CAT Inequalities Questions

There are two critical points for the inequality to consider: x=-5 and x=3/2

Region I: x is greater than 3/2
In this scenario, both the terms would be positive; cross-multiplying, we get the relation $$2x-3\le x+5$$
Giving the boundary $$x\le8$$, hence giving us the valid range as $$\frac{3}{2}<x\le8$$

Region II: $$-5<x<\frac{3}{2}$$
In this case, the right-hand side will be a negative value, and hence, the sign would change when multiplying, giving the inequality
$$2x-3\ge x+5$$
Which will give x>8, which is out of bounds for this region 

Another way is to put a value in the region to check for the validity of the inequality; by putting x=0, we could see that the inequality does not hold in this region 

Region III: x less than -5
In this scenario, both the terms are negative, essentially giving us the same boundary as region 1; we take the lower bounds, giving us that x has to be less than 5

Therefore, for the given inequality to hold true x<-5 or $$\frac{3}{2}<x\le8$$

Hence, Option A is the correct answer. 

We have two inequalities,

$$y\ \ge\ x\ +\ 4$$

$$-4\ \le\ x^2\ +\ y^2\ +\ 4\left(x\ -\ y\right)\ \le\ 0$$

The second inequality can be written as two separate inequalities,

$$-4\ \le\ x^2\ +\ y^2\ +\ 4\left(x\ -\ y\right)$$ and $$\ x^2\ +\ y^2\ +\ 4\left(x\ -\ y\right)\ \le\ 0$$

The first inequality can be written as,

$$\ x^2\ +\ y^2\ +\ 4x\ -\ 4y\ +4\ \ge\ 0$$

$$\ x^2+\ 4x\ +4\ +\ y^2-\ 4y\ +4\ -4\ge\ 0$$

$$\left(x\ +\ 2\right)^2\ +\ \left(y\ -\ 2\right)^2\ \ge\ 4$$

The second inequality can be written as,

$$\ x^2\ +\ y^2\ +\ 4x\ -\ 4y\ \ \le\ \ 0$$

$$\ x^2+\ 4x\ +4\ +\ y^2-\ 4y\ +4\ -8\ \le\ \ 0$$

$$\left(x\ +\ 2\right)^2\ +\ \left(y\ -\ 2\right)^2\ \le\ \ 8$$

Representing all three inequalities in the graph, we get the graph as shown above, and we must calculate the intersection of all three inequalities,

We can see that the line passes through the centre of both circles (-2, 2), and the area obtained from the second inequality is the area between the two circles. So, the area of intersection of all three graphs is half of the area between both the circles as the line divides the circle in half, and we must only consider the area above the line as per the given inequality. We know that the area of the bigger circle is $$\sqrt{\ 8}$$ and the area of the smaller circle is $$\sqrt{\ 4}$$ from the equations of the circles as we know that equation of circle as $$\left(x\ -\ a\right)^2\ +\ \left(y\ -\ b\right)^2\ =\ \left(radius\right)^2$$ where (a,b) is the centre of the circle.

The area of intersection 

= $$\dfrac{1}{2}$$ (Area of bigger circle - Area of smaller circle)

= $$\dfrac{1}{2}\left(\pi\ \left(\sqrt{\ 8}\right)^2\ -\pi\ \left(\sqrt{\ 4}\right)^2\right)$$

= $$\dfrac{1}{2}\left(8\pi\ \ -4\pi\right)$$

= $$\dfrac{1}{2}\left(4\pi\ \right)$$

= $$2\pi\ $$

Therefore, the correct answer is option A.

The moduli will give out only non-negative outputs, and since we are to consider only integer values of x and y, this drastically reduces the possible cases. 

We can get 2 from either 2+0 or 1+1

We get a 2+0 form when either the first term or the second term is 0
The second term is 0; this is when x=y, in this case,|2x|=2, where x can be 1 or -1; therefore, the two cases are (1,1) and (-1,-1)
The first term is 0; this is the case when x = -y, in this case, |x- (-x)|=2, giving x=1 or -1 yet again, here the two cases are (1,-1) and (-1,1)

The other way we can get 2 is through 1+1

This is possible when one of the terms is 0; if y=0, |x|+|x|=2, where x can be 1 or -1, giving two cases (1,0) and (-1,0)
Similarly,y for x=0, we get two cases, (0,1) and (0,-1)

Therefore, there are 8 pairs of (x,y) that satisfy the given equation. 

