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