Maximum and Minimum value of equation MCQ Quiz - Objective Question with Answer for Maximum and Minimum value of equation - Download Free PDF

Last updated on Apr 24, 2025

Latest Maximum and Minimum value of equation MCQ Objective Questions

Maximum and Minimum value of equation Question 1:

Find the value of x for which the expression 2 - 3x - 4x2 has the greatest value.

  1. \(-\frac{41}{16}\)
  2. \(\frac{3}{8}\)
  3. \(-\frac{3}{8}\)
  4. \(\frac{41}{16}\)

Answer (Detailed Solution Below)

Option 3 : \(-\frac{3}{8}\)

Maximum and Minimum value of equation Question 1 Detailed Solution

Given:

2 - 3x - 4x2

Concept used:

A function reaches its maximum value when its first-order derivative is zero.

Calculation:

Let, f(x) = 2 - 3x - 4x2

Performing the first-order derivative of f(x) we get,

f'(x) = 0 - 3 - (4 × 2x)

⇒ f'(x) = - 3 - 8x

For the greatest value, f'(x) = 0

So, f'(x) = - 3 - 8x

⇒ 0 = - 3 - 8x

⇒ -8x = 3

⇒ x = -3/8

⇒ x = \(-\frac{3}{8}\)

∴ The value of x for which the expression 2 - 3x - 4x2 has the greatest value is \(-\frac{3}{8}\).

Maximum and Minimum value of equation Question 2:

The integer 'k', for which the inequality x2 - 2(3k - 1)x + 8k2 - 7 > 0 is valid for every x in R, is :

  1. 2
  2. 3
  3. 4
  4. 0

Answer (Detailed Solution Below)

Option 2 : 3

Maximum and Minimum value of equation Question 2 Detailed Solution

D < 0

(2(3k - 1))2 - 4(8k2 - 7) < 0

4(9k2 - 6k + 1) - 4(8k2 - 7) < 0

k2 - 6k + 8 < 0

(k - 4) (k - 2) < 0

2 < k < 4

then k = 3

Top Maximum and Minimum value of equation MCQ Objective Questions

Maximum and Minimum value of equation Question 3:

Find the value of x for which the expression 2 - 3x - 4x2 has the greatest value.

  1. \(-\frac{41}{16}\)
  2. \(\frac{3}{8}\)
  3. \(-\frac{3}{8}\)
  4. \(\frac{41}{16}\)

Answer (Detailed Solution Below)

Option 3 : \(-\frac{3}{8}\)

Maximum and Minimum value of equation Question 3 Detailed Solution

Given:

2 - 3x - 4x2

Concept used:

A function reaches its maximum value when its first-order derivative is zero.

Calculation:

Let, f(x) = 2 - 3x - 4x2

Performing the first-order derivative of f(x) we get,

f'(x) = 0 - 3 - (4 × 2x)

⇒ f'(x) = - 3 - 8x

For the greatest value, f'(x) = 0

So, f'(x) = - 3 - 8x

⇒ 0 = - 3 - 8x

⇒ -8x = 3

⇒ x = -3/8

⇒ x = \(-\frac{3}{8}\)

∴ The value of x for which the expression 2 - 3x - 4x2 has the greatest value is \(-\frac{3}{8}\).

Maximum and Minimum value of equation Question 4:

The integer 'k', for which the inequality x2 - 2(3k - 1)x + 8k2 - 7 > 0 is valid for every x in R, is :

  1. 2
  2. 3
  3. 4
  4. 0

Answer (Detailed Solution Below)

Option 2 : 3

Maximum and Minimum value of equation Question 4 Detailed Solution

D < 0

(2(3k - 1))2 - 4(8k2 - 7) < 0

4(9k2 - 6k + 1) - 4(8k2 - 7) < 0

k2 - 6k + 8 < 0

(k - 4) (k - 2) < 0

2 < k < 4

then k = 3

Maximum and Minimum value of equation Question 5:

Find the value of x for which the expression 2 - 3x - 4x2 has the greatest value.

  1. \(-\frac{41}{16}\)
  2. \(\frac{3}{8}\)
  3. \(-\frac{3}{8}\)
  4. \(\frac{41}{16}\)
  5. None of the above/More than one of the above.