We can consider the quadrants of a graph:

First quadrant: Both x and y are positive
This would change the equation to 2x+y=15 and x=20, giving a negative value of y; hence, this is not the case. 

Second quadrant: x is negative, but y is positive
This would change the equations to y=15 and x=20, giving a positive value of x, which hence can not be the case.  

Third quadrant: Both x and y are negative
This would change the equation to y=15 and x-2y=20; this gives a positive value of y and hence can not be the case. 

Fourth quadrant: x is positive, but y is negative
This would change the equations to 2x+y=15 and x-2y=20; this gives the value of x as 10 and y as -5, which would lie in the fourth quadrant. 

The value of x-y would be 10-(-5)=15

Therefore, Option B is the correct answer. 

It is given that $$\frac{x}{y}<\ \frac{\ x+3}{y-3}$$, which can be written as $$\frac{x}{y}-\frac{\ x+3}{y-3}<0$$

=> $$\ \frac{\ x\left(y-3\right)-y\left(x+3\right)}{y\left(y-3\right)}<0$$

=> $$\ \frac{\ xy-3x-xy-3y}{y\left(y-3\right)}<0$$

=> $$\ \frac{\ -3\left(x+y\right)}{y\left(y-3\right)}<0$$

=> $$\ \frac{\ 3\left(x+y\right)}{y\left(y-3\right)}>0$$

From this inequality, we can say that, when $$y<0=>y(y-3)>0$$. Now to satisfy the given equation $$\ \frac{\ 3\left(x+y\right)}{y\left(y-3\right)}>0$$, 

$$(x+y)$$ must be greater than zero Hence, $$x>0$$ and $$\left|x\right|\ >\left|y\right|$$

Therefore, the magnitude of $$x$$ is greater than the magnitude of $$y$$. 

Hence, $$x>y$$, and $$\left|x\right|\ >\ \left|y\right|$$ =>$$-x\ <\ y$$ (Since the magnitude of $$x$$ is greater than the magnitude of $$y$$.)

The correct option is B.

It is given that there are exactly 41 numbers, which can be expressed as the power of two, and exist between $$8^m$$ and $$8^n$$, (where m, and n are positive integers, and m < n)

Hence, $$2^{3m}\ <\ \text{41 numbers }<\ 2^{3n}$$

Since, m is a positive integer, the least value of m is 1. Therefore, $$2^{3m\ }=2^3$$, hence, the 41 numbers between them are $$2^4,\ 2^5,\ 2^6,...,2^{44}$$.

Then the lowest possible value of $$8^n$$ is $$2^{45}$$. Hence, the smallest value of n is $$2^{45}\ =\ 8^n\ =>\ 2^{3n\ }=2^{45\ }=>\ n\ =\ 15$$

Hence, the smallest value of m+n is (15+1) = 16

The correct option is D

Given, $$x^{2} + (x - 2y - 1)^{2} = -4y(x + y)$$

=> $$x^2+4xy+4y^2+\left(x-2y-1\right)^2=0$$

=> $$\left(x+2y\right)^2+\left(x-2y-1\right)^2=0$$

For the L.H.S. of the equation to be 0, each of the square terms should be 0 (as squares cannot be negative)

=> x - 2y - 1 = 0 => x - 2y = 1

It is given that $$5^{n-1} < 3^{n + 1}$$, where n is a natural number. By inspection, we can say that the inequality holds when n = 1, 2, 3 4, and 5.

Now, we need to find the least integer value of m that satisfies $$3^{n+1} < 2^{n+m}$$

For, n =1, the least integer value of m is 3.

For, n = 2, the least integer value of m is 3

For, n = 3, the least integer value of m is 4.

For, n = 4, the least integer value of m is 4.

For, n= 5, the least integer value of m is 5.

Hence, the least integer value of m such that for all the values of n, the equation holds is 5.

Let us consider 3 cases:

1) x = 0, This is a solution, as both L.H.S and R.H.S will be equal (0) when x = 0. (1 solution)

2) x > 0

=> $$2x\left(x^2+1\right)=5x^2$$

=> $$2\left(x^2+1\right)=5x$$

=> $$2x^2-5x+2=0$$ => $$2x^2-4x-x+2=0$$

=> $$2x\left(x-2\right)-1\left(x-2\right)=0$$

=> $$\left(x-2\right)\left(2x-1\right)=0$$ => x = 2 or 1/2 => (1 integer solution)

3) x < 0

=> $$-2x\left(x^2+1\right)=5x^2$$

=> $$2x^2+5x+2=0$$

=> $$2x^2+4x+x+2=0$$

=> $$2x\left(x+2\right)+1\left(x+2\right)=0$$

=> $$\left(x+2\right)\left(2x+1\right)=0$$ => x = -2 or -1/2 => (1 integer solution)

So, the total number of integer solutions are 0, 2, -2 => 3.

