We know that $$x_1,x_2,x_3,x_4,x_5,x_6$$ are positive numbers.

For any two positive numbers a and b, $$(a - b)^2 \geq 0$$

=> $$a^2 + b^2 - 2ab \geq 0$$

=> $$a^2 + b^2 \geq 2ab$$

In our question, $$(x_1 - x_4)^2 \geq 0$$ => $${x_1}^2 + {x_4}^2 \geq 2 x_1x_4$$

Similarly, $$(x_2 - x_5)^2 \geq 0$$ => $${x_2}^2 + {x_5}^2 \geq 2 x_2x_5$$

and $$(x_3 - x_6)^2 \geq 0$$ => $${x_3}^2 + {x_6}^2 \geq 2 x_3x_6$$

Adding the above three equations, we get

$${x_1}^2 + {x_2}^2 + {x_3}^2 + {x_4}^2 + {x_5}^2 + {x_6}^2 \geq 2(x_1x_4 + x_2x_5 + x_3x_6)$$

=> $$840 + 840 \geq 2(x_1x_4 + x_2x_5 + x_3x_6)$$

=> $$x_1x_4 + x_2x_5 + x_3x_6 \leq 840$$

So, the maximum value is 840. But, this holds true only when $$x_1=x_4$$, $$x_2=x_5$$ and $$x_3=x_6$$. In the question, it is given that the numbers are unique. So, 840 cannot be attained. 839 is the next largest integer that can be attained by the expression $$x_1x_4 + x_2x_5 + x_3x_6$$ with all its terms unique. Hence 839 is the answer.

$$\frac{x}{(x^2 + x-56)} = \frac{x}{(x - 7)(x + 8)}$$ This is greater than 0 only if $$x(x-7)(x+8) > 0$$

From the graph we can see that, -8 < x < 0 and x > 7. Option a)

x + |y| = 5 can be divided into two lines x + y = 5 and x - y = 5, depending on whether y is positive or negative respectively.
|x| + y = 3 can be divided into two lines x + y = 3 and - x + y = 3, depending on whether x is positive or negative respectively.
The graph of these four lines is shown in the diagram.

We have to find the number of integral points in the shaded region.

The area of the shaded region can be found by finding the area of the two triangles formed and subtracting the area of the smaller triangle from the larger triangle.

The larger triangle has coordinates (0,5), (5,0) and (0,-5) => Area = $$\frac{1}{2}*10*5$$ = 25 sq units.

We know that two coordinates of the smaller triangle are (0,3) and (0,-5). To find the third coordinate, we need to find the point of intersection of x + y = 3 and x - y = 5.

These two lines intersect at (4,-1)

Hence, the area of the smaller triangle is equal to $$\frac{1}{2}*4*8$$ = 16 square units.

So, the area of the required region = 25 -16 = 9 sq units.

Now, we need to find the number of boundary points to find the number of interior integral points using Pick’s theorem.

Points on x+y=3 are (4,-1), (3,0), (2,1),(1,2),(0,3).

Points on y axis (0,4), (0,5).

Points on x - y = 5 are (4, -1) and (5,0)

Point on x + y = 5 are (5,0), (4,1), (3,2), (2,3), (1,4)

Out of these, there are 12 distinct integral points.

A = I + (12/2) - 1

=> 9 = I + (12/2) - 1

=> I = 9-6+1 = 4 => There are 4 points inside the given area.

We have to add the number of points on the boundary as the equality sign also satisfies the given conditions.

Hence, there are 4+ 12 = 16 points that satisfy the given conditions.

$$\frac{(x^2+7x-44)}{(x^2+9x-136)}$$ has two quadratic expressions that have to be factorized.

$$(x^2+7x-44)$$ can be factorized into (x+11)(x-4)

$$(x^2+9x-136)$$ can be factorized into (x+17)(x-8)

Here, -11, 4, -17 and 8 are the roots of their respective expressions.

$$\frac{(x^2+7x-44)}{(x^2+9x-136)} < 0$$

=> $$\frac{(x+11)(x-4)}{(x+17)(x-8)} < 0$$

Here the flow diagram is recommended to be used. Draw a number-line marking the roots on them as -17, -11, 4 and 8 in order of their magnitude. Draw a curve as shown in the diagram. The curve must start from right top if the sign before the quadratic expression is positive and from right bottom if the sign before the quadratic is negative. Here the sign is positive and hence the curve starts at the right top. In the region above the number-line, the value of the fraction is positive and in the region below the number line the value of the fraction is negative.

We need the integral values of x where the value of the fraction is negative.

According to the diagram, the value of the fraction is negative in the region (-17,-11) and (4,8). So the value is less than 0 at x = -16, -15, -14, -13, -12, 5, 6 and 7.

Hence there are 8 solutions in total.