Answer (Detailed Solution Below)

Option 3 : \(-\frac{3}{8}\)

Maximum and Minimum value of equation Question 5 Detailed Solution

Given:

2 - 3x - 4x2

Concept used:

A function reaches its maximum value when its first-order derivative is zero.

Calculation:

Let, f(x) = 2 - 3x - 4x2

Performing the first-order derivative of f(x) we get,

f'(x) = 0 - 3 - (4 × 2x)

⇒ f'(x) = - 3 - 8x

For the greatest value, f'(x) = 0

So, f'(x) = - 3 - 8x

⇒ 0 = - 3 - 8x

⇒ -8x = 3

⇒ x = -3/8

⇒ x = \(-\frac{3}{8}\)

∴ The value of x for which the expression 2 - 3x - 4x2 has the greatest value is \(-\frac{3}{8}\).

Maximum and Minimum value of equation Question 6:

Let

\(\rm S=\left\{\alpha: \log _{2}\left(9^{2 \alpha-4}+13\right)-\log _{2}\left(\frac{5}{2} \cdot 3^{2 \alpha-4}+1\right)=2\right\}\).

Then the maximum value of β for which the equation \(\rm x^{2}-2\left(\sum_{\alpha \in S} \alpha\right)^{2} x+\sum_{\alpha \in S}(\alpha+1)^{2} \beta=0\) has real roots, is ____. 

Answer (Detailed Solution Below) 25

Maximum and Minimum value of equation Question 6 Detailed Solution

Calculation: 

\(\log _{2}\left(9^{2 α-4}+13\right)-\log _{2}\left(\frac{5}{2} \cdot 3^{2 α-4}+1\right)=2\)

⇒ \(\frac{9^{2 α-4}+13}{\frac{5}{2} 3^{2 α-4}+1}=4\)

⇒ α = 2 or 3 

\(\sum_{\alpha \in S} \alpha=5 \text { and } \sum_{\alpha \in S}(\alpha+1)^{2}=25\)

⇒ x2 – 50x + 25β = 0 has real roots

⇒ β ≤ 25

⇒ βmax = 25

Hence, the correct answer is 25. 

Maximum and Minimum value of equation Question 7:

The value of \(3 + \frac{1}{{4 + \frac{1}{{3 + \frac{1}{{4 + \frac{1}{{3 + ........\infty }}}}}}}}\) is equal to?

  1. 1.5 + \(\sqrt3\)
  2. 2 + \(\sqrt3\)
  3. 3 + 2\(\sqrt3\)
  4. 4 + \(\sqrt3\)

Answer (Detailed Solution Below)

Option 1 : 1.5 + \(\sqrt3\)

Maximum and Minimum value of equation Question 7 Detailed Solution

Concept:

By using quadratic formula, the roots of the quadratic equation of the form ax2 + bx + c = 0, a ≠ 0 are given by,

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Explanation:

Let, y = \(3 + \frac{1}{{4 + \frac{1}{{3 + \frac{1}{{4 + \frac{1}{{3 + ........\infty }}}}}}}}\)

⇒ y =  \(3 + \frac{1}{{4 + \frac{1}{y}}}\)

⇒ y ( \(4+\frac{1}{y}\))  = 3 (\(4+\frac{1}{y}\)) + 1

⇒ 4y + 1 = 12 + \(\frac{3}{y}\)  + 1

⇒ 4y = 12 + \(\frac{3}{y}\)

⇒ 4y2 = 12y + 3 

⇒ 4y2 - 12y - 3 = 0

 Using quadratic formula, 

y = \({-(-12) ± √{(-12)^2-4\times 4\times(-3)} \over 2\times 4}\)

⇒ y =  \(\frac{12 ±√{192}}{8}\)

⇒ y = \(\frac{12 ±{8√ 3}}{8}\)

⇒ y = \(\frac{3}{2}\) ± √ 3 = 1.5 ± √ 3

The value of y cannot be negative. Therefore, y = 1.5 + √ 3 

The answer is 1.5 + √ 3.

The correct answer is option (1).

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