Applications of Derivatives MCQ Quiz - Objective Question with Answer for Applications of Derivatives - Download Free PDF

Concept:

The function f(x) is decreasing when f '(x)

Calculation:

Given, f(x) = -2x3 - 9x2 - 12x + 1

⇒ f '(x) = -6x2 - 18x - 12

⇒ f '(x) = -6(x2 + 3x + 2)

⇒ f '(x) = -6(x + 1)(x + 2)

⇒ f '(x) -6(x + 1)(x + 2)

⇒ f '(x)  0 when (x + 1)(x + 2) > 0

⇒ f(x) is decreasing for (-∞, -2) ∪ (-1, ∞)

∴ The correct answer is option (4).

Concept:

Lagrange’s Mean Value Theorem:
If a function f is defined on the closed interval [a, b] satisfying:

• The function f is continuous on the closed interval [a, b].
• The function f is differentiable on the open interval (a, b).

Then there exists a value x = c such that f'(c) = .

Calculation:

The given function f(x) =  is both differentiable and continuous in the interval [1, 3].

f'(x) = 

By the Mean Value Theorem, there exists a c ∈ [1, 3], such that:

f'(c) = 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ c = √3.

Concept Used:

A function is increasing when its derivative is greater than 0.

Calculation:

y = x2e-x

⇒

For the function to be increasing, 0\)

⇒ 0\)

Since is always positive, we need to find where 0\)

⇒  0\)

⇒ 0

The function is increasing in the interval (0, 2).

Hence option 1 is correct

Concept Used:

A tangent line parallel to the Y-axis is a vertical line, and its equation is of the form x = a.

Calculation:

Given:

Equation of the ellipse: 

To find the points where the tangent is vertical, we need to find the points on the ellipse where the slope is undefined (i.e., the derivative is infinite).

Differentiate the equation of the ellipse implicitly with respect to x:

⇒

⇒

The slope is undefined when the denominator is zero, i.e., y = 0.

Substitute y = 0 into the equation of the ellipse to find the corresponding x values:

⇒

⇒

⇒

⇒

So, the points where the tangents are vertical are and .

The equations of the tangent lines are and .

Multiplying by 4 gives 4x = 1 and 4x = -1.

Hence option 4 is correct

Concept:

The equation of the tangent to a curve y = f(x) at a point (a, b) is given by (y - b) = m(x - a), where m = y'(b) = f'(a) [value of the derivative at point (a, b)].

Calculation:

⇒ y' = f'(x) = 3x²

m = f'(1) = 3 × 1² = 3.

(y - b) = m(x - a)

⇒ (y - 1) = 3(x - 1)

⇒ y - 1 = 3x - 3

Concept:

Following steps to finding minima using derivatives.

• Find the derivative of the function.
• Set the derivative equal to 0 and solve. This gives the values of the maximum and minimum points.
• Now we have to find the second derivative: If f"(x) Is greater than 0 then the function is said to be minima

Calculation:

f(x) = x2 - x + 2

f'(x) = 2x - 1

Set the derivative equal to 0, we get

f'(x) = 2x - 1 = 0

⇒ x = 

Now, f''(x) = 2 > 0

So, we get minimum value at x =

f() = ()² - + 2 =

Hence, option (3) is correct. 

Concept:

Rate of change of 'x' is given by

Calculation:

Given that, y = 2x – x² and = 3 units/sec

Then, the slope of the curve, = 2 - 2x = m

⇒  = 0 - 2 × 

= -2(3)

= -6 units per second

Hence, the slope of the curve is decreasing at the rate of 6 units per second when x is increasing at the rate of 3 units per second.

Hence, option (2) is correct.

Concept:

|x| ≥ 0 for every x ∈ R

Calculation:

Let f(x) = |x + 3| - 2

As we know that |x| ≥ 0 for every x ∈ R

∴ |x + 3| ≥ 0

The minimum value of function is attained when |x + 3| = 0

Hence, Minimum value of f(x) = 0 – 2 = -2 

Concept:  

If f′(x) >  0 at each point in an interval, then the function is said to be strictly increasing.

