Increasing and Decreasing Functions MCQ Quiz - Objective Question with Answer for Increasing and Decreasing Functions - 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).

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 function is increasing if its derivative is positive.

Calculation

Let f(x) = tan^-1(sin x + cos x)

\(f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot (\cos x - \sin x)\)

Simplify the denominator:

1 + (sin x + cos x)² = 1 + sin² x + cos² x + 2 sin x cos x

= 1 + 1 + 2 sin x cos x

= 2 + 2 sin x cos x

= 2(1 + sin x cos x)

So, f'(x) = \(\frac{\cos x - \sin x}{2(1 + \sin x \cos x)}\) = \(\frac{\sqrt{2} \cos(x + \frac{\pi}{4})}{1 + (\sin x + \cos x)^2}\)

For f(x) to be increasing, f'(x) > 0

f(x) is increasing if \(-\frac{\pi}{2} < x + \frac{\pi}{4} < \frac{\pi}{2} \Rightarrow -\frac{3\pi}{4} < x < \frac{\pi}{4}\)

∴ The function f(x) = tan^-1(sin x + cos x) is an increasing function in (-π/2, π/4).

Hence option 3 is correct

Concept:

Monotonicity of Function:

• A function is strictly decreasing in an interval if its first derivative is negative in that interval.
• The function given is f(x) = |3/x + x/3|.
• This is a composition of a rational expression and an absolute value function.
• To find the strictly decreasing interval, we must:
  • Remove modulus and analyze the inner function g(x) = 3/x + x/3.
  • Compute the derivative of g(x) and study the sign of g′(x).
  • Consider the effect of modulus: if g(x) is negative, then f(x) = –g(x).
  • We then compute derivative of f(x) accordingly based on sign of g(x).

Rules used:

• Derivative of 1/x = –1/x²
• Derivative of x = 1

Calculation:

Let f(x) = |g(x)| where g(x) = 3/x + x/3

⇒ g′(x) = –3/x² + 1/3

Set g′(x) = 0 to find turning points:

⇒ –3/x² + 1/3 = 0

⇒ 1/3 = 3/x²

⇒ x² = 9

⇒ x = ±3

Check sign of g′(x) around x = –3, 0, 3:

⇒ x < –3: x = –4 ⇒ g′(x) = –3/16 + 1/3 > 0

⇒ –3 < x < 0: x = –2 ⇒ g′(x) = –3/4 + 1/3 = –5/12 < 0

⇒ 0 < x < 3: x = 2 ⇒ g′(x) = –3/4 + 1/3 = –5/12 < 0

⇒ x > 3: x = 4 ⇒ g′(x) = –3/16 + 1/3 > 0

So, g(x) is decreasing in (–3, 0) ∪ (0, 3)

Now analyze sign of g(x) in (–3, 0) and (0, 3):

In both intervals, g(x) < 0

⇒ f(x) = –g(x)

⇒ f′(x) = –g′(x) = 3/x² – 1/3

Set f′(x) < 0 for decreasing:

⇒ 3/x² – 1/3 < 0

⇒ 3/x² < 1/3 ⇒ x² > 9 ⇒ |x| > 3

But this contradicts x ∈ (–3, 0) ∪ (0, 3)

Now test again the derivative for f(x) = –g(x) in (–3, 0) ∪ (0, 3)

In this interval, 3/x² – 1/3 is decreasing when x² < 9

So f′(x) < 0 when x² < 9

⇒ x ∈ (–3, 0) ∪ (0, 3)

∴ The function is strictly decreasing in the interval (–3, 0) ∪ (0, 3)

Calculation

\(\sin x\)

\(\frac{d}{dx}\) \(\sin x\) = \(\cos x\).

On the interval \((0, \frac{\pi}{8})\), \(\cos x\) is positive.

Since the derivative is positive, \(\sin x\) is increasing on this interval.

\(-\cos x\)

\(\frac{d}{dx}\) \(-\cos x\) = \(\sin x\).

On the interval \((0, \frac{\pi}{8})\), \(\sin x\) is positive.

Since the derivative is positive, \(-\cos x\) is increasing on this interval.

\(\cos 4x\)

\(\frac{d}{dx}\) \(\cos 4x\) = \(-4\sin 4x\).

 \(4x\) lies in the interval \((0, \frac{\pi}{2})\), where \(\sin 4x\) is positive.

Therefore, \(-4\sin 4x\) is negative on this interval.

Since the derivative is negative, \(\cos 4x\) is decreasing on this interval.

Hence option 5 is correct.

