Application of Derivatives: Chapter 6 Links
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Application of Derivatives 6.2 introduces the concepts of increasing and decreasing functions. The derivative \frac{dy}{dx} helps us determine whether the value of a function increases or decreases as the independent variable changes.
\displaystyle f'(x)\gt 0 \Rightarrow f(x)\text{ is increasing}
\displaystyle f'(x)\lt 0 \Rightarrow f(x)\text{ is decreasing}
The sign of the first derivative provides valuable information about the behaviour of a function on a given interval. By examining whether the derivative is positive or negative, we can identify intervals where the function increases or decreases.
In this exercise, we will learn how to use derivatives to determine increasing and decreasing intervals of functions and apply these concepts to solve NCERT questions.
Key Concepts
1. Increasing and Decreasing Functions
A function f(x) is said to be increasing on an interval I if for any x_1,x_2\in I,
\displaystyle x_1\lt x_2 \Rightarrow f(x_1)\leq f(x_2)
Similarly, f(x) is said to be decreasing on I if
\displaystyle x_1\lt x_2 \Rightarrow f(x_1)\geq f(x_2)
Thus, larger values of x correspond to equal or larger values of f(x) for an increasing function, and to equal or smaller values of f(x) for a decreasing function.
2. Strictly Increasing and Strictly Decreasing Functions
A function f(x) is strictly increasing on an interval I if
\displaystyle x_1\lt x_2 \Rightarrow f(x_1)\lt f(x_2)
Similarly, f(x) is strictly decreasing on I if
\displaystyle x_1\lt x_2 \Rightarrow f(x_1)\gt f(x_2)
In a strictly increasing function, larger values of x always correspond to larger values of f(x). Similarly, in a strictly decreasing function, larger values of x always correspond to smaller values of f(x).
3. Constant Function
A function is said to be constant on an interval I if
\displaystyle f(x)=c
for every x\in I, where c is a constant.
The value of a constant function remains unchanged throughout the interval.
4. Derivative Test for Increasing and Decreasing Functions
If a function f(x) is continuous on [a,b] and differentiable on (a,b), then:
- f'(x)\gt 0 for all x\in(a,b) ⇒ f(x) is increasing on [a,b].
- f'(x)\lt 0 for all x\in(a,b) ⇒ f(x) is decreasing on [a,b].
- f'(x)=0 for all x\in(a,b) ⇒ f(x) is constant on [a,b].
5. Finding Increasing and Decreasing Intervals
- Find the first derivative f'(x).
- Determine the critical points by solving f'(x)=0 or where f'(x) is undefined.
- Check the sign of f'(x) in each interval.
- Positive derivative implies increasing behaviour, while negative derivative implies decreasing behaviour.
This procedure helps determine the intervals on which a function is increasing or decreasing.
6. Useful Shortcuts for Exams
The following observations can help determine the sign of f'(x) more quickly and reduce the need for testing additional points in examination problems.
Important: If f'(x) is expressed as a product of distinct linear factors, then the sign of f'(x) changes alternately while crossing successive critical points arranged in increasing order. This observation often helps determine increasing and decreasing intervals without testing additional points.
For example, consider
\displaystyle f'(x)=(x-1)(x-3)(x-5)
The critical points are x=1,3,5. Since each factor is linear and distinct, the sign of f'(x) changes whenever one of these points is crossed.
− + − +
───●──────●──────●──────
1 3 5
Thus, the sign of f'(x) alternates across successive critical points.
Important: If a critical point corresponds to an even-powered factor such as (x-a)^2, the sign of f'(x) does not change while crossing that point. Therefore, such a point may not correspond to a change from increasing to decreasing or vice versa.
For example, consider
\displaystyle f'(x)=(x+1)^2(x-2)
The critical points are x=-1 and x=2. Since (x+1)^2 is a repeated factor, the sign of f'(x) remains unchanged while crossing x=-1. However, the sign changes at x=2 because (x-2) is a linear factor.
− − +
──────●──────────●──────
-1 2
Thus, there is no sign change at x=-1, whereas the sign changes from negative to positive at x=2.
