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

Concept:

• \(\text{Related Rates}: \)  Problems where two or more quantities change with respect to time.
• \(\text{Volume of Cone } \) The formula is \(V = \frac{1}{3}\pi r^2 h\), where \(V\) is volume, \(r\) is radius, and \(h\) is height.
• \(\text{Constant Ratio Condition} : \) Given \(h = \frac{1}{6}r\), the radius and height are proportional, so express \(r\) in terms of \(h\).
• Differentiation: Differentiate \(V\) with respect to time \(t\) to find how fast the height changes as the volume increases.

Calculation:

Given,

\(\frac{dV}{dt} = 12 \, \text{cm}^3/\text{s}\)

Height and radius relation: \(h = \frac{1}{6}r\) so \(r = 6h\)

Volume of cone:

\(V = \frac{1}{3}\pi r^2 h\)

⇒ Substitute \(r = 6h\):

\(V = \frac{1}{3}\pi (6h)^2 h\)

⇒ \(V = \frac{1}{3}\pi \times 36h^2 \times h\)

⇒ \(V = 12\pi h^3\)

Differentiate both sides w.r.t \(t\):

\(\frac{dV}{dt} = 36\pi h^2 \frac{dh}{dt}\)

Substitute \(\frac{dV}{dt} = 12\) and \(h = 4\):

⇒ \(12 = 36\pi \times (4)^2 \times \frac{dh}{dt}\)

⇒ \(12 = 36\pi \times 16 \times \frac{dh}{dt}\)

⇒ \(12 = 576\pi \frac{dh}{dt}\)

⇒ \(\frac{dh}{dt} = \frac{12}{576\pi} = \frac{1}{48\pi}\)

The height increases at a rate of \(\frac{1}{48\pi} \, \text{cm/s}\). 

⇒ 96πh = 2

Hence 2 is the correct answer. 

Calculation: 

f(x) = 2x³ + (2p – 7)x² + 3(2p – 9)x – 6

⇒ f'(x) = 6x² + 2(2p – 7)x + 3(2p – 9)

⇒ f'(0) < 0

∴ 3(2p – 9) < 0

⇒ \(\mathrm{p}<\frac{9}{2}\)

⇒ \(\mathrm{p} \in\left(-\infty, \frac{9}{2}\right)\)

Hence, the correct answer is Option 3.

Concept:

Local Extrema (Maxima and Minima):

• The local maxima of a function occur where the derivative changes from positive to negative.
• The local minima of a function occur where the derivative changes from negative to positive.
• To find the points of local maxima or minima, we calculate the first derivative of the function and set it equal to zero to find critical points.
• The second derivative test can further confirm whether the critical points correspond to maxima or minima:
  • If \( f''(x) > 0 \), the function has a local minimum at that point.
  • If \( f''(x) < 0 \), the function has a local maximum at that point.
  • If \( f''(x) = 0 \), the test is inconclusive.

Given Function:

• The function is defined as follows:
  • For \( x \neq 0 \), \( f(x) = \frac{6x + \sin x}{2x + \sin x} \)
  • For \( x = 0 \), \( f(x) = \frac{7}{3} \)

Graph Behavior:

• The function is a rational function, and its behavior at critical points and over intervals needs to be analyzed.
• We analyze the function in the interval \( [\pi, 6\pi] \) and \( [2\pi, 4\pi] \) for local minima and maxima.

Calculation:

Step 1: Finding the first derivative of the function.

The given function is \( f(x) = \frac{6x + \sin x}{2x + \sin x} \), and we need to find its first derivative.

\( f'(x) = \frac{(2x + \sin x)(6 - \cos x) - (6x + \sin x)(2 - \cos x)}{(2x + \sin x)^2} \)

Step 2: Solving for critical points.

To find the critical points, we solve \( f'(x) = 0 \). This will give us the points where the function has potential maxima or minima.

Step 3: Second derivative test for confirmation.

The second derivative \( f''(x) \) is used to confirm the nature of the critical points (whether they are maxima or minima).

\( f''(x) = \frac{d}{dx} \left[ \frac{(2x + \sin x)(6 - \cos x) - (6x + \sin x)(2 - \cos x)}{(2x + \sin x)^2} \right] \)

Step 4: Local Maxima and Minima.

The results of the calculations reveal that:

• Statement A: The point \( x = 0 \) is a point of local maxima of \( f \).
• Statement B: The point \( x = 0 \) is a point of local minima of \( f \).
• Statement C: The number of points of local maxima of \( f \) in the interval \( [\pi, 6\pi] \) is 3.
• Statement D: The number of points of local minima of \( f \) in the interval \( [2\pi, 4\pi] \) is 1.

