CAT Question Bank  Maxima & Minima

Q19) If minimum value of A is 1 then find the minimum value of A + (1/A)

Q20) If f(x) = min(1  3x, 2x  1), find the maximum value of f(x).

Q21) Find the minimum value of (p + q + r)(1/p + 1/q + 1/r) where p, q and r are natural numbers.

Q22) Let x, y, z be three nonnegative integers such that x + y + z = 10. The maximum possible value of xyz + xy + yz + zx is

Q23) which of the following is the maximum value of f(x) = x / (x^2  x + 1) ? ( given x is a real number )
a) 1/3
b) 2
c) 1
d) 1/3

Q24) A function f(x) is defined as f(x) = min{4x + 1, x + 2, 2x + 4}
Then maximum value of f(x) is
a) 2
b) 5/2
c) 8/3
d) 17/6
e) 3

Q25) A function f(x) is defined for all real values of x as min(–x^2, x – 20, –x – 20).
What is the maximum value of f(x)?
(a) –16
(b) –20
(c) –25
(d) None of these

Q26) a + b + c = 0
a^2 + b^2 + c^2 = 1
If a, b, c be real numbers, find the maximum possible value of (9abc)^2

Q27) Find the minimum value of f(x) = (6/x) + x^2

Q28) Find the maximum and minimum values of 3sinx + 4cosx

Q29) If the equation 3x + 4y = k has 15 positive integral solutions then find the maximum and minimum values of k

Q30) If a, b, c, d, e and f are non negative real numbers such that a + b + c + d + e + f = 1, then the maximum value of (ab + bc + cd + de + ef) is


@rowdyrathore 243




Funda : If the product of positive numbers is a constant then sum is minimum when all numbers are equal.
here a * b * c = 90
So sum is minimum when a = b = c = (90)^1/3
Now this won't be a natural number solution. So we need place them as close as possible
90 = 3^2 * 5 * 2
90 = 3 * 5 * 6
minimum of a + b + c = 3 + 5 + 6 = 14

No distinct real roots means D = 0
b^2 = 24a
a = b^2/24
3a + b = b^2/8 + b = (b^2 + 8b)/8
Complete the square, (b^2 + 8b + 16  16)/8 = (b + 4)^2/8  2
as (b + 4)^2 is always non negative, minimum value is 2 when b =  4

@SoumyaRanjanPatra Answer is root 2. Explanations below
Method 1 :
Let x^2+y^2 = k = k * 1
=> x^2 + y^2 = k(x^2 + 2xy  y^2)/2
Divide throughout by xy
=> 2 (x/y) + 2 (y/x) = k (x/y) + 2k  k (y/x)
Let x/y = t
=> (2k)t^2 + 2kt +(2+k)=0
since x/y is real, so D ≥ 0
=> 4k^2  4 (2k)(2+k) ≥ 0
4k^2  4(4k^2) ≥ 0
8k^2 16 ≥ 0
k^2 ≥ 2 and k should be +ve only
so k ≥ root2
so min value= sqrt (2)
[Credits : Bruce Wayne]Method 2:
x^2 + 2xy  y^2 = 2
x = asinQ
y = acosQ
so x^2 + y^2 = a^2
a^2sin^Q + 2a^2sinQ * cosQ  a^2cos^2Q = 2
a^2((sin^2Q  cos^2Q + 2sinQ * coSQ)=2
a^2((sin2Q  cos2Q) = 2
a^2 = 2/(sin2Qcos2Q)
now max value of sin2Q  cos2Q=root2
so a^2 = 2/root2
so a^2 = root2
so min value will be root2
[Credits : @hemant_malhotra ]