**Maxima and Minima Questions for CAT:**

Download important CAT Maxima and Minima Questions PDF based on previous asked questions in CAT and other MBA exams. Top 25 Maxima and Minima functions questions for CAT quantitative aptitude.

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**Question 1: **If y is a real number, what is the difference in the maximum and minimum values obtained by $\frac{y+5}{y^2+5y+25}$ ?

a) 2/15

b) 4/15

c) 1/5

d) 1/15

**Question 2: **What is the maximum value of g(x) = 22 – |x+8| and for what value of x is this reached?

a) 22, 8

b) 14, 0

c) 14, -8

d) None of these

**Question 3: **If 2 $\leq$ x $\leq$ 3 and 4 $\leq$ y $\leq$ 5, what are the minimum and maximum values of $\frac{y+x}{y+2x}$

a) 9/10, 7/8

b) 10/11, 8/9

c) 7/10, 7/9

d) None of these

**Question 4: **If $x^{2} – 5x + 4 <= 0$ and $y^{2} -6y+5 <= 0$, what are the maximum and minimum values of $\frac{y+7x}{y+4x}$

a) 29/17 , 12/9

b) 23/13, 11/8

c) 25/11, 14/11

d) None of these

**Question 5: **If X and Y are positive real numbers and 3X+4Y = 14, what is the maximum value of $X^{3}*Y^4$ ?

a) 64

b) 128

c) 2187

d) None of these

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**Question 6: **What is the maximum value of g(x) = |34-x| – |x + 11|?

a) 34

b) 45

c) 23

d) None of these

**Question 7: **If X and Y are positive and X+Y = 10, what is the maximum value of $X^3*{Y^2}$ ?

a) 3125

b) 3456

c) 3600

d) None of these

**Question 8: **If f(x) = min(3x-4, 11-7x). What is the maximum value of f(x)?

a) 3/2

b) -3/2

c) 1/2

d) Infinite

**Question 9: **For how many real values does the function $F(X) = 28 – |X^2+6|$ attain its maximum value

a) 1

b) 2

c) 3

d) None of these

**Question 10: **If ‘a’ and ‘b’ are two whole numbers such that |a-2|+|b-3| = 4, what is the maximum value of a*b ?

a) 24

b) 18

c) 14

d) 20

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**Question 11: **If X,Y and Z are real numbers such that $-6 \leq X \leq -2$ and $-4 \leq Y \leq 4$ and $3 \leq Z \leq 7$. If $W=\frac{Y}{XZ}$, which of the following is necessarily true?

a) $0 \leq W \leq 2$

b) $-1/2 \leq W \leq 1/2$

c) $-2/3 \leq W \leq 1/3$

d) $-2/3 \leq W \leq 2/3$

**Question 12: **Two natural numbers, X and Y, satisfy the equation 2X+Y = 33. Find the maximum value of XY?

a) 135

b) 140

c) 136

d) 144

**Question 13: **Two natural numbers, A and B, satisfy the equation 3A+2B = 64. Find the maximum value of ${A*B}$?

a) 160

b) 162

c) 170

d) 180

**Question 14: **If a,b and c are three whole numbers such that |a-1|+|b-2|+|c-3|=3, what is the maximum value of (a+1)(b+2)(c+3)?

a) 120

b) 144

c) 112

d) 105

**Question 15: **If a,b and c are positive real numbers, what is the minimum value of a*(1/b+1/c)+b*(1/c+1/a)+c*(1/a+1/b)

a) 9

b) 6

c) 3

d) 27

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**Question 16: **If $x^2 – 5x + 4 \leq 0$ and $y^2 -6y +5 \leq 0$, then what are the maximum and minimum values of $(y+7x)/(y+4x)$.

a) 29/17, 12/9

b) 23/13, 11/8

c) 25/11, 14/11

d) 29/11, 11/9

e) None of these

**Question 17: **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) The minimum possible value of $a^2 + b^2 + c^2 + d^2$ is $4m^2-2m+1$

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

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

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

**Question 18: **If x, y, z are distinct positive real numbers the $(x^2(y+z) + y^2(x+z) + z^2(x+y))/xyz$ would be

a) greater than 4

b) greater than 5

c) greater than 6

d) None of the above

**Question 19: **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) 4

b) 1

c) 16

d) 18

**Question 20: **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) 12

b) 20

c) 12.5

d) 13.3

**Question 21: **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)?[CAT 2001]

a) 4

b) 1

c) 16

d) 18

**Question 22: **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[CAT 2001]

a) 12

b) 20

c) 12.5

d) 13.3

**Question 23: **If a, b, c, d are positive numbers, what is the minimum value of (a+b+c+d)*(1/a+1/b+1/c+1/d)?

a) 9

b) 16

c) 25

d) 1

**Question 24: **If | r – 6 | = 11 and | 2q – 12 | = 8, what is the minimum possible value of q / r?

a) -2/5

b) 2/17

c) 10/17

d) None of these

**Question 25: **If ‘$p$’ and ‘$q$’ are two positive rational numbers such that $p+q=1$, which of the following statements is true?

a) The minimum value of $p^{\frac{1}{4}}+q^{\frac{1}{4}}$ is $2^\frac{1}{2}$

b) The maximum value of $p^{\frac{1}{4}}+q^{\frac{1}{4}}$ is $2^\frac{1}{2}$

c) The maximum value of $p^{\frac{1}{3}}+q^{\frac{1}{3}}$ is $2^\frac{2}{3}$

d) The minimum value of $p^{\frac{1}{3}}+q^{\frac{1}{3}}$ is $2^\frac{2}{3}$

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**Answers & Solutions:**

**1) Answer (B)**

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.

