# CAT Inequalities Questions PDF (Most Important with Solutions)

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Inequalities form an important part of CAT Algebra. You can expect close to 2-3 questions in the 22-question format of the paper. Inequalities is often one of the most feared topics among the candidates. In this article, we will look into some Inequalities Questions for the CAT Exam. If you want to practice these questions, you can download the PDF, which is completely Free.
Inequalities-based questions appear in the CAT test almost every year. Inequalities are one of the simplest topics to solve and score on the CAT; therefore questions with inequalities should not be avoided.

• CAT Inequalities – Tip 1: The concept of Inequalities questions covered in the CAT Algebra in the Quantitative Aptitude portion of the CAT Exam are: Range of Inequalities, Modulus functions, Possible solutions, and so on.
• CAT Inequalities – Tip 2: The subject also covers linear and quadratic equations, root finding, polynomials, functions, and other topics.
• CAT Inequalities – Tip 3: For CAT, inequalities are considered slightly challenging in inequalities topic. A strong foundation in this topic will aid a student in answering questions about Functions as well. Every year 2-3 questions are asked from Inequalities in the CAT Exam.

You can check the answers below the questions, with the detailed solutions

Question 1:Â The number of distinct pairs of integers (m,n), satisfying $\mid1+mn\mid<\mid m+n\mid<5$ is:

Question 2:Â For a real number x the condition $\mid3x-20\mid+\mid3x-40\mid=20$ necessarily holds if

a)Â $10<x<15$

b)Â $9<x<14$

c)Â $7<x<12$

d)Â $6<x<11$

Question 3:Â $f(x) = \frac{x^2 + 2x – 15}{x^2 – 7x – 18}$ is negative if and only if

a)Â -5 < x < -2 or 3 < x < 9

b)Â x < -5 or -2 < x < 3

c)Â -2 < x < 3 or x > 9

d)Â x < -5 or 3 < x < 9

Question 4:Â The number of integers n that satisfy the inequalities $\mid n – 60 \mid < \mid n – 100 \mid < \mid n – 20 \mid$ is

a)Â 21

b)Â 19

c)Â 18

d)Â 20

Question 5:Â The minimum value of $\frac{3(6+x)(x+12)}{2(4+x)}$, Where $x>-4$ is

a)Â 18

b)Â 27

c)Â 36

d)Â 45

Question 6:Â Find z, if it is known that:
a: $-y^2 + x^2 = 20$
b: $y^3 – 2x^2 – 4z \geq -12$ and
c: x, y and z are all positive integers

a)Â Any integer greater than 0 and less than 24

b)Â 24

c)Â We need one more equation to find z

d)Â 6

e)Â 1

Question 7:Â if x and y are positive real numbers satisfying $x+y=102$, then the minimum possible valus of $2601(1+\frac{1}{x})(1+\frac{1}{y})$ is

Question 8:Â For real x, the maximum possible value of $\frac{x}{\sqrt{1+x^{4}}}$ is

a)Â $\frac{1}{2}$

b)Â $1$

c)Â $\frac{1}{\sqrt{3}}$

d)Â $\frac{1}{\sqrt{2}}$

Question 9:Â The number of pairs of integers $(x,y)$ satisfying $x\geq y\geq-20$ and $2x+5y=99$

Question 10:Â Among 100 students, $x_1$ have birthdays in January, $X_2$ have birthdays in February, and so on. If $x_0=max(x_1,x_2,….,x_{12})$, then the smallest possible value of $x_0$ is

a)Â 8

b)Â 9

c)Â 10

d)Â 12

Question 11:Â Consider the four variables A, B, C and D and a function Z of these variables, $Z = 15A^2 – 3B^4 + C + 0.5D$ It is given that A, B, C and D must be non-negative integers and thatall of the following relationships must hold:
i) $2A + B \leq 2$
ii) $4A + 2B + C \leq 12$
iii) $3A + 4B + D \leq 15$
If Z needs to be maximised, then what value must D take?

