Mathematics

Maxima and Minima

191 Questions

Maxima and minima problems involve finding the highest and lowest values of mathematical functions within given intervals. These calculus questions require differentiation and analytical logic. They are common in civil service and state level mathematics examinations.

Extreme function valuesComplex number minimizationMinimum variables calculationMaximum matrix analysisCalculus optimization

Maxima and Minima Questions

Multiple choice

Minimum SOP of $\bar X \bar Y \bar Z + \bar X \bar Y Z + \bar XYZ + X \bar YZ+ XYZ$ is
  1. $XY+\bar XZ$
  2. $\bar X \bar Y+ YZ + XZ$
  3. $\bar X \bar Y + Z$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The minimum Sum of Products (SOP) form is X̄Ȳ + Z. Using Boolean algebra simplification or Karnaugh map: X̄ȲZ̄ + X̄ȲZ + X̄YZ + XȲZ + XYZ = X̄Ȳ(Z̄+Z) + X̄YZ + XȲZ + XYZ = X̄Ȳ + Z(X̄Y + XȲ + XY) = X̄Ȳ + Z(X̄Y + X(Ȳ+Y)) = X̄Ȳ + Z(X̄Y + X) = X̄Ȳ + Z(X + Y) = X̄Ȳ + XZ + YZ = X̄Ȳ + XZ + YZ. Further analysis shows X̄Ȳ + Z is the minimal form.

Multiple choice

In the following figure, the minimum value of the constant “C”, which is to be added to y1 (t) such that y1 (t) and y2 (t) are different, is
  1. $\Delta$
  2. $\dfrac{\Delta}{2}$
  3. $\dfrac{\Delta ^ 2}{12}$
  4. $\dfrac{\Delta}{L}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When $\Delta / 2$ is added to $y(t)$ then signal will move to next quantization level. Otherwise if they have step size less $\dfrac{\Delta}{2}$ then they will be on the same quantization level.

Multiple choice

If AC is minimum

  1. MC > AC

  2. MC > AC

  3. MC = AC

  4. none of these

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When average cost is at its minimum point, marginal cost equals average cost. This is the mathematical relationship: MC intersects AC from below at AC's minimum. Before the minimum, MC < AC (pulling average down). After the minimum, MC > AC (pulling average up). At the minimum itself, MC = AC.

Multiple choice

If the number of conditions in a decision table is n, then the maximum possible number of rules (column) is

  1. 2n

  2. 5n

  3. 7n/2

  4. log2n
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A decision table with n conditions can create 2^n possible combinations of condition outcomes. Each combination becomes a rule (column) in the table. For example, 3 conditions yield 8 rules, 4 conditions yield 16 rules. This exponential growth makes decision tables complex for many conditions.

Multiple choice

The smallest positive integer x, for which 945x2 = N3, where N is a natural number, is

  1. 49

  2. 25

  3. 35

  4. 9
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

945 = 3³ × 5 × 7. For 945x² to be a perfect cube N³, x² must provide the missing factors to make all exponents multiples of 3. We need one more 3, one more 5, and two more 7s. So x² = 3 × 5 × 7² = 735, giving x = √735 = 35. Verification: 945 × 35² = 945 × 1225 = 1157625, and 1157625^(1/3) = 105.

Multiple choice

The allocation matrix of the values - Max matrix Need Matrix A B C A B C 7 5 3 7 4 3 3 2 2 1 2 2 9 0 2 6 0 0 2 2 2 0 1 1 4 3 3 4 3 1 is

  1. A B C 0 1 0 2 0 0 3 0 2 2 1 1 2 1 2 0 0 2

  2. Allocation matrix A B C 0 1 0 2 0 0 3 0 3 2 1 1 2 1 1 0 0 2

  3. Allocation matrix A B C 0 1 0 2 0 0 3 1 2 2 1 1 2 1 1 0 0 2

  4. Allocation matrix A B C 0 1 0 2 0 0 3 0 2 2 1 1 2 1 1 0 1 2

  5. Allocation matrix A B C 0 1 0 2 0 0 3 0 2 2 1 1 2 1 1 0 0 2
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

The formula to calculate the need matrix is - Need = Max - Allocation. But we are given max and need matrices, so the formula to calculate allocation will be - Allocation = Max - Need. So, Max matrix Need Matrix A B C A B C A B C 75 3 - 7 4 3 = 0 1 0, 3 2 2 - 1 2 2 = 2 0 0, 9 0 2 - 6 0 0 = 3 0 2, 2 2 2 - 0 1 1 = 2 1 1, 4 3 3 - 4 3 1 = 0 0 2. Hence, the correct answer is - Allocation matrix A B C 0 1 0 2 0 0 3 0 2 2 1 1 2 1 1 0 0 2.