Applications Of Derivatives Assessment Test

1. The function 2tan x – 3tan x + 12tanx + 3, x Î is...

Decreasing

Increasing

None of the above

I don't know

The given function is a polynomial function in terms of tan x. Since the coefficient of the highest power of tan x is positive, and all the other terms are also positive, the function will increase as the value of tan x increases. Therefore, the correct answer is "Increasing".

Explanation

The given function is a polynomial function in terms of tan x. Since the coefficient of the highest power of tan x is positive, and all the other terms are also positive, the function will increase as the value of tan x increases. Therefore, the correct answer is "Increasing".

2. An absolute maximum or minimum must occur at a critical point or at an endpoint.

True

False

Maybe

Depends on some factors

An absolute maximum or minimum refers to the highest or lowest value of a function over its entire domain. In order for this maximum or minimum to occur, it must either be at a critical point (where the derivative of the function is either zero or undefined) or at an endpoint of the domain. This is because these are the points where the function can potentially reach its extreme values. Therefore, the statement that an absolute maximum or minimum must occur at a critical point or at an endpoint is true.

Explanation

An absolute maximum or minimum refers to the highest or lowest value of a function over its entire domain. In order for this maximum or minimum to occur, it must either be at a critical point (where the derivative of the function is either zero or undefined) or at an endpoint of the domain. This is because these are the points where the function can potentially reach its extreme values. Therefore, the statement that an absolute maximum or minimum must occur at a critical point or at an endpoint is true.

3. Let f (x) = (4–x2)2/3 , then f has a...

A local maxima at x = 0

A local maxima at x = 2

A local maxima at x = –1

None of the above

The function f(x) = (4-x^2)^(2/3) has a local maximum at x = 2. This can be determined by analyzing the behavior of the function around x = 2. As x approaches 2 from the left, f(x) increases, and as x approaches 2 from the right, f(x) decreases. This indicates that f(x) reaches a local maximum at x = 2.

Explanation

The function f(x) = (4-x^2)^(2/3) has a local maximum at x = 2. This can be determined by analyzing the behavior of the function around x = 2. As x approaches 2 from the left, f(x) increases, and as x approaches 2 from the right, f(x) decreases. This indicates that f(x) reaches a local maximum at x = 2.

4. Newton's method is used to approximate the extrema of a function.

Always true

Often true

Occasionally true

Never true

Newton's method is a numerical algorithm used to find the roots or extrema of a function. It involves iteratively refining an initial guess by using the function's derivative. Since the method is specifically designed for finding extrema, it is always true that Newton's method can be used to approximate the extrema of a function.

Explanation

Newton's method is a numerical algorithm used to find the roots or extrema of a function. It involves iteratively refining an initial guess by using the function's derivative. Since the method is specifically designed for finding extrema, it is always true that Newton's method can be used to approximate the extrema of a function.

5. The absolute minimum value of x4-x2-2x+5...

Is equal to 5

Is equal to 3

Is equal to 7

Does not exist

The absolute minimum value of the given expression can be found by taking the derivative of the expression and setting it equal to zero. By doing this, we can find the critical points of the function. In this case, the critical point is x = 1. Plugging this value back into the original expression, we get a minimum value of 3. Therefore, the absolute minimum value of the expression is equal to 3.

Explanation

The absolute minimum value of the given expression can be found by taking the derivative of the expression and setting it equal to zero. By doing this, we can find the critical points of the function. In this case, the critical point is x = 1. Plugging this value back into the original expression, we get a minimum value of 3. Therefore, the absolute minimum value of the expression is equal to 3.

6. If a, b, c, d are four positive real numbers such that abcd =1, then minimum value of (1+ a) (1 + b) (1 + c) (1 + d) is...

16

13

17

18

The minimum value of (1+ a) (1 + b) (1 + c) (1 + d) can be found by applying the AM-GM inequality. According to the inequality, the arithmetic mean of a set of positive real numbers is always greater than or equal to the geometric mean of the same set. In this case, we can rewrite the expression as (1/2)(2 + 2a)(2 + 2b)(2 + 2c)(2 + 2d), which simplifies to (2 + 2ab)(2 + 2cd). By applying the AM-GM inequality to each term, we get (2 + 2ab)(2 + 2cd) ≥ 4√(ab)(cd) = 4√(abcd) = 4. Therefore, the minimum value of the expression is 4 * 4 = 16.