a)Â 15

b)Â 12

c)Â 0

d)Â 10

e)Â 5

Question 12:Â The inequality of p$^2$ + 5 < 5p + 14 can be satisfied if:

a)Â p â‰¤ 6, p > âˆ’1

b)Â p = 6, p = âˆ’2

c)Â p â‰¤ 6, p â‰¤ 1

d)Â p â‰¥ 6, p = 1

Question 13:Â For what range of values of ‘x’, will be the inequality $15x – \left(\frac{2}{x}\right) > 1$?

a)Â $x > 0.4$

b)Â $x < \frac{1}{3}$

c)Â $\frac{-1}{3} < x < 0.4, X > \frac{15}{2}$

d)Â $\frac{-1}{3} < x < 0, X > \frac{2}{5}$

Question 14:Â Consider the equation :
$\mid x-5\mid^2+5\mid x-5\mid-24=0$
The sum of all the real roots of the above equationis :

a)Â 2

b)Â 3

c)Â 8

d)Â 10

Question 15:Â If x is a real number, then $\sqrt{\log_{e}{\frac{4x – x^2}{3}}}$ is a real number if and only if

a)Â $1 \leq x \leq 3$

b)Â $1 \leq x \leq 2$

c)Â $-1 \leq x \leq 3$

d)Â $-3 \leq x \leq 3$

Question 16:Â If $\mid \frac{x + 1}{x – 1} \mid > \frac{x + 1}{x – 1}$, then

a)Â $-1 \leq x \leq -1$

b)Â $-1 < x < 1$

c)Â $x > 1$

d)Â $x < -1$

Question 17:Â If $y^2$ + 3y – 18 â‰¥ 0, which of the following is true?

a)Â y â‰¤ 3 or y â‰¥ 0

b)Â y > – 6 or y < 3

c)Â -6 â‰¤ y â‰¤ 3

d)Â y â‰¥ 3 or y â‰¤ – 6

Question 18:Â Consider the function f(x) = (x + 4)(x + 6)(x + 8) â‹¯ (x + 98). The number of integers x for which f(x) < 0 is:

a)Â 23

b)Â 26

c)Â 24

d)Â 48

Question 19:Â The smallest integer $n$ such that $n^3-11n^2+32n-28>0$ is

Question 20:Â If 2 â‰¤ |x – 1| Ã— |y + 3| â‰¤ 5 and both x and y are negative integers, find the number of possible combinations of x and y.

a)Â 4

b)Â 5

c)Â 6

d)Â 8

e)Â 10

Question 21:Â p, q and r are three non-negative integers such that p + q + r = 10. The maximum value of pq + qr + pr + pqr is

a)Â $\geq40$and$<50$

b)Â $\geq50$and$<60$

c)Â $\geq60$and$<70$

d)Â $\geq70$and$<80$

e)Â $\geq80$and$<90$

Question 22:Â Consider the expression $\frac{(a^2+a+1)(b^2+b+1)(c^2+c+1)(d^2+d+1)(e^2+e+1)}{abcde}$, where a,b,c,d and e are positive numbers. The minimum value of the expression is

a)Â 3

b)Â 1

c)Â 10

d)Â 100

e)Â 243

Question 23:Â The minimum possible value of the sum of the squares of the roots of the equation $x^2+(a+3)x-(a+5)=0$ is

a)Â 1

b)Â 2

c)Â 3

d)Â 4

Question 24:Â For how many integers n, will the inequality $(n – 5) (n – 10) – 3(n – 2)\leq0$ be satisfied?

Question 25:Â If a and b are integers of opposite signs such that $(a + 3)^{2} : b^{2} = 9 : 1$ and $(a -1)^{2}:(b – 1)^{2} = 4:1$, then the ratio $a^{2} : b^{2}$ is

a)Â 9:4

b)Â 81:4

c)Â 1:4

d)Â 25:4

Question 26:Â If $x$ and $y$ are real numbers, the least possible value of the expression $4(x – 2)^{2} + 4(y – 3)^{2} – 2(x – 3)^{2}$ is :

a)Â – 8

b)Â – 4

c)Â – 2

d)Â 0

e)Â 2

Question 27:Â The smallest integer x for which the inequality $\frac{x-7}{x^2 + 5x-36}$ > 0 is given by