It is given that the population of the town in 2020 was 100000. The population decreased by y% from the year 2020 to 2021 and increased by x% from the year 2021 to 2022, where x and y are two natural numbers.

Hence, the population in 2021 is $$100000\left(\ \frac{\ 100-y}{100}\right)$$.

The population in 2022 is $$100000\left(\ \frac{\ 100-y}{100}\right)\left(\ \frac{\ 100+x}{100}\right)$$

It is also given that the population in 2022 was greater than the population in 2020 and the difference between x and y is 10.

Hence, 

$$100000\left(\ \frac{\ 100-y}{100}\right)\left(\ \frac{\ 100+x}{100}\right)>\ 100000$$, and (x-y) = 10

=> $$100000\left(\ \frac{\ 100-y}{100}\right)\left(\ \frac{\ 110+y}{100}\right)>\ 100000$$

=> $$\ \frac{\ 100-y}{100}\left(\ \frac{\ 110+y}{100}\right)>\ 1$$

To get the minimum possible value of 2021, we need to increase the value of y as much as possible.

Hence, $$\left(\ \ 100-y\right)\left\{\left(\ \ 100+y\right)+10\right\}>\ 10000$$

=> $$10000-y^2+1000-10y>\ 10000$$

=> $$y^2+10y<1000$$

=> $$y^2+10y+25<1025$$

=> $$\left(y+5\right)^2=1024\ <\ 1025$$

=> $$\left(y+5\right)^2=32^2$$

=> $$y=27$$

Hence, the population in 2021 is 100000*(100-27) = 73000

The correct option is C

Given that the average marks of 4 girls and 6 boys is 24.

Let us assume 'b' is the marks scored by a boy and 'g' is the marks scored by a girl.

=> 4g + 6b = 10*24 = 240 ---(1)

Given that, $$b\le g\le2b$$

We need to find the distinct possible values of 2g + 6b = 2g + 240 - 4g = 240 - 2g.

From (1)

when b = g => 10g = 240 => g = 24

when b = g/2 => 7g = 240 => g = 240/7

=> 240 - 2g varies from 240 - 2*24 to 240 - 2*240/7

=> 171.42 to 192 => Integer values of 172 to 192 => 21 values.

It is given that if a certain amount of money is divided equally among n persons, each one receives Rs 352. Hence, the total amount of money is (352*n) = 352n ... Eq(1)

It is also known that if two persons receive Rs 506 each and the remaining amount is divided equally among the other persons, each of them receives less than or equal to Rs 330

Hence, the maximum amount of money with them = 506*2 + (n-2)*330 = 1012+330n-660 = 352+330n

Now, $$352+330n\ \ge\ 352n\ $$

=> $$22n\ \le\ 352$$ 

=> $$n\ \le\ 16$$

Hence, the maximum value is 16

Given that 2pq - 20 = 52 - 2pq => 4pq = 72 => pq = 18 ----(1)

Now, $$p^2+q^2-29=2pq-20$$ => $$p^2+q^2-2pq=9$$ => $$\left(p-q\right)^2=9$$ => $$p-q=\pm\ 3$$

Also, $$p^2+q^2-29=2pq-20$$ => $$p^2+q^2=2pq+9=2\left(18\right)+9=45$$

Now, $$p^3-q^3=\left(p-q\right)\left(p^2+pq+q^2\right)$$ = $$\left(p-q\right)\left(45+18\right)$$ = $$\left(p-q\right)\left(63\right)$$

=> When p-q = -3 => The value is 63(-3) = -189 and when p-q = 3 => The value is 63(3) = 189.

=> The difference = 189 - (-189) = 378.