Alternate Explanation:

The equation can be factorized as $$\frac{(x-4)(x+11)}{(x-8)(x+17)}$$

The equation is false for all x>8. Between 4 and 8, both values excluded, the inequality will hold true. Between -11 to 4, the inequality will not hold.

Between -11 and -17, again both values excluded, the inequality will be true.

Hence the set of values for which this is true is {-16, -15, -14, -13, -12, 5, 6, 7}.

Hence number of integral values is 8.

Since, we need to minimise the number of gold coins received by the child who received the maximum number of gold coins and it is known that each child received distinct number of gold coins, therefore the number of gold coins received by the children have to be consecutive.
Assume that the number of gold coins received by seven children be (n - 3), (n - 2), (n - 1), (n), (n + 1), (n + 2) and (n + 3)
Given that any four of her children together received more number of gold coins than remaining three children together.
So, if (n - 3) + (n - 2) + (n - 1) + n > (n + 1) + (n + 2) + (n + 3), then the condition given above will be always true.
Or, 4n - 6 > 3n + 6. Therefore, n > 12. So min n=13
Therefore, least number of gold coins that could have been received by the child who received maximum number of gold coins is equal to (n + 3) = 16.

||x - |x - 2| + 3| - 4| < 3
=> -3 < |x - |x - 2| + 3| - 4 < 3
=> 1 < |x - |x - 2| + 3| < 7
If x >=2, the inequality becomes: 1 < | x - x + 2 + 3 | < 7
=> 1 < 5 < 7, which is true. So, for all x >=2, the inequality holds good

For x < 2, the equation becomes: 1 < | x + x - 2 + 3 | < 7
=> 1 < |2x + 1| < 7
For x >= -1/2, the inequality becomes 1 < 2x + 1 < 7 => 0 < 2x < 6 => 0 < x < 3
So, the intersection is 0 < x < 2
For x < -1/2, the inequality becomes 1 < -2x - 1 < 7 => 2x < -2 and 2x > -8
=> x < -1 and x > - 4
So, the intersection is -4 < x < -1
Let us check for the boundary cases: if x = -4, the expression doesn’t hold good
For x = -1 also, the expression doesn’t hold good. So, x takes 2 negative integral values which are -2 and -3.

Let $$\frac{y+5}{y^2+5y+25} = k$$

So, $$ky^2+5ky+25k-y-5=0$$

For the above quadriatic equation to have real roots, its discriminant should be greater than or equal to zero.

$$(5k-1)^2 \geq 4*k*5*(5k-1)$$
$$25k^2-10k+1\ge20k\left(5k-1\right)$$
$$25k^2-10k+1\ge100k^2-20k$$
$$75k^2-10k-1\le0$$

solving we get
$$(15k+1)(5k-1)\leq 0 $$
So $$-1/15 \leq k \leq 1/5$$

So, the difference in the maximum and minimum value is 4/15

$$\frac{(x-5)(x-3)}{(x+2)(x+6)} < 0$$
Divide the number line into five parts as follows: x < -6, -6 < x < -2, -2 < x < 3, 3 < x < 5 and x > 5
The expression $$\frac{(x-5)(x-3)}{(x+2)(x+6)}$$ is less than 0 and greater than 0 in alternate intervals.
The expression is greater than 0 for x > 5. So, it is less than 0 for 3 < x < 5, greater than 0 for -2 < x < 3, less than 0 for -6 < x < -2 and it is greater than 0 for x < -6
So, the integral values of ‘x’ that satisfy the inequality are -5, -4, -3 and 4.
So, four values satisfy the inequality.

If you carefully look at the denominator, it is {(x-9)}^2 which is always positive.
Hence numerator would be negative.
$$x^2-13|x|+30<0$$
(|x|-10)(|x|-3)<0
3<|x|<10
x = (3,10)U(-10,-3)

There are total of 12 integral solutions.

But this solution set contains 9 and denominator becomes zero at this value. So the total solutions are 11

a*(1/b+1/c)+b*(1/c+1/a)+c*(1/a+1/b) equals (a/b+b/a)+(b/c+c/b)+(a/c+c/a).
All these terms are at their minimum when a = b = c and the minimum value of the sum is 6

Option b) is the correct answer.

$$Y^2+4Y+5$$ is always greater than 0.

So, the inequality becomes $$|Y^2+6Y+2| > 4(Y^2+4Y+5)$$

Condition 1

If $$Y^2+6Y+2 > 0$$, 

we can write the inequality as $$Y^2+6Y+2 > 4(Y^2+4Y+5)$$.

The inequality becomes $$0>3Y^2+10Y+18$$, which has no real roots as $$\sqrt{\ b^2-4ac}$$<0.

Condition 2

If $$Y^2+6Y+2 < 0$$, {the range  of the equation to satisfy the condition y {-0.354, -5.6 }

we can write the inequality as $$-Y^2-6Y-2 > 4(Y^2+4Y+5)$$.

the inequality becomes $$5Y^2+22Y+22<0$$, we can find the range of y {-1.53, -2.8 } which has only one integral root which is -2.

