Question

1.The solution of
x+ 1/x-2 = 2+ 1/x-1
2. The total area enclosed by lines y=|x| , |x|=1 and y=0 is
3. The minimum value of 3tan^2θ + 12cot^2θ is
4. If midpoint of join (x,y+1) and (x+1,y+2) is {3/2,5/2} then the midpoint of join of
(x-1,y+1) and (x+1, y-1) is
5. If pand q are roots of equation x^2+px+q=0, then
two values for p are
6.if the roots of x^2-ax+b=0 differ by unity then:
a. B^2=1+4a b. a^2 =1+4b c. if x belongs to real no.s, a^2>4b d. a^2 +4b =1
7. Total no. of solutions of
2^x+3^x+4^x-5^x=0 is
If a,b, c are three positive real numbers, then mnimum value of expression
b+c/a + a+c/b + a+b/c is

Answer 1

$x + \frac{1}{x-2}=2+\frac{1}{x-1}\\\\\frac{x^2-2x+1}{x-2}=\frac{2x-2+1}{x-1}\\\\\frac{(x-1)^2}{x-2}=\frac{2x-1}{x-1}\\\\(x-1)^3=(x-2)(2x-1)\\\\x^3-3x^2+3x-1=2x^2-x-4x+2\\\\x^3-5x^2+8x-3=0\\\\x=0.5344$
The above equation seems to have one real irrational root, and two imaginary roots.
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There are two triangles for the area given. O(0,0), (1,0) and (1,1) form one triangle.  O(0,0), (-1,0), and (-1,1) form another triangle. Are a of both is same = 1/2.  Total area = 1.
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$3tan^2\theta+12cot^2\theta=3(tan\theta+2cot\theta)^2-12\\\\\frac{d(tan\theta+2cot\theta)}{d\theta}=sec^2\theta-2cosec^2\theta=0\ for\ minimum\\\\sin^2\theta=2cos^2\theta\\\\tan^2\theta=2\\\\Substitute\ to\ get,\ 3*2+12/2=12$
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(x+x+1)/2=3/2  => x = 1    and  (y+1+y+2)/2 = 5/2      =>  y = 1
       mid point of (2,2) and (2, 0) is    (2, 1)
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as p is a root,  substitute in place of x .
  p² +p*p + q = 0    => 2p² = - q
  q² + p q + q = 0    =>  4p⁴ - 2 p³ -2p²  = 0 ,   
one possibility ,  p = 0, and q = 0.  cancel 2 p² in above equation.
  2 p² - p -1 = 0
    p =  [ 1 + - sqroot(1 +8) ] / 4  =  1  or  -1/2
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let roots be = s+1/2 and  s - 1/2
   sum of roots =  a = 2s    =>  s = a/2
       product of roots  =  b = s² - 1/4  = (a²-1)/4  =>  a² = 1+4b
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   if x is real, then discriminant >= 0,,  so a² - 4b >= 0 
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$Y=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\\\\As\ all\ are\ positive\ and\ Y\ is\ symmetric\ in\ a,b,\ and\ c,\\\\Minimum\ will\ be\ when\ a=b=c\\\\Y\ minimum=2+2+2=6\\\\Alternately,\ differentiate\ wrt\ a,\ treating\ b\ and\ c\ as\ constants.\\\\-(b+c)/a^2+1/b+1/c=0\\\\solve\ to\ get\ a^2=bc\\\\similarly\ b^2=ca,\ \ c^2=ab\\then\ a=b=c$