Case 1: When $$x^2-3x-10=0$$ and $$x^2-10\ne\ 0$$

$$x^2-3x-10=0\ $$or, $$(x-5)(x+2) = 0$$

or, x= 5 or -2

Case 2: $$x^2-10=1$$

$$x^2-11=0$$

No integer solutions

Case 3: $$x^2-10=-1\ and\ x^2-3x-10\ is\ even$$

$$x^2-9=0$$

or, (x+3)(x-3)=0

or, x= -3 and 3

for x= -3 and +3 $$x^2-3x-10$$ is even

In total 4 values of x satisfy the equations

Let $$\ \log_2n=y$$

$$\ \ \dfrac{\ 4-y}{3-\dfrac{y}{2}}<0$$

$$\ \ \left(4-y\right)\left(3-\dfrac{y}{2}\right)<0$$

$$\ \ \left(4-y\right)\left(6-y\right)<0$$

$$\ \ \left(y-4\right)\left(y-6\right)<0$$

$$4 < y < 6$$

$$4<\log_2n<6$$

$$2^4<n<2^6$$

$$16<n<64$$

n can take values from 17 to 63(inclusive).

The number of n values possible = 47

Let us break this up into 2 inequations [ Let us assume x as m and y as n ]

| 1 + mn | < | m + n |

| m + n | < 5

Looking at these expressions, we can clearly tell that the graphs will be symmetrical about the origin.

Let us try out with the first quadrant and extend the results to the other quadrants.

We will also consider the +X and +Y axes along with the quadrant.

So, the first inequality becomes,

1 + mn < m + n

1 + mn - m - n < 0

1 - m + mn - n < 0

(1-m) + n(m-1) < 0

(1-m)(1-n) < 0

(m - 1)(n - 1) < 0

Let us try to plot the graph.

If we consider only mn < 0, then we get 

But, we have (m - 1)(n - 1) < 0, so we need to shift the graphs by one unit towards positive x and positive y.

So, we have,

But, we are only considering the first quadrant and the +X and +Y axes. Hence, if we extend, we get the following region.

So, if we look for only integer values, we get

(0,2), (0,3),.......

(0,-2), (0, -3),......

(2,0), (3,0), ......

(-2,0), (-3,0), .......

Now, let us consider the other inequation as well, in which |x + y| < 5

Since one of the values is always zero, the modulus of the other value is less than or equal to 4.

Hence, we get 

(0,2), (0,3), (0,4)

(0,-2), (0, -3), (0, -4)

(2,0), (3,0), (4,0)

(-2,0), (-3,0), (-4,0)

Hence, a total of 12 values.

We have $$\mid n - 60 \mid < \mid n - 100 \mid < \mid n - 20 \mid$$.

Now, the difference inside the modulus signified the distance of n from 60, 100, and 20 on the number line. This means that when the absolute difference from a number is larger, n would be further away from that number.

Example: The absolute difference of n and 60 is less than that of the absolute difference between n and 20. Hence, n cannot be $$\le\ 40$$, as then it would be closer to 20 than 60, and closer on the number line would indicate lesser value of absolute difference. Thus we have the condition that n>40.

The absolute difference of n and 100 is less than that of the absolute difference between n and 20. Hence, n cannot be $$\le\ 60$$, as then it would be closer to 20 than 100. Thus we have the condition that n>60.

The absolute difference of n and 60 is less than that of the absolute difference between n and 100. Hence, n cannot be $$\ge80$$, as then it would be closer to 100 than 60. Thus we have the condition that n<80.

The number which satisfies the conditions are 61, 62, 63, 64......79. Thus, a total of 19 numbers.

Alternatively

as per the given condition : $$\mid n - 60 \mid < \mid n - 100 \mid < \mid n - 20 \mid$$.

Dividing the range of n into 4 segments. (n < 20, 20<n<60, 60<n<100, n > 100 )

1) For n < 20.

|n-20| = 20-n, |n-60| = 60- n, |n-100| = 100-n 

considering the inequality part :$$\left|n-100\right|<\ \left|n\ -20\right|$$

100 -n < 20 -n,

No value of n satisfies this condition.

2) For 20 < n < 60.

|n-20| = n-20, |n-60| = 60- n, |n-100| = 100-n.

60- n < 100 - n and 100 - n < n - 20

For 100 -n < n - 20.

120 < 2n and n > 60. But for the considered range n is less than 60.

3) For 60 < n < 100

|n-20| = n-20, |n-60| = n-60, |n-100| = 100-n

n-60 < 100-n and 100-n < n-20.

For the first part 2n < 160 and for the second part 120 < 2n.

n takes values from 61 ................79.

