

= lim_{h ->0}  (k-1) sin (-πh) - 0 / (-h)                          if k is even ,   here we used sin πk = 0 and sin (2π-x) = sin (-x) 

= π(k-1)                   here we used sin (-πx)/(-πx) = 1

second case

= lim_h->0 (k-1) sin (πh) - 0 / (-h)                            if k is odd,    here we used sin πk = 0 and sin (3π-x) = sin x

= -π(k-1) 

Left hand derivative of given function at x = k (k is an integer) is

   lim_{h -> 0} {f(k-h) - f(k)}/ {k-h-k}  

= lim_{h -> 0} {[k-h] sin π(k-h) - [k] sin πk}/{-h}                  now 2 cases arise for even and odd k

first case

= lim_{h ->0}  (k-1) sin (-πh) - 0 / (-h)                          if k is even ,   here we used sin πk = 0 and sin (2π-x) = sin (-x) 

= π(k-1)                   here we used sin (-πx)/(-πx) = 1

second case

= lim_h->0 (k-1) sin (πh) - 0 / (-h)                            if k is odd,    here we used sin πk = 0 and sin (3π-x) = sin x

= -π(k-1) 

so for even and odd k answer will be different.

I think the right question is  

f(x)  = ₀∫^x ( t²-t+2 )²⁰⁰⁵( t²-t-2 )²⁰⁰⁷( t²-t-6 )²⁰⁰⁹( t²-t-12 )²⁰¹¹( t²-3t+2 )²⁰¹²  dt    

then sum of values of x,  where maxima of f(x) will be obtained is

for maxima f '(x) = 0

by using Newton Lebnitz formula we get

f ^'(x) =  (x²-x+2 )²⁰⁰⁵( x²-x-2 )²⁰⁰⁷( x²-x-6 )²⁰⁰⁹( x²-x-12 )²⁰¹¹( x²-3x+2 )²⁰¹²

comparing it with 0 and making factors we will get x = -1, 2, 3, -2, 4, -3, 1

now obtain the maxima point then sum these values

n = (1223334444)₇ = 1*7⁹ + 2*7⁸ + 2*7⁷ + 3*7⁶ + 3*7⁵ + 3*7⁴ + 4*7³ + 4*7² + 4*7¹ + 4*7⁰

so if we will divide n! by 7^x we will get 1*7⁸ + 2*7⁷ + 2*7⁶ + 3*7⁵ + 3*7⁴ + 3*7³ + 4*7² + 4*7¹ + 4*7⁰ as x

A light example

if we divide (1*7² + 2*7¹ + 3*7⁰)! by 7^x we get x = 1*7¹ + 2*7⁰ = 9     

let f(x) = x^3-3x+k

so f^'(x) = 3 x^2 - 3

on comparing with 0, x = +1, -1

on double differentiating we get that function f(x) is minimum at +1 and maximum at -1

so for getting the only +ve real root plot a graph such that it could intersect positive x axis at any point and nowhere else. From the graph it is clear that k will be negative.

at x = +1,  f(x) = k-2

at x = -1, f(x) = k+2

so on the basis of given conditions k-2<0 or k<2  and  k+2<0 or k<-2 

finally k<-2 so [|k|] has minimum value 2

x^7 - 3x^4+2x^3-k=0

according to sign change rule there are 3 pair of sign + -, - +, + - ,       see the sign before coefficient of different powers

so 3 positive real roots are confirmed

now take  x = -x

we get -x^7 - 3x^4 - 2x^3 - k = 0, there is no sign change so no negative root

total real roots = 3

so imaginary roots = 4

(1 + 0.0001) ¹⁰⁰⁰⁰ = 1 + 10000*0.0001 + other terms = greater then 2

besides it lim _x->0 (1+x)^1/x = e = less then 3

it means the given expression is more then 2 but less then 3 

so 2 will be the answer

1)   (10/12)(9/11)(8/10)(7/9)

2)   4(10/12)(9/11)(8/10)(2/9)

3)   (4*3/2)(10/12)(9/11)(2/10)(1/9)

4)   1 - (10/12)(9/11)(8/10)(7/9)

Please send the complete question

(1+λ)ⁿ = C₀ + C₁x + ..................................C_nxⁿ

similarly for (1+μ)ⁿ and 

(λ+μ)ⁿ = C₀λⁿ + C₁λ^n-1μ + ........................ + C_nμⁿ

On multiplying these 3, we will get the coefficient of  λⁿμⁿ = ∑C_n³      (Here you should remember that C₀ = C_n, C₁ = C_n-1 and etc)

For getting the value of  ∑C_n³ :

multiply the expansions of (1+x)ⁿ, (1-x)ⁿ and [1-(1/x²)]ⁿ and calculate the coefficient of x⁰, we get ∑C_n³

and if we multiply (1+x)ⁿ, (1-x)ⁿ and [1-(1/x²)]ⁿ, then we get (-1)ⁿ(1-x²)²ⁿ/x²ⁿ so coefficient of x⁰ in this expression will be (-1)^n 2nC_n (-1)ⁿ  = ²ⁿC_n

Let x = (5√5 + 11)²ⁿ⁺¹ and y = (5√5 - 11)²ⁿ⁺¹

so x*y = 4²ⁿ⁺¹ = even number

similarly x - y = integer

if we consider the value of y , it will be (5*2.3-5)²ⁿ⁺¹ = (0.5)²ⁿ⁺¹  = fraction number less than 1   (approx)

so definitely y will be the fractional part of x, and difference of x and y is even so integer part of x will be even