a)Â -12

b)Â 9

c)Â -9

d)Â -8

Question 28:Â If x satisfies the inequality $|x – 1| + |x – 2| + |x – 3| \geq 6$, then:

a)Â 0 â‰¤ x â‰¤ 4

b)Â x â‰¤ 0 or x â‰¥ 4

c)Â x â‰¤ -2 or x â‰¥ 3

d)Â None of the above

Question 29:Â If $x=(9+4\sqrt{5})^{48} = [x] +f$, where [x] is defined as integral part of x and f is a fraction, then x (1 – f) equals

a)Â 1

b)Â Less than 1

c)Â More than 1

d)Â Between 1 and 2

e)Â None of the above

Question 30:Â a, b, c are integers, |a| â‰  |b| â‰ |c| and -10 â‰¤ a, b, c â‰¤ 10. What will be the maximum possible value of [abc – (a + b + c)]?

a)Â 524

b)Â 693

c)Â 731

d)Â 970

e)Â None of the above

Question 31:Â If | r – 6 | = 11 and | 2q – 12 | = 8, what is the minimum possible value of q / r?

a)Â -2/5

b)Â 2/17

c)Â 10/17

d)Â None of these

Question 32:Â The number of positive integer valued pairs (x, y), satisfying 4x – 17 y = 1 and x < 1000 is:

a)Â 59

b)Â 57

c)Â 55

d)Â 58

InstructionsFor these questions the following functions have been defined.

$la(x, y, z) = min (x+y, y+z)$
$le(x, y, z) = max(x -y, y-z)$
$ma (x, y, z) = \frac{1}{2} (le (x, y, z) + la (x, y, z))$

Question 33:Â For x=15, y=10 and z=9 , find the value of le(x, min(y, x-z), le(9, 8, ma(x, y, z)).

a)Â 5

b)Â 12

c)Â 9

d)Â 4

Question 34:Â What is the value of ma(10, 4, le((la10, 5, 3), 5, 3))?

a)Â 7

b)Â 6.5

c)Â 8

d)Â 7.5

Question 35:Â Given that $x >y> z> 0$. Which of the following is necessarily true?

a)Â $la(x, y, z) < le(x, y, z)$

b)Â $ma(x, y, z) < la(x, y, z)$

c)Â $ma(x, y, z) < le(x, y, z)$

d)Â None of these

Question 36:Â Which of the following values of x do not satisfy the inequality $(x^2 – 3x + 2 > 0)$ at all?

a)Â $1\leq x \leq 2$

b)Â $-1\geq x \geq -2$

c)Â $0 \leq x \leq 2$

d)Â $0\geq x \geq -2$

Question 37:Â x, y and z are three positive integers such that x > y > z. Which of the following is closest to the product xyz?

a)Â (x-1)yz

b)Â x(y-1)z

c)Â xy(z-1)

d)Â x(y+1)z

Question 38:Â The number of integers n satisfying -n+2Â â‰¥Â 0 and 2n â‰¥Â 4 is

a)Â 0

b)Â 1

c)Â 2

d)Â 3

Question 39:Â From any two numbers $x$ and $y$, we define $x* y = x + 0.5y – xy$ . Suppose that both $x$ and $y$ are greater than 0.5. Then
$x* x < y* y$ if

a)Â 1 > x > y

b)Â x > 1 > y

c)Â 1 > y > x

d)Â y > 1 > x

Question 40:Â If pqr = 1, the value of the expression $1/(1+p+q^{-1}) + 1/(1+q+r^{-1}) + 1/(1+r+p^{-1})$

a)Â p+q+r

b)Â 1/(p+q+r)

c)Â 1

d)Â $p^{-1} + q^{-1} + r^{-1}$

Question 41:Â If u, v, w and m are natural numbers such that $u^m + v^m = w^m$, then which one of the following is true?

a)Â m >= min(u, v, w)

b)Â m >= max(u, v, w)

c)Â m < min(u, v, w)

d)Â None of these

Question 42:Â If $x^2 + 5y^2 + z^2 = 2y(2x+z)$, then which of the following statements is(are) necessarily true?
A. x = 2y B. x = 2z C. 2x = z

a)Â Only A

b)Â B and C

c)Â A and B

d)Â None of these

Question 43:Â If x, y and z are real numbers such that x + y + z = 5 and xy + yz + zx = 3, what is the largest value that x can have?

