48 - Algebra Questions Answers

Domain of the given function is (-∞, -1) υ [0, ∞)

Now at x=0, y=1

and at x=∞, y=0

similarly left limiting value of y = -e

and for x=-∞, y=-∞

so range is (-∞, -e] υ (0, 1]

(a+b)/2 = √ab + 3/2                   (1)                        or         a+b = 2√ab + 3

√ab      = 2ab/(a+b)  +  6/5       (2)                        or         5(a+b)√ab  =  10ab + 6(a+b)      

on solving these 2 we get               5[2√ab + 3]√ab = 10ab + 6[2√ab + 3]

                                           or               10ab + 15√ab  =  10ab + 12√ab + 18

                                           or                3√ab = 18     or      ab = 36          and      a+b = 15 = α

 so a-b = √(225 - 144)     =    √81     =     9 = β

According to the given conditions

α+β=2p      (1)                 α*β=2        (2)

β+γ=2q       (3)                 β*γ=3        (4)

γ+α=2r       (5)                  γ*α=6        (6)

so α+β+γ = p+q+r              (7)

and α*β*γ=6                       (8)

dividing eq. (8) by (2), (4) , (6)  one by one we will get α, β, γ    hence we will get p+q+r

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)