Important: A function that is either increasing or decreasing throughout an interval is called a monotonic function on that interval.
Let us now solve all the NCERT questions step by step in this exercise 6.2 of Application of Derivatives.
Question 1: Application of Derivatives 6.2
1. Show that the function given by f(x)=3x+17 is increasing on \mathbb{R}.
Solution
Method 1: Using the Definition of Increasing Function
Let x_1 and x_2 be any two real numbers such that x_1\lt x_2.
Given
f(x)=3x+17
Multiplying both sides of x_1\lt x_2 by 3, we get
3x_1\lt 3x_2
Adding 17 to both sides, we get
3x_1+17\lt 3x_2+17
That is,
f(x_1)\lt f(x_2)
Hence, f(x)=3x+17 is strictly increasing on \mathbb{R}.
Method 2: Using Derivatives
Given
f(x)=3x+17
Differentiating with respect to x, we get
f'(x)=3
Since
f'(x)=3\gt 0 for all x\in\mathbb{R},
therefore, f(x)=3x+17 is increasing on \mathbb{R}.
Question 2: Application of Derivatives 6.2
2. Show that the function given by f(x)=e^{2x} is increasing on \mathbb{R}.
Solution
Given
f(x)=e^{2x}
Differentiating with respect to x,
We get, \displaystyle f'(x)=\frac{d}{dx}\left(e^{2x}\right)
\displaystyle =e^{2x}\cdot\frac{d}{dx}(2x)
\displaystyle =2e^{2x}
Since e^{2x}\gt 0 for every real number x, we have
f'(x)=2e^{2x}\gt 0 for all x\in\mathbb{R}.
Therefore, the function f(x)=e^{2x} is increasing on \mathbb{R}.
You may already be following Maths Better for NCERT Solutions for the topics like
- Matrices
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Likewise this Exercise 6.2 of Application of Derivatives for Class 12 Maths is designed to strengthen your concepts and improve step-by-step problem-solving skills. Now, let’s proceed to the next question.
Question 3: Increasing and Decreasing Functions
3. Show that the function given by f(x)=\sin x is (a) increasing in \left(0,\frac{\pi}{2}\right) (b) decreasing in \left(\frac{\pi}{2},\pi\right) (c) neither increasing nor decreasing in (0,\pi).
Solution
Given
f(x)=\sin x
Differentiating with respect to x, we get
f'(x)=\cos x
(a) Increasing in \left(0,\frac{\pi}{2}\right)
For every x\in\left(0,\frac{\pi}{2}\right) i.e. in 1st Quadrant, we have
\cos x\gt 0
Therefore,
f'(x)\gt 0 for all x\in\left(0,\frac{\pi}{2}\right).
Hence, f(x)=\sin x is increasing in \left(0,\frac{\pi}{2}\right).
(b) Decreasing in \left(\frac{\pi}{2},\pi\right)
For every x\in\left(\frac{\pi}{2},\pi\right) i.e. in 2nd Quadrant, we have
\cos x\lt 0
Therefore,
f'(x)\lt 0 for all x\in\left(\frac{\pi}{2},\pi\right).
Hence, f(x)=\sin x is decreasing in \left(\frac{\pi}{2},\pi\right).
(c) Neither Increasing nor Decreasing in (0,\pi)
From parts (a) and (b), we observe that \sin x is increasing in \left(0,\frac{\pi}{2}\right) and decreasing in \left(\frac{\pi}{2},\pi\right).
Thus, the function does not remain increasing throughout (0,\pi), nor does it remain decreasing throughout (0,\pi).
Hence, f(x)=\sin x is neither increasing nor decreasing in (0,\pi).
| Interval | Sign of f'(x) | Nature of Function |
|---|---|---|
| \left(0,\frac{\pi}{2}\right) | (+) | Increasing |
| \left(\frac{\pi}{2},\pi\right) | (-) | Decreasing |
| (0,\pi) | (+),(-) | Neither Increasing nor Decreasing |
Thus, \sin x is increasing in \left(0,\frac{\pi}{2}\right), decreasing in \left(\frac{\pi}{2},\pi\right), and neither increasing nor decreasing in (0,\pi).
