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) < 0

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) < 0 when -6(x + 1)(x + 2) < 0

⇒ f '(x) < 0 when (x + 1)(x + 2) > 0

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

∴ The correct answer is option (4).

\(=\displaystyle \underset{h\rightarrow 0}{\lim}\frac{f(3) \cdot f(h)-f(3)}{h}\) ......... As \(f(x+y)=f(x) \cdot f(y)\)

\(\Rightarrow f'(3)=3\,\underset{h \rightarrow 0}{\lim}\dfrac{f(h)-1}{h}\) \([\because f(3)=3]\)

Use L'Hospital rule,

\(\Rightarrow f'(3) = 3\,\underset{h\rightarrow 0}{\lim}\dfrac{f'(h)}{1}\)

\(=3f'(0)\) \([\because f'(0)=11]\)

\(=3\times 11=33\)

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) = \(\rm \frac{f(b)-f(a)}{b-a}\).

Calculation:

The given function f(x) = \(\rm x+\frac1x\) is both differentiable and continuous in the interval [1, 3].

f'(x) = \(\rm 1-\frac1{x^2}\)

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

f'(c) = \(\rm \frac{f(b)-f(a)}{b-a}\)

⇒ \(\rm 1-\frac1{c^2}=\frac{f(3)-f(1)}{3-1}\)

⇒ \(\rm 1-\frac1{c^2}=\frac{\left(3+\frac13\right)-\left(1+\frac11\right)}{2}\)

⇒ \(\rm 1-\frac1{c^2}=\frac{\frac43}{2}=\frac23\)

⇒ \(\rm \frac1{c^2}=1-\frac23=\frac13\)

⇒ c = √3.

Concept Used:

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

Calculation:

y = x2e-x

⇒ \(\frac{dy}{dx} = 2xe^{-x} + x^2(-e^{-x}) = e^{-x}(2x - x^2)\)

For the function to be increasing, \(\frac{dy}{dx} > 0\)

⇒ \(e^{-x}(2x - x^2) > 0\)

Since \(e^{-x}\) is always positive, we need to find where \(2x - x^2 > 0\)

⇒ \(x(2-x) > 0\)

⇒ 0 < x < 2.

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: \(16x^2 + 25y^2 = 1\)

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 \(\frac{dy}{dx}\) is infinite).

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

⇒ \(32x + 50y\frac{dy}{dx} = 0\)

⇒ \(\frac{dy}{dx} = -\frac{32x}{50y} = -\frac{16x}{25y}\)

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:

⇒ \(16x^2 + 25(0)^2 = 1\)

⇒ \(16x^2 = 1\)

⇒ \(x^2 = \frac{1}{16}\)

⇒ \(x = \pm\frac{1}{4}\)

So, the points where the tangents are vertical are \((\frac{1}{4}, 0)\) and \((-\frac{1}{4}, 0)\).

The equations of the tangent lines are \(x = \frac{1}{4}\) and \(x = -\frac{1}{4}\).

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 = \(\frac12\)

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

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