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heureka

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Questions 17
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 #2
avatar+26396 
+4

i need the answer and solution preferably please.

 

if  tan(120x)=sin(120)sin(x)cos(120)cos(x), where 0<x<180, compute x

 

Let s=sin(120)=sin(60)=32Let c=cos(120)=cos(60)=12

LHS:

tan(120x)=sin(120x)cos(120x)=sin(120)cos(x)cos(120)sin(x)cos(120)cos(x)+sin(120)sin(x)=scos(x)csin(x)ccos(x)+ssin(x)

 

RHS:

sin(120)sin(x)cos(120)cos(x)=2cos(120+x2)sin(120x2)2sin(120+x2)sin(120x2)=cos(120+x2)sin(120+x2)=cos(60+x2)sin(60+x2)=cos(60)cos(x2)sin(60)sin(x2)sin(60)cos(x2)+cos(60)sin(x2)=sin(60)sin(x2)cos(60)cos(x2)sin(60)cos(x2)+cos(60)sin(x2)=ssin(x2)+ccos(x2)scos(x2)csin(x2)

 

scos(x)csin(x)ccos(x)+ssin(x)=ssin(x2)+ccos(x2)scos(x2)csin(x2)(scos(x)csin(x))(scos(x2)csin(x2))=(ccos(x)+ssin(x))(ssin(x2)+ccos(x2))s2cos(x)cos(x2)sccos(x)sin(x2)scsin(x)cos(x2)+c2sin(x)sin(x2)=c2cos(x)cos(x2)+sccos(x)sin(x2)+scsin(x)cos(x2)+s2sin(x)sin(x2)cos(x)cos(x2)(s2c2)sin(x)sin(x2)(s2c2)2sccos(x)sin(x2)2scsin(x)cos(x2)=0(s2c2)(cos(x)cos(x2)sin(x)sin(x2)=cos(x+x2))2sc(cos(x)sin(x2)+sin(x)cos(x2)=sin(x+x2))=0(s2c2)cos(x+x2)2scsin(x+x2)=02scsin(x+x2)=(s2c2)cos(x+x2)2scsin(3x2)=(s2c2)cos(3x2)tan(3x2)=s2c22sc|s2c2=3414=12tan(3x2)=122sc|2sc=232(12)=32tan(3x2)=1232tan(3x2)=333x2=arctan(33)+n180nZx=23arctan(33)+23n180x=23(30)+n120x=20+n120n=1(0<x<180)x=20+120x=100

 

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Feb 22, 2019
 #2
avatar+26396 
+4

Find all real values of a for which the equation (x2+a)2+a=x has four real roots.

 

My attempt:

I assume there are 4 distinct real roots. So there are 3 local maxima/minima.

We find the local maxima/minima by differentiation. Maxima/minima occur when f'(x) = 0

 

1. Differentiation:

y=(x2+a)2x+ay=2(x2+a)2x10=4x(x2+a)14x(x2+a)=1|:4x(x2+a)=14x3+ax14=0

 

The zero set of discriminant of the cubic  Ax3+Bx2+Cx+D has discriminant
  B2C24AC34B3D27A2D2+18ABCD.
The discriminant is zero if and only if at least two roots are equal.
If the coefficients are real numbers, and the discriminant is not zero,
the discriminant is positive if the roots are three distinct real numbers,
and negative if there is one real root and two complex conjugate roots.

 

Ax3+Bx2+Cx+Dx3+ax14=0|A=1, B=0, C=a, D=14

B2C24AC34B3D27A2D2+18ABCD>0|there are three distinct real numbers4a327(14)2>04a32716>04a3>2716|:4a3<2764a3<3343a<34

 

(x2+a)2+a=x has four distinct real roots if  a<34

 

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Feb 22, 2019
 #1
avatar+26396 
+3

If a, b, c and d are positive real numbers such that
log_a  b= 8/9,
log_b  c= -(3/4),
log_c d = 2,
find the value of log_d (abc)

 

loga(b)=logd(b)logd(a)=logb(b)logb(a)logd(b)logd(a)=logb(b)logb(a)|logb(b)=1logd(b)logd(a)=1logb(a)|loga(b)logb(a)=1logd(b)logd(a)=loga(b)logd(a)logd(b)=1loga(b)logd(a)=logd(b)loga(b)(1)logb(c)=logd(c)logd(b)=logc(c)logc(b)logd(c)logd(b)=logc(c)logc(b)|logc(c)=1logd(c)logd(b)=1logc(b)|logb(c)logc(b)=1logd(c)logd(b)=logb(c)logd(b)logd(c)=1logb(c)logd(b)=logd(c)logb(c)(2)logc(d)=logd(d)logd(c)|logd(d)=1logc(d)=1logd(c)logd(c)=1logc(d)(3)

 

logd(abc)=logd(a)+logd(b)+logd(c)|logd(a)=logd(b)loga(b)(1)=logd(b)loga(b)+logd(b)+logd(c)=logd(b)(1loga(b)+1)+logd(c)|logd(b)=logd(c)logb(c)(2)=logd(c)logb(c)(1loga(b)+1)+logd(c)=logd(c)(1logb(c)(1loga(b)+1)+1)|logd(c)=1logc(d)(3)logd(abc)=1logc(d)(1logb(c)(1loga(b)+1)+1)

 

logd(abc)=1logc(d)(1logb(c)(1loga(b)+1)+1)=12(134(189+1)+1)=12(43(18+1)+1)=12(43(178)+1)=12(176+1)=12(116)=1112

 

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Feb 22, 2019
 #2
avatar+26396 
+3

 For how many integers n between 1 and 11 (inclusive) n/12 is a repeating decimal?

 

prime number only factorizeprime numbers2 or 5nn12cancelthe denominatorin the denominatorwill terminate111211212232, 32212161232, 33312141222terminate441213133551251212232, 3661212122terminate771271272232, 38812232339912343222terminate101012565232, 31111121112112232, 3

 

see: https://gmatclub.com/forum/math-number-theory-88376.html

 

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Feb 22, 2019
 #6
avatar+26396 
+3
Feb 22, 2019