$(5k-1)^2 \geq 4*k*5*(5k-1)$

$(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

**2) Answer (D)**

g(x) reaches its maximum when |x+8| =0 or x = -8. So, maximum value of g(x) is 22

**3) Answer (C)**

$\frac{y+x}{y+2x}$ = 1- $\frac{x}{y+2x}$, which equals 1 – $\frac{1}{\frac{y}{x}+2}$. So, the function is at its maximum when y/x is at its highest (x=2, y= 5) and is at its minimum when y/x is lowest (x=3, y= 4).

**4) Answer (A)**

$\frac{y+7x}{y+4x}$ = 1 + $\frac{3x}{y+4x}$, which equals 1 – $\frac{3}{y/x+4}$. This reaches its maximum when $\frac{y}{x}$ is at its minimum and vice versa. So, the minimum value of $\frac{y+7x}{y+4x}$ is $\frac{5+7}{5+4}$ = 4/3 and the maximum is $\frac{1+28}{1+16}$ = 29/17

**5) Answer (B)**

$X+X+X+Y+Y+Y+Y = 14.$ So,$(X+X+X+Y+Y+Y+Y)/7 \geq (X*X*X*Y*Y*Y*Y)^{1/7}$ So, $(X^{3}*Y^{4})^{1/7} \leq 2$. So, the maximum value of $X^{3}*Y^4$ is 128

**6) Answer (B)**

If $x \leq -11$, g(x) = 45.

If $-11 < x \leq 34$, g(x) = 23-2x which peaks when x= -11.

If $x> 34$, g(x) = -45.

So, the maximum value of g(x) is 45

**7) Answer (B)**

$X/3 + X/3 +X/3 + Y/2 + Y/2 = 10.$ So, $X^{3}*Y^{2} \leq 2^{5} * 3^{3}*2^{2} = 3456.$ The equality happens when X=6 and Y=4

**8) Answer (C)**

f(x) reaches its maximum value when 3x-4 = 11-7x or if x = 3/2 So, the maximum value of f(x) is 1/2

**9) Answer (A)**

F(x) attains its maximum value when X = 0. So, the number of real values for which F(X) attains its minimum value is 1

**10) Answer (D)**

There are twelve integral solutions to the equation |a-2|+|b-3| = 4 The product a*b is maximized when a=4 and b=5 and equals 20

**11) Answer (D)**

The minimum value of W is when Y=4, X = -2 and Z=3 and equals -2/3 The maximum value of W is when Y=-4, X=-2 and Z=3 and equals 2/3

**12) Answer (C)**

The maximum value occurs when 2X and Y are closest to each other. So, when X=8 and Y=17, XY reaches its peak of 136.

**13) Answer (C)**

The maximum value occurs when 3A and 2B are closest to each other. So, when A=10 and B=17, AB reaches its peak of 170.

**14) Answer (A)**

There are fifteen solutions to the equation |a-1|+|b-2|+|c-3|=3.

Let |a-1| be x, |b-2| be y and |c-3| be z.

x, y and z are whole numbers.

x + y + z = 3.

Therefore, the number of solutions to this equation will be n+r-1 C r-1 = 3+3-1C2 = 5C2 = 10.

There are 10 solutions to the equation x+y+z = 3.

After enumerating them, we find that (a+1)(b+2)(c+3) is maximum when a=b=c=3 and its maximum value is 120.

**15) Answer (B)**

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.

**16) Answer (A)**

$\frac{y+7x}{y+4x}$ = 1 + $\frac{3x}{y+4x}$, which equals 1 + $\frac{3}{y/x+4}$.

This reaches its maximum when $\frac{y}{x}$ is at its minimum and vice versa.

So, the minimum value of $\frac{y+7x}{y+4x}$ is $\frac{5+7}{5+4}$ = 4/3 and the maximum is $\frac{1+28}{1+16}$ = 29/17

**17) Answer (B)**

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

**18) Answer (C)**

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

**19) Answer (C)**

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.

**20) Answer (C)**

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

**21) Answer (C)**

Since the product is constant, the expression (1+a)(1+b)(1+c)(1+d) is least when a = b = c = d.

So, a = b = c = d = 1 and the minimum value of the expression is 2*2*2*2 = 16.

**22) Answer (C)**

The expression attains the minimum value when x = y

x=y=1/2

So, the value = $2.5^2 * 2 = 12.5$

**23) Answer (B)**

AM >= HM

So, (a+b+c+d)/4 >= 4/(1/a+1/b+1/c+1/d) => (a+b+c+d)*(1/a+1/b+1/c+1/d) >= 4*4 = 16

**24) Answer (D)**

| 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

**25) Answer (C)**

The mean of $\frac{1}{3}$ power of two numbers is less than $\frac{1}{3}$ power of the mean of the numbers.

Hence, $\frac{p^{\frac{1}{3}}+q^{\frac{1}{3}}}{2} \leq {(\frac{p+q}{2})}^{\frac{1}{3}}$

So, $p^\frac{1}{3}+q^\frac{1}{3} \leq 2^\frac{2}{3}$