A total of 19 values

4) For n > 100

|n-20| = n-20, |n-60| = n-60, |n-100| = n-100

n-60 < n - 100.

No value of n in the given range satisfies the given inequality.

Hence a total of 19 values satisfy the inequality.

Case 1 : $$x\ge\frac{40}{3}$$
we get 3x-20 +3x-40 =20 
6x=80
x=$$\frac{80}{6}$$=$$\frac{40}{3}$$=13.33
Case 2 :$$\frac{20}{3}\le\ x<\frac{40}{3}\ $$
we get 3x-20+40-3x =20
we get 20=20
So we get x$$\in\ \left[\frac{20}{3},\frac{40}{3}\right]$$
Case 3 $$x<\frac{20}{3}$$
we get 20-3x+40-3x =20
40=6x
x=$$\frac{20}{3}$$
but this is not possible
so we get from case 1,2 and 3
$$\frac{20}{3}\le\ x\le\frac{40}{3}$$
Now looking at options
we can say only option C satisfies for all x .
Hence 7<x<12.

$$f(x) = \frac{x^2 + 2x - 15}{x^2 - 7x - 18}$$<0

$$\frac{\left(x+5\right)\left(x-3\right)}{\left(x-9\right)\left(x+2\right)}<0$$

We have four inflection points -5, -2, 3, and 9.

For x<-5, all four terms (x+5), (x-3), (x-9), (x+2) will be negative. Hence, the overall expression will be positive. Similarly, when x>9, all four terms will be positive.

When x belongs to (-2,3), two terms are negative and two are positive. Hence, the overall expression is positive again.

We are left with the range (-5,-2) and (3,9) where the expression will be negative.

$$(\sqrt[7]{10})(\sqrt[7]{10})^{2}...(\sqrt[7]{10})^{n}>999$$

$$(\sqrt[7]{10})^{1+2+...+n}>999$$

$$10^{\frac{1+2+...+n}{7}}>999$$

For minimum value of n,

$$\frac{1+2+...+n}{7}=3$$

1 + 2 + ... + n = 21

We can see that if n = 6, 1 + 2 + 3 + ... + 6 = 21.

$$x_0=max(x_1,x_2,....,x_{12})$$

$$x_0$$ will be minimum if x1,x2...x12 are close to each other

100/12=8.33

.'. max$$(x_1,x_2,....,x_{12})$$ will be minimum if $$(x_1,x_2,....,x_{12})$$=(9,9,9,9,8,8,8,8,8,8,8,8,)

.'. Option B is correct.

We have 2x + 5y = 99 or $$x=\frac{\left(99-5y\right)}{2}$$

Now $$x\ge\ y\ \ge\ -20$$ ; So $$\frac{\left(99-5y\right)}{2}\ge\ y\ ;\ 99\ge7y\ or\ y\le\ \approx\ 14$$

So $$-20\le y\le14$$. Now for this range of "y", we have to find all the integral values of "x". As the coefficient of "x" is 2,

then (99 - 5y) must be even, which will happen when "y" is odd. However, there are only 17 odd values of "y" be -20 and 14.

Hence the number of possible values is 17.

Now $$\frac{x}{\sqrt{\ 1+x^4}}=\ \frac{\ 1}{\sqrt{\ \ \frac{\ 1+x^4}{x^2}}}=\frac{1}{\sqrt{\ \frac{1}{x^2}+x^2}}$$

Applying A.M>= G.M.

$$\frac{\left(\frac{1}{x^2}+x^2\right)}{2}\ge\ 1\ or\ \ \frac{1}{x^2}+x^2\ge\ 2$$ Substituting we get the maximum possible value of the equation as $$\frac{1}{\sqrt{\ 2}}$$

Now we have $$2601\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=2601\left(\frac{xy+y+x+1}{xy}\right)$$