a)Â 5/3

b)Â $\sqrt{19}$

c)Â 13/3

d)Â None of these

Question 44:Â If x>2 and y>-1,then which of the following statements is necessarily true?

a)Â xy>-2

b)Â -x<2y

c)Â xy<-2

d)Â -x>2y

Question 45:Â Let x and y be two positive numbers such that $x + y = 1.$
Then the minimum value of $(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$ is

a)Â 12

b)Â 20

c)Â 12.5

d)Â 13.3

Question 46:Â If a, b, c and d are four positive real numbers such that abcd = 1, what is the minimum value of (1 + a)(1+b)(1+c)(1+d)?

a)Â 4

b)Â 1

c)Â 16

d)Â 18

Question 47:Â $m$ is the smallest positive integer such that for any integer $n \geq m$, the quantity $n^3 – 7n^2 + 11n – 5$ is positive. What is the value of $m$?

a)Â 4

b)Â 5

c)Â 8

d)Â None of these

Question 48:Â x and y are real numbers satisfying the conditions 2 < x < 3 and – 8 < y < -7. Which of the following expressions will have the least value?

a)Â $x^2y$

b)Â $xy^2$

c)Â $5xy$

d)Â None of these

Question 49:Â If x > 5 and y < -1, then which of the following statements is true?

a)Â (x + 4y) > 1

b)Â x > -4y

c)Â -4x < 5y

d)Â None of these

Question 50:Â What values of x satisfy $x^{2/3} + x^{1/3} – 2 <= 0$?

a)Â $-8 \leq x \leq 1$

b)Â $-1 \leq x \leq 8$

c)Â $1 \leq x \leq 8$

d)Â $1 \leq x \leq 18$

e)Â $-8 \leq x \leq 8$

Question 51:Â The number of solutions of the equation 2x + y = 40 where both x and y are positive integers and x <= y is:

a)Â 7

b)Â 13

c)Â 14

d)Â 18

e)Â 20

Question 52:Â If x, y, z are distinct positive real numbers the $(x^2(y+z) + y^2(x+z) + z^2(x+y))/xyz$ would always be

a)Â Less than 6

b)Â greater than 8

c)Â greater than 6

d)Â Less than 8

Question 53:Â Given that $-1 \leq v \leq 1, -2 \leq u \leq -0.5$ and $-2 \leq z \leq -0.5$ and $w = vz /u$ , then which of the following is necessarily true?

a)Â $-0.5 \leq w \leq 2$

b)Â $-4 \leq w \leq 4$

c)Â $-4 \leq w \leq 2$

d)Â $-2 \leq w \leq -0.5$

Question 54:Â Let a, b, c, d be four integers such that a+b+c+d = 4m+1 where m is a positive integer. Given m, which one of the following is necessarily true?

a)Â The minimum possible value of $a^2 + b^2 + c^2 + d^2$ is $4m^2-2m+1$

b)Â The minimum possible value of $a^2 + b^2 + c^2 + d^2$ is $4m^2+2m+1$

c)Â The maximum possible value of $a^2 + b^2 + c^2 + d^2$ is $4m^2-2m+1$

d)Â The maximum possible value of $a^2 + b^2 + c^2 + d^2$ is $4m^2+2m+1$

## CAT Inequalities Questions Answers & Solutions:

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)Â < 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.

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.

We haveÂ $\mid n – 60 \mid < \mid n – 100 \mid < \mid n – 20 \mid$.

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.Â 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 :Â $\mid n – 60 \mid < \mid n – 100 \mid < \mid n – 20 \mid$.

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,Â |n-60| = 60- n,Â |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)Â 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)Â 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)Â 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 rangeÂ satisfies the given inequality.

Hence a total of 19 values satisfy the inequality.

