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.
Answer (Detailed Solution Below)
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 :
Answer (Detailed Solution Below)
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.
Answer (Detailed Solution Below)
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 :
Answer (Detailed Solution Below)
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.
Answer (Detailed Solution Below)
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?
Answer (Detailed Solution Below)
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).