JEE 2014 :: Advanced 2014 :: Mathematics 2014

• Advanced 2014 - Physics 2014
• Advanced 2014 - Chemistry 2014
• Advanced 2014 - Mathematics 2014

 In a triangle, the sum of two sides is x  and the product of the same two sides is y. If x²-  c²  = y,  where c is the third side of the triangle, then the ratio of the inradius to the circumradius  of the triangele is 

Answer:

Explanation:

  Plan

    (i)  $\cos C=\frac{a^{2}+b^{2}-c^{2}}{2ab}$

(ii)    R= $\frac{abc}{4\triangle},r=\frac{\triangle}{s}$

  where, R,r,$\triangle$  denote the circumradius , inradius and area of traingle , respectively

 Let the sides of triangle  be a,b, and c 

 Given  x=a+b+c

     y=ab

   x²-c²=y

     $\Rightarrow$         (a+b)²-c²=y

$\Rightarrow$      a²+b²+2ab-c²=ab

$\Rightarrow$     a²+b²-c²=-ab

$\Rightarrow   $     $\frac{a^{2}+b^{2}-c^{2}}{2ab}=-\frac{1}{2}$

                                  = $\cos 120^{0}$

$\Rightarrow$                   $\angle  C= \frac{2\pi}{3}$

$\therefore$                        R= $\frac{abc}{4\triangle},r=\frac{\triangle}{s}$

$\Rightarrow$     $  \frac{r}{R}=\frac{4\triangle^{2}}{s(abc)}$

$=\frac{4\left[\frac{1}{2}ab\sin\left(\frac{2\pi}{3}\right)\right]^{2}}{\frac{x+y}{2}.y.c}$

     $\therefore $    $ \frac{r}{R}=\frac{3y}{2c(x+c)}$

Three boys and two girls stand in a queue. The probability  that the number of boys ahead of every girl is at least  one more than the number of girls ahead  of her, is 

Answer:

Explanation:

Plan (i)  The number of arrangement of m distinct objects is given by n!.

 (ii) Probability =

                      NUmber of favourable outcomes/ Number of total out comes

  Total number of ways  to arrange 3 boys and 2 girls  are 5!

 According to given condition , following cases may be

   BGGBB

   GGBBB

   GBGBB

   GBBGB

   BGBGB

  So, number of favourable ways

      = 5! x 3! x 2!

  = 60

 $\therefore$   Required probability

                    =    $\frac{60}{120}=\frac{1}{2}$

Box I contains three cards bearing numbers 1,2,3: box II  contains five cards bearing numbers 1,2,3,4,5:  and box III  contained seven cards bearing numbers 1,2,3,4,5, 6,7. A card is drawn from each of the boxes. Let x_i be the numbers on the cars drawn from i th box i=1,2,3.

The probability that x₁,x₂,x₃  are in an arithmetic progression is

Answer:

Explanation:

 Plan   x₁,x₂,x₃    atr in A.P

 then x₂-x₁=x₃-x₂

  $\Rightarrow$     2 x₂= x₁+x₃

 $\therefore$   x₁,x₂,x₃  are in A.P

  x₁₊ x₃   =2x₂

  So, x₁ +x₃   should be even number

 Either both x₁ and x₃ are odd or both are even.

$\therefore$   Required probability 

       $=\frac{^{2}C_{1}\times^{4}C_{1}+^{1}C_{1}\times^{3}C_{1}}{^{3}C_{1}\times^{5}C_{1}\times^{7}C_{1}}$

     = $\frac{11}{105}$

Box I contains three cards bearing numbers 1,2,3: box II  contains five cards bearing numbers 1,2,3,4,5  and box III  contained seven cards bearing numbers 1,2,3,4,5, 6,7. A card is drawn from each of the boxes. Let x_i be the numbers on the card drawn from i th box i=1,2,3.

The probability that x₁ +x₂+x₃  is odd is

Answer:

Explanation:

 Plan 

  Probability =  Number of favourable outcomes /  Number of total outcomes

  As   x₁ +x₂+x₃  is odd 

 So, all may be odd or one  of them is odd and other two are even

 $\therefore$ Required probability

$\frac{^{2}C_{1}\times^{3}C_{1}\times^{4}C_{1}+^{1}C_{1}\times^{2}C_{1}\times^{4}C_{1}+^{2}C_{1}\times^{2}C_{1}\times^{3}C_{1}+^{1}C_{1}\times^{3}C_{1}\times^{3}C_{1}}{^{3}C_{1}\times^{5}C_{1}\times^{7}C_{1}}$

        = $\frac{24+8+12+9}{105}=\frac{53}{105}$

Let $n\geq2$   be an integer .Take n distinct points on a circle and join each pair of points by a line segment. Colour the line segment joining every pair of adjacent points by blue and the rest by red. If the numbers  of red  and blue line segments are equal, then the  value of n is 

Answer:

Explanation:

Plan Number of line segment joining pair of adjacent point =n

 Number of red k=line segment = $^{n}C_{2}-n$

 Number of blue line segments= n

   $\therefore$    $^{n}C_{2}-n=n$

$\Rightarrow$              $\frac{n(n-1)}{2}=2n$

$\Rightarrow$           n=5

• Advanced 2014 - Physics 2014
• Advanced 2014 - Chemistry 2014
• Advanced 2014 - Mathematics 2014