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kvpy2013solution

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kvpy2013solution

4) Let a,b be non-zero real numbers. Which of the following statements about the following quadratic equation 
                 ax² + (a+b)x + b = 0
is necessarily true ?
i) It has at least one negative root.
ii) It has at least one positive root.
iii) Both its roots are real.
     (A) (i) and (ii) only              (B) (i) and (iii) only
     (C) (ii) and (iii) only             (D) All of them

Solution:
         First we have to factor by grouping given equation.
               ax² + (a+b)x + b = 0
               ax² + ax + bx + b = 0
               ax(x+1) + b(x+1) = 0
               (x+1)(ax+b) = 0
          Equating (x+1) and (ax+b) with zero, we get,
               x+1 = 0      or     ax+b = 0
               x = -1        or     x = \(\frac{-b}{a}\)
      Now we can conclude that one root is negative and the other is real but we cannot say that the other root is positive only because a and b are non-zero numbers. Hence the correct option is
(B).

5) Let x,y,z be non-zero real numbers such that \(\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 7\) and \(\frac{y}{z} + \frac{z}{x} + \frac{x}{z} = 9\) then \(\frac{x³}{y³} + \frac{y³}{z³} + \frac{z³}{x³}\)-3 is equal to
       (A) 152         (B) 153      (C) 154       (D) 155

Solution:
      Let \(\frac{x}{y} = a, \frac{y}{z} = b, \frac{z}{x} = c\).
      Then,   a + b + c = 7 and \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\) = 9.
      Now we have to find (a+b+c)².
               (a+b+c)² = a²+b²+c²+2ab+2bc+2ca 
               7² = a²+b²+c²+2(ab+bc+ca)   -------(1)
       Now we have to find the value of ab+bc+ca.
               ab+bc+ca = \(\frac{x}{y}*\frac{y}{z} + \frac{y}{z}*\frac{z}{x} + \frac{x}{y}*\frac{z}{x}\)
               ab+bc+ca = \(\frac{x}{z} + \frac{y}{x} + \frac{z}{y}\)
               ab+bc+ca = 9
        Now plug the value of ab+bc+ca in equation (1), we get,
               49 = a²+b²+c²+2(9)
               a²+b²+c² = 49-18
               a²+b²+c² = 31
         Now we have to find the value of abc.
               abc = \(\large \frac{x}{y}*\frac{y}{z}*\frac{z}{x}\)
               abc = 1
         Now,
               a³+b³+c³-3abc = (a+b+c)(a²+b²+c²-ab-bc-ca)
               a³+b³+c³-3 = 7(31-(9)) 
               a³+b³+c³-3 = 7(22)
               a³+b³+c³-3 = 154
               \(\large \frac{x³}{y³}+\frac{y³}{z³}+\frac{z³}{x³}\)-3 = 154 

    Hence the correct option is (C).

6) In a triangle ABC with \(\angle A\) < \(\angle B\) < \(\angle C\), points D,E,F are the interior of segments BC,CA,AB respectively. Which of the following triangles CANNOT BE similar to ABC.
    A) Triangle ABD              B) Triangle BCE
    C) Triangle CAF              D) Triangle DEF

Solution :
   Here we see that \(\angle A\) is the smallest angle in the triangle ABC so if we draw a rough figure for this it may look like this.


    First let us see if the first option that is \(\triangle ABD\) is similar or not with the \(\triangle ABC\).
    We know that in similar triangles all the corresponding angles are equal. But in \(\triangle ABC\) and \(\triangle ABD\), \(\angle BAC\) \(\neq \angle BAD\). Hence \(\triangle ABC\) is not similar with \(\triangle ABD\). So the correct option is (A).


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