Let $x+4=y$
So, the expressionÂ $\frac{3\ \left(x+6\right)\left(x+12\right)}{2\left(x+4\right)}$ equals $\frac{3*(y+2)(y+8)}{2*y}$
This equals $\frac{3}{2} \times (y+16/y+10)$
The minimum value of $y+16/y$ is 8 when $y$ equals 4
So, the minimum value of the expression is 3/2*18 = 27

SinceÂ $x^2-y^2=20$ and x,y,z are positive integers,

(x+y)*(x-y) = 20, Hence x-y, x+y are factors of 20.

Since x, y are positive integers, x+y is always positive, and for the product of (x+y)*(x-y) Â to be positive x-y must be positive.

x, y are positive integers and x-y is positiveÂ x must be greater thanÂ y.

The possible cases are : (x+y = 10, x-y = 2), (x+y = 5, x-y = 4).

The second case fails because we get x =9/2, y = 1/2 but x, y are integral values

For case one x = 6, y = 4.

$y^3 – 2x^2 – 4z \geq -12$

Substituting Â the values of x and y, we have :

64 – 72 – 4*zÂ $\ge\ -12$

-8 – 4*zÂ $\ge\ -12$

z$\le\ 1$

Since x, y, z are positive integers, the only possible value for z is 1.

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

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 xyÂ  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

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

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

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 asÂ $\frac{1}{\sqrt{\ 2}}$

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

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

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 –Â 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.

$x_0=max(x_1,x_2,….,x_{12})$

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

100/12=8.33

.’.Â max$(x_1,x_2,….,x_{12})$ will be minimum ifÂ $(x_1,x_2,….,x_{12})$=(9,9,9,9,8,8,8,8,8,8,8,8,)

.’. Option B is correct.

To maximize Z, B has to be minimized and A,C,D are to be maximised.

The value of B can be 0 or 1.

Case 1:

B=0 => A=1=> C=8 and D=12.

Z= 15+8+6=29.

Case 2:

B=1 => A=0=> C=12 and D=15.

Z=-3+12+7.5=16.5.

We have,Â p$^2$ + 5 < 5p + 14

=>Â p$^2$ – 5p – 9 < 0

=> p< $\ \frac{\ 5\ +\ \sqrt{\ 61}}{2}$ or p>Â $\ \frac{\ 5\ -\ \sqrt{\ 61}}{2}$

=> p<6.4 or p>-1.4

Hence, p â‰¤ 6, p > âˆ’1 will satisfy the inequalities

We have
$15x-\frac{2}{x}>1$
â‡’Â $15x^2-2>x$
â‡’$15x^2-x-2>0$
â‡’$15x^2-6x+5x-2>0$
â‡’ (3x+1)(5x-2)>0
we get x>2/5 or x<-1/3

Let’s considerÂ x-5 as ‘p’

Case 1: $p \ge\$ 0

$\mid x-5\mid^2+5\mid x-5\mid-24=0$

$p^2\ +5p\ -24\ =0\$

$p^2\ +8p\ -3p\ -24\ =0\$

$p\left(p+8\right)-3\left(p+8\right)\ =0\$

$\left(p+8\right)\left(p-3\right)$ = 0

p=-8 and p=3

x-5=3,x=8 This is a real root since x is greater than 5.

x-5=-8, x=-3. This root can be negated because x is not greater than 5.

Case 1: $p <Â$ 0

$p^2\ -5p\ -24\ =0\$

$p^2\ -8p\ +3p\ -24\ =0\$

p=8, -3

x-5=8, x=13. This root can be negated because x is not less than 5

x-5=-3, x=2. This is a real root because x is less than 5.

The sum of the real roots = 8+2=10

$\sqrt{\log_{e}{\frac{4x – x^2}{3}}}$ will be real ifÂ $\log_e\ \frac{\ 4x-x^2}{3}\ \ge\ 0$

$\frac{\ 4x-x^2}{3}\ >=\ 1$

$\ 4x-x^2-3\ >=\ 0$

$\ x^2-4x+3\ =<\ 0$

1=< x=< 3

$\mid \frac{x + 1}{x – 1} \mid will be greater thanÂ \frac{x + 1}{x – 1}$ only ifÂ  $\frac{x + 1}{x – 1}$ < 0

$\therefore$ x $\in$ (-1,1)

Hence B is the correct answer.