Now we know that x+y=102. Substituting it in the above equation

$$2601\left(\frac{xy+y+x+1}{xy}\right)=2601\left(\frac{103}{xy}+1\right)$$

Maximum value of xy  can be found out by AM>= GM relationship

$$\ \frac{\ x+y}{2}\ge\ \sqrt{xy}\ or\ \ \sqrt{\ xy}\le\ 51\ or\ xy\le\ 2601$$

Hence the maximum value of "xy" is 2601. Substituting in the above equation we get

$$2601\left(\ \frac{\ 103+2601}{2601}\right)=2704$$

$$\sqrt{\log_{e}{\dfrac{4x - x^2}{3}}}$$ will be real if $$\log_e\ \dfrac{\ 4x-x^2}{3}\ \ge\ 0$$

$$\dfrac{\ 4x-x^2}{3}\ \ge\ 1$$

$$\ 4x-x^2-3\ \ge\ 0$$

$$\ x^2-4x+3\ \le\ 0$$

$$1\le\ x\le\ 3$$

We can see that at n = 2, $$n^3-11n^2+32n-28=0$$ i.e. (n-2) is a factor of $$n^3-11n^2+32n-28$$

$$\dfrac{n^3-11n^2+32n-28}{n-2}=n^2-9n+14$$

We can further factorize n^2-9n+14 as (n-2)(n-7).

$$n^3-11n^2+32n-28=(n-2)^2(n-7)$$

$$\Rightarrow$$ $$n^3-11n^2+32n-28>0$$

$$\Rightarrow$$ $$(n-2)^2(n-7)>0$$

Therefore, we can say that n-7>0

Hence, n$$_{min}$$ = 8

Since the square root can be positive or negative we will get two cases for each of the equation.

For the first one,

a + 3 = 3b .. i

a + 3 = -3b ... ii

For the second one,

a - 1 = 2(b -1) ... iii

a - 1 = 2 (1 - b) ... iv

we have to solve i and iii, i and iv, ii and iii, ii and iv.

Solving i and iii,

a + 3 = 3b and a = 2b  - 1, solving, we get a = 3 and b = 2, which is not what we want.

Solving i and iv

a + 3 = 3b and a = 3 - 2b, solving, we get b = 1.2, which is not possible.

Solving ii and iii

a + 3 = -3b and a = 2b - 1, solving, we get b = 0.4, which is not possible.

Solving ii and iv,

a + 3 = -3b and a = 3 - 2b, solving, we get a = 15 and b = -6 which is what we want.

Thus, $$\frac{a^2}{b^2} = \frac{25}{4}$$

Let the roots of the equation $$x^2+(a+3)x-(a+5)=0 $$ be equal to $$p,q$$

Hence, $$p+q = -(a+3)$$ and $$p \times q = -(a+5)$$

Therefore, $$p^2+q^2 = a^2+6a+9+2a+10 = a^2+8a+19 = (a+4)^2+3$$

As $$(a+4)^2$$ is always non negative, the least value of the sum of squares is 3

$$(n - 5) (n - 10) - 3(n - 2)\leq0$$
=> $$ n^2 - 15n + 50 - 3n + 6 \leq 0$$
=> $$n^2 - 18n + 56 \leq 0$$
=> $$(n - 4)(n - 14) \leq 0$$
=> Thus, n can take values from 4 to 14. Hence, the required number of values are 14 - 4 + 1 = 11.

y = 38 => x = 1

y = 36 => x = 2

...

...

y = 14 => x = 13

y = 12 => x = 14 => Cases from here are not valid as x > y.

Hence, there are 13 solutions.

Try to solve this type of questions using the options.

Subsitute 0 first => We ger -2 <=0, which is correct. Hence, 0 must be in the solution set.

Substitute 8 => 4 + 2 - 2 <=0 => 6 <= 0, which is false. Hence, 8 must not be in the solution set.

=> Option 1 is the answer.

Consider 3 family. Let 1st one have 2 A , 1B and 2 G , 2nd one have 2 A , 2B and 1 G, 3rd one have 2 A , 2B and 1 G . SO total A-6 , B - 5 , g - 4 , F - 3 . Hence minimum is 3 .

Taking lowest possible positive value of m i.e. 1 . Such that a+b+c+d=5 , so atleast one of them must be grater than 1 ,

take a=b=c=1 and d=2

we get $$a^2 + b^2 + c^2 + d^2 = 7$$ which is equal to $$4m^2+2m+1$$ for other values it is greater than $$4m^2+2m+1$$ . so option B

We know $$w = vz /u$$ so taking max value of u and min value of v and z to get min value of w which is -4.