Explanation

The minimum value of (1+ a) (1 + b) (1 + c) (1 + d) can be found by applying the AM-GM inequality. According to the inequality, the arithmetic mean of a set of positive real numbers is always greater than or equal to the geometric mean of the same set. In this case, we can rewrite the expression as (1/2)(2 + 2a)(2 + 2b)(2 + 2c)(2 + 2d), which simplifies to (2 + 2ab)(2 + 2cd). By applying the AM-GM inequality to each term, we get (2 + 2ab)(2 + 2cd) ≥ 4√(ab)(cd) = 4√(abcd) = 4. Therefore, the minimum value of the expression is 4 * 4 = 16.

7. The minimum value of ax + by, where xy = r , is (r, ab >0)...

2r √ab

2ab√r

–2r √ab

None of the above

The given expression is ax + by, where xy = r. To find the minimum value of this expression, we can use the AM-GM inequality. According to the AM-GM inequality, the arithmetic mean of two non-negative numbers is always greater than or equal to their geometric mean. Applying this concept, we can rewrite the expression as (ax + by)/2 ≥ √(ax * by). Substituting xy = r, we get (ax + by)/2 ≥ √(abr). To minimize the expression, we need to make the left side equal to the right side. This occurs when ax = by and both are equal to √(abr). Solving these equations, we find that ax = by = √(abr). Substituting this back into the original expression, we get 2√(abr), which is equivalent to 2r√ab. Therefore, the correct answer is 2r√ab.

Explanation

The given expression is ax + by, where xy = r. To find the minimum value of this expression, we can use the AM-GM inequality. According to the AM-GM inequality, the arithmetic mean of two non-negative numbers is always greater than or equal to their geometric mean. Applying this concept, we can rewrite the expression as (ax + by)/2 ≥ √(ax * by). Substituting xy = r, we get (ax + by)/2 ≥ √(abr). To minimize the expression, we need to make the left side equal to the right side. This occurs when ax = by and both are equal to √(abr). Solving these equations, we find that ax = by = √(abr). Substituting this back into the original expression, we get 2√(abr), which is equivalent to 2r√ab. Therefore, the correct answer is 2r√ab.

8. The number of value of k for which the equation x3 – 3x + k = 0 has two different roots lying in the interval (0, 1) are...

Infinitely many

2

3

0

The equation x3 – 3x + k = 0 is a cubic equation. In order for the equation to have two different roots lying in the interval (0, 1), the graph of the equation must intersect the x-axis twice within this interval. Since the interval (0, 1) is infinite, there are infinitely many values of k for which this condition is satisfied. Thus, the answer is infinitely many.

Explanation

The equation x3 – 3x + k = 0 is a cubic equation. In order for the equation to have two different roots lying in the interval (0, 1), the graph of the equation must intersect the x-axis twice within this interval. Since the interval (0, 1) is infinite, there are infinitely many values of k for which this condition is satisfied. Thus, the answer is infinitely many.

9. Rolle's theorem holds for the function x3 + bx2 + cx, 1 < x < 2 at the point 4/3, the value of b and c are...

B = 8, c = - 5

B = -5, c = 8

B = 5, c = -8

B = -5, c = -8

Rolle's theorem states that if a function is continuous on a closed interval [a, b], differentiable on the open interval (a, b), and f(a) = f(b), then there exists at least one point c in the open interval (a, b) where the derivative of the function is equal to zero. In this case, the function x^3 + bx^2 + cx satisfies the conditions of Rolle's theorem on the interval 1

Explanation

Rolle's theorem states that if a function is continuous on a closed interval [a, b], differentiable on the open interval (a, b), and f(a) = f(b), then there exists at least one point c in the open interval (a, b) where the derivative of the function is equal to zero. In this case, the function x^3 + bx^2 + cx satisfies the conditions of Rolle's theorem on the interval 1

10. L'Hopital's rule states that the limit of a function of the form f(x) / g(x) is equal to the limit of the derivative off(x) / g(x).

Always true

Often true

Occasionally true

Never true

L'Hopital's rule states that the limit of a function of the form f(x) / g(x) is equal to the limit of the derivative of f(x) / g(x). However, this rule is not always true. There are certain conditions that need to be met for L'Hopital's rule to be applicable. For example, both f(x) and g(x) must approach either 0 or infinity as x approaches the limit point, and the limit of the derivative of f(x) / g(x) must exist. If these conditions are not met, L'Hopital's rule cannot be used, and therefore, it is never true in all cases.

Explanation

L'Hopital's rule states that the limit of a function of the form f(x) / g(x) is equal to the limit of the derivative of f(x) / g(x). However, this rule is not always true. There are certain conditions that need to be met for L'Hopital's rule to be applicable. For example, both f(x) and g(x) must approach either 0 or infinity as x approaches the limit point, and the limit of the derivative of f(x) / g(x) must exist. If these conditions are not met, L'Hopital's rule cannot be used, and therefore, it is never true in all cases.