$y^2 + 3y – 18 \geq 0$

$\Rightarrow y^2 + 6y – 3y – 18 \geq 0$

$\Rightarrow y (y+6) – 3 (y+6) \geq 0$

$\Rightarrow (y-3) (y+6) \geq 0$

$\Rightarrow y\geq3Â andÂ y\leq-6$

The critical points of the function are -4, -6, -8, … , -98 ( 48 points).

For all integers less than -98 Â and greater than -4 f(x) > 0 always .

for x= -5, f(x) < 0

Similarly, for x= -9, -13, …., -97 (This is an AP with common difference -4)

Hence, in total there are 24 such integers satisfying f(x)< 0.

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 factorizeÂ n^2-9n+14 as (n-2)(n-7).

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

$\Rightarrow$Â $n^3-11n^2+32n-28>0$

$\Rightarrow$Â $(n-2)^2(n-7)>0$

Therefore, we can say that n-7>0

Hence, n$_{min}$ = 8

2 â‰¤ |x – 1| Ã— |y + 3| â‰¤ 5
The product of two positive number lies between 2 and 5.
As x is a negative integer, the minimum value of |x – 1| will be 2 and the maximum value of |x – 1| will be 5 as per the question.

When, |x – 1| = 2, |y + 3| can be either 1 or 2
So, for x =Â  -1, y can be – 4 or – 2 or – 5 or -1.
Thus, we get 4 pairs of (x, y)

When |x – 1| = 3, |y + 3| can be 1 only
So, for x = – 2, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)

When |x – 1| = 4, |y + 3| can be 1 only
So, for x = – 3, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)

When |x – 1| = 5, |y + 3| can be 1 only
So, for x = – 4, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)

Therefore, we get a total of 10 pairs of the values of (x, y)
Hence, option E is the correct answer.

The product of 2 numbers A and B is maximum when A = B.
If we cannot equate the numbers, then we have to try to minimize the difference between the numbers as much as possible.
pq will be maximum when p=q.
qr will be maximum when q=r.
qr will be maximum when r=p.
Therefore, p, q, and r should be as close to each other as possible.
We know that p,q,and r are integers and p+q+r=10.
=> p,q, and r should be 3,3, and 4 in any order.
Substituting the values in the expression, we get,
pq+qr+pr+pqr = 3*3 + 3*4 +Â 3*4 + 3*3*4
= 9 + 12 +Â 12 + 36
= 69
Therefore, option C is the right answer.

The given expression can be written as $\frac{a^2+a+1}{a}*\frac{b^2+b+1}{b}*\frac{c^2+c+1}{c}*\frac{d^2+d+1}{d}*\frac{e^2+e+1}{e}$.
$\frac{a^2+a+1}{a}=a+\frac{1}{a}+1$
We know that for positive values, AM $\geq$ GM.
$\frac{1+\frac{1}{a}}{2} \geq \sqrt{a*\frac{1}{a}}$
$a+\frac{1}{a} \geq 2$
The least value that $a+\frac{1}{a}$ can take is $2$.
Therefore, the least value that the term $a+\frac{1}{a}+1$ can take is $3$.
Similarly, the least value that the other terms can take is also $3$.
=> The least value of the given expression = $3*3*3*3*3$ = $243$.
Therefore, optionÂ E is the right answer.

Let the roots of the equationÂ $x^2+(a+3)x-(a+5)=0$ 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 +Â 1 = 11.

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

$4(x – 2)^{2} + 4(y – 3)^{2} – 2(x – 3)^{2}$
$y$ is an independent variable. The value of $y$ is unaffected by the value of $x$. Therefore, the least value that the expression $4(y-3)^2$ can take is $0$ (at $y=3$).

Let us expand the remaining terms.
$4(x-2)^2-2(x-3)^2$=4*$(x^2-4x+4)-2*(x^2-6x+9)$
$=2x^2-4x-2$
=$2(x^2-2x-1)$
=$2(x^2-2x+1-2)$
=$2((x-1)^2-2)$
The least value that the expression $(x-1)^2$ can take is $0$ (at $x$ = $1$)
Therefore, the least value that the expressionÂ $2((x-1)^2-2$ can take is $2*(0-2)=2*(-2) = -4$
Therefore, optionÂ B is the right answer.