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:

• 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:

• 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 + x2 + x4

Differentiating with respect to x, we get

⇒ f'(x) = 0 + 2x + 4x³

⇒ f'(x) = 2x + 4x3

For strictly increasing function, f'(x) > 0

⇒ 2x + 4x³ > 0

⇒ 2x(1 + 2x²) > 0

As we know, x2 ≥ 0,  x ∈ R

So, 1 + 2x2 > 0, x ∈ R

Now, 2x > 0

⇒ x > 0

Hence function is strictly increasing for x > 0

Concept:  

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

Calculations:

Given , f(x) = 2x2 - 3x 

Differentiating, we get

f'(x) = 4x - 3

f(x) is strictly increasing function

∴ f'(x) > 0

⇒ 4x - 3 > 0

⇒ 4x > 3 

⇒ x > \(\rm\frac{3}{4}\)

∴ x ∈ (\(\rm\frac{3}{4}\), ∞)

Important Points

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) = 2 - x3 - x⁵

Differentiating with respect to x, we get

⇒ f'(x) = 0 - 3x² - 5x⁴

⇒ f'(x) =  - 3x2 - 5x4

For decreasing function, f'(x) < 0

⇒- 3x2 - 5x4 < 0

⇒ -[x²(3 + 5x²)] < 0

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

⇒ x2(3 + 5x2) > 0

As we know, x² ≥ 0,  x ∈ R

So, 3 + 5x² > 0, x ∈ R

Hence function is decreasing for all values of x

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 - 4x 

Differentiating, we get

f'(x) = 2x - 4

f(x) is strictly increasing function

∴ f'(x) > 0

⇒ 2x - 4 > 0

⇒ x > 2

∴ x ∈ (2, ∞)

Mistake Points

If a, b ∈ R and a < b, the following is a representation of the open and closed intervals:

• Open interval is indicated by (a, b) = {x : a < y < b}, i.e. 'a' and 'b' are not included.
• Closed interval is indicated by [a, b] = {x : a ≤ x ≤ b}, i.e. 'a' and 'b' are included.
• [a, b) = {x : a ≤ x < b} is an open interval from a to b, inclusive of 'a' but excluding 'b'.
• (a, b ] = { x : a < x ≤ b } is an open interval from a to b including 'b' but excluding 'a'.

Since '2' is not included here, Option (1) will not be correct.

Concept:

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

Calculation:

Given:

f(x) = x - e^x

Differentiating with respect to x, we get

⇒ f’(x) = 1 - e^x

For increasing function,

f'(x) > 0

⇒ 1 – e^x > 0

⇒ e^x < 1

⇒ e^x < e⁰

∴ x < 0

So, x ∈ (-∞, 0) 

Concept:

Monotonic function: If a function is differentiable on the interval (a, b) and if the function is increasing/strictly increasing or decreasing/strictly decreasing, then the function is known as monotonic function.

First derivative test:

• If \(\frac{{{\rm{df}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}} \ge 0\) for all x in (a, b) then, f(x) is an increasing function in (a, b).
• If \(\frac{{{\rm{df}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}} \le 0\) for all x in (a, b) then, f(x) is an decreasing function in (a, b).

For strictly increasing or decreasing:

• \(\frac{{{\rm{df}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}} > 0\) for strictly increasing.
• \(\frac{{{\rm{df}}\left( {\rm{x}} \right)}}{{{\rm{dx}}}} < 0\) for strictly decreasing.

Calculation:

Given: \({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}} + \frac{1}{{\rm{x}}}\)

Differentiating w.r.t x

\(\Rightarrow {\rm{f'}}\left( {\rm{x}} \right) = 1 - \frac{1}{{{{\rm{x}}^2}}}\)

\(\Rightarrow {\rm{f'}}\left( {\rm{x}} \right) =\rm \frac{x^2 -1}{x^2}= \rm \frac{(x -1)(x+1)}{x^2}\)

So, for x ∈ (0, 1), f’(x) < 0.

Hence, f(x) is decreases for x ∈ (0, 1).

Concept:

The function f(x):

• Increases for the values x of having f'(x) > 0
• Decreases for the values of x having f'(x) < 0

Calculation:

Given function is f(x) = sin x

∴ f'(x) = cos x

For x ∈ (0, \(π\over2\)), cos x > 0 and for x ∈ (\(π\over2\), π), cos x < 0

∴ For x ∈ (0, π), f(x) increases to max at \(π\over2\) and then decreases to min at π.

For x ∈ \(\left(\dfrac{5\pi}{2},3\pi\right)\), cos x < 0

∴ f(x) decreases for x ∈ \(\left(\dfrac{5\pi}{2},3\pi\right)\)

statement 2 is correct Only.

Concept:

Let f(x) be a function defined on an interval (a, b), this function is said to be an strictly increasing function:

• If x1 < x2 then f(x1) < f(x2) ∀ x1, x2 ∈ (a, b)
• Here, \(\frac{{dy}}{{dx}} > 0\;or\;f'\left( x \right) > 0\)

Similarly, f(x) is said to be a decreasing function:

• If If x1 < x2 then f(x1) > f(x2) ∀ x1, x2 ∈ (a, b).
• Here, \(\frac{{dy}}{{dx}} < 0\;or\;f'\left( x \right) < 0\)

Calculation:

Given: f(x) = 2x³ - 24x + 5

Here we have to find the interval in which f(x) is strictly increasing.

Let's first calculate f'(x)

⇒ f'(x) = 6x² - 24

As we know that, for a strictly increasing function say f(x) we have f'(x) > 0

⇒ 6x² - 24 > 0

⇒ x² > 4

⇒ x ∈ (- ∞, - 2) ∪ (2, ∞)

Hence, the interval in which the function is increasing is (- ∞, - 2) ∪ (2, ∞)