Similarly taking min value of u and max value of v and z to get max value of w which is 4

Take v = 1, z = -2 and u = -0.5, we get w = 4

Take v = -1, z = -2 and  u = -0.5, we get w = -4

For the given expression value of x,y,z are distinct positive integers . So the value of expression will always be greater than value when all the 3 variables are equal . substitute x=y=z we get minimum value of 6 .

$$(x^2(y+z) + y^2(x+z) + z^2(x+y))/xyz$$ = x/z + x/y + y/z + y/x + z/y + z/x

Applying AM greater than or equal to GM, we get minimum sum = 6

The given equations are  x + y + z = 5 -- (1) , xy + yz + zx = 3 -- (2)
xy + yz + zx = 3
x(y + z) + yz = 3
=> x ( 5 -x ) +y ( 5 - x - y) = 3
=> $$ - y^2 - y (5 -x) - x ^2 + 5x = 3$$
=> $$ y^2 + y (x-5) + ( x ^2 - 5x +3)  = 0 $$
The above equation should have real roots for y, => Determinant >= 0

=>$$b^2-4ac>0$$

=> $$ ( x - 5)^2 - 4(x ^2 - 5x +3 ) \geq 0$$
=>  $$ 3x^2 -10x - 13 \leq 0$$
=>  $$ -1 \leq x \leq \frac{13}{3}$$
Hence maximum value x can take is $$\frac{13}{3}$$, and the corresponding values for y,z are $$\frac{1}{3},\frac{1}{3}$$

The equation is not satisfied for only x = 2y.

Using statements B and C, i.e., x = 2z and 2x = z, we see that the equation is not satisfied.

Using statements A and B, i.e., x = 2y and x = 2z, i.e., z = y = x/2, the equation is satisfied.

Option c) is the correct answer.

Substitute value of u = v = 2, w = 4 and m = 1. Here the condition holds and options A and B are false. Hence, we can eliminate options A and B.

Substitute u = v = 1, w=2 and m= 1. Here m=min(u, v, w). Hence, option C also does not hold. Hence, we can eliminate option C.

Option d) is the correct answer.

Let p = q = r = 1

So, the value of the expression becomes 1/3 + 1/3 + 1/3 = 1

If we substitute these values, options a), b) and d) do not satisfy.

Option c) is the answer.

$$xy^2$$ will have it's least value when y=-7 and x=2 and equals 98.
So $$xy^2>98$$

$$x^2y$$ will have it's least value when y=-8 and x=3 and equals -72.
So, $$x^2y > -72$$

$$5xy$$ will have it's least value when y=-8 and x=3 and equals -120
So, $$5xy > -120$$

So, of the three expressions, the least possible value is that of 5xy
 

$$n^3 - 7n^2 + 11n - 5 = (n-1)(n^2 - 6n +5) = (n-1)(n-1)(n-5)$$
This is positive for n > 5
So, m = 6

Since the product is constant, 

(a+b+c+d)/4 >= $$(abcd)^{1/4}$$

We know that abcd = 1.

Therefore, a+b+c+d >= 4

$$(a+1)(b+1)(c+1)(d+1)$$ 
= $$1+a+b+c+d+ ab + ac + ad + bc + bd + cd+ abc+ bcd+ cda+ dab+abcd$$
We know that $$abcd = 1$$
Therefore, $$a = 1/bcd, b = 1/acd, c = 1/bda$$ and $$d = 1/abc$$
Also, $$cd = 1/ab, bd = 1/ac, bc = 1/ad$$
The expression can be clubbed together as $$1 + abcd + (a + 1/a) + (b+1/b) + (c+1/c) + (d+1/d) + (ab + 1/ab) + (ac+ 1/ac) + (ad + 1/ad)$$
For any positive real number $$x$$, $$x + 1/x \geq 2$$
Therefore, the least value that $$(a+1/a), (b+1/b)....(ad+1/ad)$$ can take is 2.
$$(a+1)(b+1)(c+1)(d+1) \geq 1 + 1 + 2 + 2 + 2+ 2 + 2 + 2 + 2$$
=> $$(a+1)(b+1)(c+1)(d+1) \geq 16$$
The least value that the given expression can take is 16. Therefore, option C is the right answer.

Approach 1:

The given expression is symmetric in x and y and the limiting condition (x+y=1) is also symmetric in x and y.

=>This means that the expression attains the minimum value when x = y
x=y=1/2
So, the value = $$(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$$ = $$(2+\frac{1}{2})^2+(2+\frac{1}{2})^2$$ =12.5

Approach 2:

$$(x+1/x)^2$$ + $$(y+1/y)^2$$ = $$(x+1/x+y+1/y)^2$$ - $$2*(x+1/x)(y+1/y)$$

Let x+1/x and y+1/y be two terms. Thus (x+1/x+y+1/y)/2 would be their Arithmetic Mean(AM) and $$\sqrt{(x+1/x)(y+1/y)}$$ would be their Geometric Mean (GM).

