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MATHMAA

# 16th KVJ Solutions3-4

## Probability Introduction :

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3Q) If squares of the roots of $$x^{4}+bx^{2}+cx+d=0$$ are $$\alpha ,\beta ,\gamma ,\delta$$, then prove that:

$$64 \alpha \beta \gamma \delta -[4\sum\alpha\beta -(\sum\alpha)^{2}]^{2} = 0$$.

Solution:

$$x^{4}+bx^{2}+cx+d=0$$......(1)

Let y be the root of  transformed equation.

y=x^2

$$x^{4}+bx^{2}+d=-cx$$ since bring all even powers one side and odd powers other side.

Squaring on both sides, we get

$$(x^{4}+bx^{2}+d)^{2}=c^{2}x^{2}$$

$$(y^{2}+by+d)^{2}=c^{2}y$$ since  $$y=x^{2}$$

$$y^{4}+b^{2}y^{2}+d^{2}+2by^{3}+2bdy+2dy^{2} =c^{2}y$$

$$y^{4}+2by^{3}+(b^{2}+2d)y^{2}+(2bd-c^{2})y+d^{2}=0$$...(2)

It is the equation whose roots are squares of roots of $$x^{4}+bx^{2}+cx+d=0$$

now for (2)$$\alpha, \beta, \gamma, \delta$$ are the roots.

$$\sum \alpha$$ =-2b

$$\sum \alpha \beta$$=$$b^{2} +2d$$

$$\sum \alpha \beta \gamma$$ =-(2bd-c^2)

$$\alpha \beta \gamma \delta$$ = $$d^{2}$$

Now plug the values in required function.

$$64 \alpha \beta \gamma \delta -[4\sum\alpha\beta -(\sum\alpha)^{2}]^{2}$$ = $$64d^{2} -(4(b^{2}+2d)- (-2b)^{2})^{2}$$

=$$64d^{2} - (4b^{2} +8d -4b^{2})^{2}$$

=$$64d^{2}-(8d)^{2}$$

=$$64d^{2}- 64d^{2}$$

= 0

Hence it is proved.

16th KVJ Solutions3-4

4Q) a) Let a, b,c be the length of the sides of a triangle and r be the in-radius. Show that:

r< $$\frac{a^{2}+b^{2}+c^{2}}{3(a+b+c)}$$.

Solution:

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b) A family consists of a grand father, 6 sons & daughters and 4 grand children. They are to be seated in a row for a dinner. The grand children wish to occupy the two seats at each end and the grand father refuses to have a grand child on either side of him. Determine the number of ways in which they can be seated for the dinner.

Solution b) :

Family consists of

Grand father  =1

Sons and daughters = 6

Grand children  =4

Total =11 members

- - - - - - - - - - -   11 places to arrange them in row.

Grand children wish to occupy end seats .

First two and last two seats are occupied by grand children.

They can be arranged in 4x3x2x1 ways =24ways.

grand father refuses to sit behind grand children means he has to be seated 5 places in between as he cannot sit in first 3 and last 3 places.

He can be seated in 5 ways.

Remaining Sons and daughters(6) can be arranged in 6! ways=720 ways

Total number of arrangements =24 x5x720 =86400 ways.

Therefore all members can be seated for dinner in 86,400 ways.

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