Hence, the answer is 1

[x]+[10-x] = 10 + [x]+[-x]. [x]+[-x] is 0 when x is an integer and is equal to -1 if x is not an integer. 

10 is a constant and any variation in [10-x] will result due to x. Therefore, we can write [10-x] as 10+[-x].

 So, [x]+[10-x] $$\geq$$ 10 when x is an integer. 

So, the number of values is 19

Let the initial number of red, blue and green marbles be 2x, 3x and 7x respectively.
Let the new number of red, blue and green marbles be 4y, 5y and 9y
We know that the number of green marbles has decreased.
7x > 9y => x > 9y/7
We also know that number of red and blue marbles have increased. So
4y > 2x => x < 2y
and
5y > 3x => x < 5y/3
Combining the three equations, we get
9y/7 < x < 5y/3
For minimizing the number of red marbles, we have to minimize (4y- 2x)
This can be written as 2(2y- x)
Hence we have to minimize 2y - x, subject to the constraints 9y/7 < x < 5y/3
We know that x and y have to be integers.
For y = 1, we have 9/7 < x < 5/3
There is no integral value of x which lies in this range.
When y = 2, we can get x = 3
Hence y = 2, x = 3 are the values at which the number of red marbles will be minimized
Hence, the minimum number of marbles to be added = 4*2 - 2*3 = 2

3x+2y+z = 3x+y+y+(z/3+z/3+z/3)

We know that AM $$\geq$$ GM
$$[3x+y+y+(z/3+z/3+z/3)]/6\geq (3xy^2z^3/27)^{1/6}$$
$$[3x+y+y+(z/3+z/3+z/3)]/6\geq (2^{12}*9/9)^{1/6}$$
$$[3x+y+y+(z/3+z/3+z/3)]/6\geq 4$$
$$[3x+y+y+(z/3+z/3+z/3)] \geq 24$$

For x<2, the equation becomes 2-x < 3-x => 2 < 3 which is true for all values between 0 and 2.

For 2<x<3, the equation becomes x-2 < 3-x => x<2.5. So the equation is true for all values between 2 and 2.5

For x> 3, the equation becomes x-2<x-3 => -2 < -3 which is false. So the equation has no solution from 3 to 4.

Thus the equation is valid for (0, 2) and (2, 2.5) inside the interval (0, 4) which makes the probability = $$\frac{2.5}{4}=\frac{5}{8}$$.

$$\frac{6x^2-19x+8}{x^2+x+7}<0$$
Now the denominatior $$x^2+x+7$$ is always greater than 0 for any value of x.
It means the numerator should be less than zero
$$6x^2-19x+8<0$$
$$6x^2-16x-3x+8<0$$
(2x-1)(3x-8)<0

The only integers satisfying this inequality are 1 and 2.
Sum = 1+2 = 3

We have $$\log_{2}\frac{3x-7}{2x+3} \leq 0$$
=> $$\frac{3x - 7}{2x + 3}\leq 1$$
=> 3x - 7 ≤ 2x + 3
=> x ≤ 10
=> We know that logarithm is defined only for positive integers, so
=> $$\frac{3x - 7}{2x + 3}\geq 0$$
=> x > 7/3 or x < -3/2 but we have been asked only about the positive solutions so we can ignore the negative x. Hence x > 7/3
So x can take all integer values between 7/3 and 10 (10 included).
Hence required values are 3,4,5,6,7,8,9 and 10
So correct answer is 8.

For x<2, the equation becomes 2-x < 3-x => 2 < 3 which is true for all values between 0 and 2.

For 2<x<3, the equation becomes x-2 < 3-x => x<2.5. So the equation is true for all values between 2 and 2.5

For x> 3, the equation becomes x-2<x-3 => -2 < -3 which is false. So the equation has no solution from 3 to 4.

Thus the equation is valid for (0, 2) and (2, 2.5) inside the interval (0, 4) which makes the probability = $$\frac{2.5}{4}=\frac{5}{8}$$.

If p = $$y^2+\frac{1}{y^2}$$, then $$p^2=y^4+\frac{1}{y^4}+2$$

K = $$\frac{p^2-1}{p+1}=p-1$$
So, $$K=y^2+\frac{1}{y^2} -1=(y-\frac{1}{y})^2+1$$
So, the least possible value of K is 1 for any real value of y.
Hence, K doesn't take the value of 0.5

 

$$x^2 + (2a+1)x + a^2+1 = (x+a+1/2)^2 + 3/4-a \geq 0 $$
The above inequality is always true if $$a \leq 3/4 $$
So, the largest integral value of 'a' is 0

The inequality is satisfied if $$14 < X^2 < 26$$. The integral values solving that inequality are -4,4,-5 and 5 which is a total of 4 integers.