Therefore, we can express the above equation as $$(x+1/x)^2$$ + $$(y+1/y)^2$$ = $$4AM^2$$ - $$GM^2$$. As AM >= GM, the minimum value of expression would be attained when AM = GM.

When AM = GM, both terms are equal. That is x+1/x = y +1/y.

Substituting y=1-x we get

x+1/x = (1-x) + 1/(1-x)

On solving we get 2x-1 = (2x-1)/ x(1-x)

So either 2x-1 = 0 or x(1-x) = 1

x(1-x) = x * y

As x and y are positive numbers whose sum = 1, 0<= x, y <=1. Hence, their product cannot be 1.

Thus, 2x-1 = 0 or x=1/2

=> y = 1/2

So, the value = $$(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$$ = $$(2+\frac{1}{2})^2+(2+\frac{1}{2})^2$$ = 12.5

Substitute x=6 and y=-6 ,

x+4y = -18

x = 6, -4y = 24

-4x = -24, 5y = -30

So none of the options out of a,b or c satisfies .

This kind of questions must be solved using the counter example method.

x = 100 and y = -1/2 rules out option a)
x = 3 and y = 0 rules out options c) and d)
Option b) is correct.
 

y = $$\frac{4x-1}{17}$$

The integral values of x for which y is an integer are 13, 30, 47,......

The values are in the form 17n + 13, where $$ n \geq 0$$

17n + 13 < 1000

=> 17n < 987

=> n < 58.05

=> n can take values from 0 to 58 => Number of values = 59

| r-6 | = 11 => r = -5 or 17

| 2q - 12 | = 8 => q = 10 or 2

So, the minimum possible value of q/r = 10/(-5) = -2

Best approach to these type question remain assuming values and checking

Case - 1.x=8 ; y=7 ; z = 5

la (x,y,z) = 12

le (x,y,z) = 2

ma (x,y,z) = 7

Case -2: Let us try to find values for which la(x,y,z) and le(x,y,z) would be equal. In such a case, ma(x,y,z) would also be the same.

So max(x-y,y-z)= min(x+y, y+z)

As x>y>z>0, min(x+y, y+z) = y+z

So max(x-y, y-z) =y+z

Either x-y=y+z or y-z = y+z

So x=2y+z or z=0

But z cannot be 0 according to given condition.

So, x=2y+z

Let us assume y=2 and z=1

So x=5

la (x,y,z)= 3

le (x,y,z) = 3

ma (x,y,z)= 3

based on these two cases we can deduce that non of the given options holds true.

So the correct option to choose is D - None of these.

$$Ma(10, 4, le((la10, 5, 3), 5, 3))$$
Or $$Ma(10, 4, le(8, 5, 3))$$
Or $$Ma (10,4,3)$$
Or $$\frac{1}{2} (6+7) = 6.5$$

Given expression can be reduced to
le(15, min(10,15-9) , le(9,8,12))
Or le(15,6,1) = 9

After solving given equation, we will have inequality resolved to:

(x-1)(x-2)>0

Or we can say range of x will be as follows:

x<1;  x>2

Hence, option A has a set of values which don't lie in the possible range of x.

So the answer will be A.

-n+2 >= 0
or n<=2
and 2n>=4
or n>=2
So we can take only one value of n i.e. 2

The expressions in the four options can be expanded as

xyz-yz; xyz-xz; xyz-xy and xyz+xz

The closest value to xyz would be xyz-yz, as yz is the least value among yz, xz and xy.

Option a) is the correct answer.

$$x*x < y*y$$
or $$x + 0.5x - x^2 < y + 0.5y - y^2$$
$$y^2 - x^2 + 1.5x - 1.5y < 0$$
$$(y - x)(y + x) - 1.5 (y - x) < 0$$
$$(y - x)(y + x -1.5) < 0$$
$$(x - y)(1.5 - (x + y)) < 0$$
Now there will be two possibilities
$$x < y$$ and $$(x + y) < 1.5$$ ...........(i)
or $$x > y$$ and $$(x + y) > 1.5$$ ............(ii)
Among all options only option B satisfies (ii).
Hence, option B is the correct answer.