$\frac{x-7}{x^2 + 5x-36}$ > 0

$\frac{x-7}{(x+9)(x-4)}$ > 0

So $x \epsilon (-9,4) U (7, \infty$)

So smallest integer in this range is $x = -8$.

Given thatÂ  $|x – 1| + |x – 2| + |x – 3| \geq 6$.

Case 1: When x > 3

$(x – 1) + (x – 2) + (x – 3) \geq 6$

$x \geq 4$

Therefore, the value of x $\in$ [4, $\infty$)

Case 2: When 2 < x < 3

$(x – 1) + (x – 2) – (x – 3) \geq 6$

$x \geq 6$

Therefore, no possible value of x in this domain.

Case 3: When 1 < x < 2

$(x – 1) – (x – 2) – (x – 3) \geq 6$

$x \leq -2$

Therefore, no possible value of x in this domain.

Case 3: When x < 1

$-(x – 1) – (x – 2) – (x – 3) \geq 6$

$x \leq 0$

Therefore, the value of x $\in$ (-$\infty$, 0]

Therefore, the value of x that will satisfy this inequality: x $\in$ (-$\infty$, 0] $\cup$ [4, $\infty$).

Hence, option B is the correct answer.

It is given that $x=(9+4\sqrt{5})^{48}$ … (1)

Let us assume that $y=(9-4\sqrt{5})^{48}$ … (2)

We can see thatÂ  0 < y < 1.

Also, x +Â y = 2($48C0*(9)^{48}$+$48C2*(9)^{46}*(4\sqrt{5})^2$+$48C4(9)^{44}*(4\sqrt{5})^4$+…+$48C48*(4\sqrt{5})^{48}$)

We can see that x +Â y is an integer therefore we can say that y + f = 1. Hence, y = (1 – f)

We can see that =Â x(1 – f) = x*y

$\Rightarrow$Â $(9+4\sqrt{5})^{48}(9-4\sqrt{5})^{48}$

$\Rightarrow$Â $(9^2-(4\sqrt{5})^2)^{48}$

$\Rightarrow$Â $(81-80)^{48}$ = 1

Hence, option A is the correct answer.

|a| â‰  |b| â‰ |c| and -10 â‰¤ a, b, c â‰¤ 10

ExpressionÂ : $[abc – (a + b + c)]$

For maximum value, two of a,b and c should be negative, as all three negative will make abc negative.

Thus, max value will occur if $a = -10 , b = -9 , c = 8$

=> Max value = $[(-10 \times -9 \times 8) – (-10 -9 + 8)]$

= $720 + 11 = 731$

| 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

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

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

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

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.

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; Â x>2

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

So theÂ answer will be A.

The expressions in the four options can be expanded as

xyz-yz; xyz-xz; xyz-xyÂ and xyz+xz

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

Option a) is the correct answer.

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

$x*x < y*y$
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.

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.

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.

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.

The given equations are Â 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
=> $– y^2 – y (5 -x) – x ^2 + 5x = 3$
=> $y^2 + y (x-5) + ( x ^2 – 5x +3) Â = 0$
The above equation should have real roots for y, => Determinant >= 0

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

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

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.

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 = $(x+\frac{1}{x})^2+(y+\frac{1}{y})^2$ =Â $(2+\frac{1}{2})^2+(2+\frac{1}{2})^2$ =12.5

Approach 2:

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

Let x+1/x and y+1/y be two terms. ThusÂ (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 asÂ $(x+1/x)^2$ +Â $(y+1/y)^2$ = $4AM^2$ – $GM^2$. 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 = yÂ +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, yÂ <=1. Hence, their product cannot be 1.

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

=> y = 1/2

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

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, optionÂ C is the right answer.

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

$xy^2$ 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

Substitute x=6 and y=-6 ,

x+4y = -18

x = 6, -4y = 24

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

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

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.

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.

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

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 Â u = -0.5, we get w = -4

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