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

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 $-\frac{3}{8}$.

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

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 $-\frac{3}{8}$.

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

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 $-\frac{3}{8}$.

Concept:

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

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Explanation:

Let, y = $3+\frac{1}{4+\frac{1}{3+\frac{1}{4+\frac{1}{3+........\mathrm{\infty }}}}}$

⇒ 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 = $\frac{-(-12)\pm \surd (-12{)}^2-4\times 4\times (-3)}{2\times 4}$

⇒ y =  $\frac{12\pm \surd 192}{8}$

⇒ y = $\frac{12\pm 8\surd 3}{8}$

⇒ 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).