Conclusion:

Hence, the correct answers are:

• Statement B: The point \( x = 0 \) is a point of local minima of \( f \).
• Statement C: The number of points of local maxima of \( f \) in the interval \( [\pi, 6\pi] \) is 3.
• Statement D: The number of points of local minima of \( f \) in the interval \( [2\pi, 4\pi] \) is 1.

Calculation: 

⇒ f'(x) = 8x³ - 36x + 8 = 4(2x³ - 9x + 2) 

f'(x) = 0 

∴ \(x=\frac{\sqrt{6}-2}{2}\)

Now

⇒ \(f(x)=\left(x^{2}-2 x-\frac{9}{2}\right)\left(2 x^{2}+4 x-1\right)+24 x+7.5\)

∴ \(f\left(\frac{\sqrt{6}-2}{2}\right)=M=12 \sqrt{6}-\frac{33}{2} \)

Hence, the correct answer is Option 1.

Concept:

• The question involves testing continuity of rational functions, minima of polynomial functions, and differentiability of composite trigonometric and floor functions.
• Continuity requires the numerator and denominator to cancel out the points where the denominator becomes zero.
• For finding the minima, the first derivative is used to locate critical points, and the second derivative test confirms whether the point is a minima.
• Non-differentiability occurs where absolute functions like sin |x - k| cause sharp turns (cusps).

Calculation:

P) Let k(x) = 10x³ - 45x² + 60x + 35

⇒ k′(x) = 30x² - 90x + 60 = 30(x - 1)(x - 2)

⇒ k(x) is continuous in [1, 2]

⇒ [k(1), k(2)] are same integer for all x ∈ [1, 2]

⇒ Minimum value of n = 9     

(P → 2)

Q) g(x) = (2n² - 13n - 15) / (n² + 3x)

⇒ (2n² - 13n - 15) / (n² + 3x) ≥ 0 for g(x) ≥ 0 for g(x) = mix of n & x

⇒ Minimum value of n is 5     

(Q → 1)

R) h(x) = (x² - 9)²(x² + 2x + 3)

⇒ h(x) has a local minima at x = 3 for n = 6

⇒ (3 + δ) h(3 + δ) (δ is a small positive real number)

⇒ has local minimum at x = 3 for n = 6

⇒ (R → 4)

S) g(x) = sin |x - k| + cos |x - k - 1/2| + sin |x - k - 1| + cos |x - k - 3/2| + ⋯ + sin |x - 4| + cos |x - 9/2|

as sin |x - k| is non-differentiable at x = k    

but, cos |x - λ| is differentiable at x = λ

⇒ g(x) is non-differentiable at x₀ = 0, 1, 2, 3, 4 (5 points)

⇒ (S → 3)

∴ The correct matching is P → 2, Q → 1, R → 4, S → 3.

Hence, Option 2 is the correct answer.

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 = \(\frac12\)

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

So, we get minimum value at x = \(\frac12\)

f(\(\frac12\)) = (\(\frac12\))² - \(\frac12\) + 2 = \(\frac74\)

Hence, option (3) is correct. 

Concept:

Rate of change of 'x' is given by \(\rm \frac {dx}{dt}\)

Calculation:

Given that, y = 2x – x² and \(\rm \frac {dx}{dt}\) = 3 units/sec

Then, the slope of the curve, \(\rm \frac {dy}{dx}\) = 2 - 2x = m

⇒\(\rm \frac {dm}{dt}\)  = 0 - 2 × \(\rm \frac {dx}{dt}\)

= -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) < 0.
• 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, the local maximum value of the function occurs at x = 0.

Concept:

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

⇒ -1 - 3x2 < 0

⇒ -(1 + 3x2) < 0

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) < 0.
• 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' = \(\rm\frac{d}{dx}\left(-x^3 + 3x^2 + 2x - 27\right)\) = -3x² + 6x + 2.

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

Now, m' = 0.

⇒ \(\rm\frac{d}{dx}\left(-3x^2 + 6x+2\right)\) = -6x + 6 = 0.

⇒ x = 1.

And m'' = -6 < 0.

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: \(\rm V = 4π r^3\)

Differentiating with respect to r, we get

\( \frac{dV}{dr} = 4π \frac{dr^3}{dr}\)

\(= 4π \times 3r^2\)

= 12πr² 

When r = 2

\(\rm \Rightarrow \frac{dV}{dr} = 12π \times 4 = 48π\)