Calculations:

Given , f(x) = x2 - 2x 

Differentiating, we get

f'(x) = 2x - 2

f(x) is strictly increasing function

∴ f'(x) > 0

⇒ 2x - 2 > 0

⇒ x > 1

∴ x ∈ (1, ∞)

Concept:

For a function y = f(x):

• Relative (local) maxima are the points where the function f(x) changes its direction from increasing to decreasing.
• Relative (local) minima are the points where the function f(x) changes its direction from decreasing to increasing.
• At the points of relative (local) maxima or minima, f'(x) = 0.
• At the points of relative (local) maxima, f''(x) .
• At the points of relative (local) minima, f''(x) > 0.

Calculation:

For the given function f(x) = 3x4 + 4x3 - 12x2 + 12, first let's find the points of local maxima or minima:

f'(x) = 12x³ + 12x² - 24x = 0

⇒ 12x(x² + x - 2) = 0

⇒ x(x + 2)(x - 1) = 0

⇒ x = 0 OR x = -2 OR x = 1.

f''(x) = 36x² + 24x - 24.

f''(0) = 36(0)2 + 24(0) - 24 = -24.

f''(-2) = 36(-2)2 + 24(-2) - 24 = 144 - 48 - 24 = 72.

f''(1) = 36(1)2 + 24(1) - 24 = 36 + 24 - 24 = 36.

Since, at x = 0 the value f''(0) = -24 0.

Concept:

• If f′(x) > 0 then the function is said to be strictly increasing.
• If f′(x) decreasing.

Calculation:

Given: f(x) = 1 - x - x3

Differentiating with respect to x, we get

⇒ f'(x) = 0 - 1 - 3x2

⇒ f'(x) =  - 1 - 3x2

For decreasing function, f'(x)

⇒ -1 - 3x2 

⇒ -(1 + 3x2)

As we know, Multiplying/Dividing each side of an inequality by a negative number reverses the direction of the inequality symbol.

⇒ (1 + 3x2)  > 0

As we know, x2 ≥ 0,  x ∈ R

So, 1 + 3x2 > 0, x ∈ R

Hence, the function is decreasing for all values of x

Concept:

Following steps to finding maxima and minima using derivatives.

• Find the derivative of the function.
• Set the derivative equal to 0 and solve. This gives the values of the maximum and minimum points.
• Now we have find second derivative.

1. f``(x) is less than 0 then the given function is said to be maxima
2. If f``(x) Is greater than 0 then the function is said to be minima

Calculation:

Given:

f(x) = e^x

Differentiating with respect to x, we get

⇒ f’(x) = e^x

For maximum value f’(x) = 0

∴ f’(x) = e^x = 0

Exponential function can never assume zero for any value of x, therefore function does not have local maxima or minima.

Concept:

Slope of a curve y = f(x) at a point is the value of its first derivative at that point.

i.e. m = f'(x).

Maxima/Minima of a function y = f(x):

• Relative (Local) maxima are the points where the function f(x) changes its direction from increasing to decreasing.
• Relative (Local) minima are the points where the function f(x) changes its direction from decreasing to increasing.
• At the points of relative (local) maxima or minima, f'(x) = 0.
• At the points of relative (local) maxima, f''(x)
• At the points of relative (local) minima, f''(x) > 0.

Calculation:

The slope of the curve y = -x3 + 3x2 + 2x - 27 will be given by:

m = y' =  = -3x² + 6x + 2.

In order to maximize the slope, we must have m' = 0 and m''

Now, m' = 0.

⇒  = -6x + 6 = 0.

⇒ x = 1.

And m'' = -6

Therefore, the slope is maximum at x = 1.

Concept;

If y = f(x), then dy/dx denotes the rate of change of y with respect to x.

Decreasing rate is represented by a negative sign whereas the increasing rate is represented by a positive sign

Calculation:

Given: 

Differentiating with respect to r, we get

= 12πr² 

When r = 2