Question 4: Nature of Functions: Increasing or Decreasing
4. Find the intervals in which the function f(x)=2x^2-3x is (a) increasing (b) decreasing.
Solution
Given
f(x)=2x^2-3x
Differentiating with respect to x, we get
f'(x)=4x-3
To find the critical points, let f'(x)=0. Then
4x-3=0
\Rightarrow x=\frac{3}{4}
Thus, the critical point x=\frac{3}{4} divides the number line into two disjoint intervals, namely \left(-\infty,\frac{3}{4}\right) and \left(\frac{3}{4},\infty\right).
| Interval | Sign of f'(x)=4x-3 | Nature of Function |
|---|---|---|
| \left(-\infty,\frac{3}{4}\right) | (-) | Decreasing |
| \left(\frac{3}{4},\infty\right) | (+) | Increasing |
Hence, the function f(x)=2x^2-3x is
- Increasing in \left(\frac{3}{4},\infty\right).
- Decreasing in \left(-\infty,\frac{3}{4}\right).
Question 5: Application of Derivatives 6.2
5. Find the intervals in which the function f(x)=2x^3-3x^2-36x+7 is (a) increasing (b) decreasing.
Solution
Given
f(x)=2x^3-3x^2-36x+7
Differentiating with respect to x,
We get, f'(x)=6x^2-6x-36
\displaystyle =6(x^2-x-6)
\displaystyle =6(x-3)(x+2)
To find the critical points, let f'(x)=0. Then
6(x-3)(x+2)=0
\Rightarrow x=3,\,-2
Thus, the critical points x=-2 and x=3 divide the number line into three disjoint intervals, namely (-\infty,-2), (-2,3) and (3,\infty).
| Interval | Sign of f'(x)=6(x-3)(x+2) | Nature of Function |
|---|---|---|
| (-\infty,-2) | (-)(-)\gt 0 | Increasing |
| (-2,3) | (-)(+)\lt 0 | Decreasing |
| (3,\infty) | (+)(+)\gt 0 | Increasing |
Hence, the function f(x)=2x^3-3x^2-36x+7 is
- Increasing in (-\infty,-2)\cup(3,\infty).
- Decreasing in (-2,3).
Important: If a function has n distinct critical points, then these points divide the number line into at most n+1 disjoint intervals for sign analysis.
Question 6: Sign of Derivative
6. Find the intervals in which the following functions are strictly increasing or decreasing:
(a) f(x)=x^2+2x-5
Solution
Given
f(x)=x^2+2x-5
Differentiating with respect to x, we get
f'(x)=2x+2
\displaystyle =2(x+1)
To find the critical points, let f'(x)=0. Then
2(x+1)=0
\Rightarrow x=-1
Thus, the critical point x=-1 divides the number line into two disjoint intervals, namely (-\infty,-1) and (-1,\infty).
| Interval | Sign of f'(x)=2(x+1) | Nature of Function |
|---|---|---|
| (-\infty,-1) | (-) | Strictly Decreasing |
| (-1,\infty) | (+) | Strictly Increasing |
Hence, the function f(x)=x^2+2x-5 is
- Strictly increasing in (-1,\infty).
- Strictly decreasing in (-\infty,-1).
(b) f(x)=10-6x-2x^2
Solution
Given
f(x)=10-6x-2x^2
Differentiating with respect to x, we get
f'(x)=-6-4x
\displaystyle =-2(2x+3)
To find the critical points, let f'(x)=0. Then
-2(2x+3)=0
\Rightarrow x=-\frac{3}{2}
Thus, the critical point x=-\frac{3}{2} divides the number line into two disjoint intervals, namely \left(-\infty,-\frac{3}{2}\right) and \left(-\frac{3}{2},\infty\right).
| Interval | Sign of f'(x)=-2(2x+3) | Nature of Function |
|---|---|---|
| \left(-\infty,-\frac{3}{2}\right) | (-)(-)\gt 0 | Strictly Increasing |
| \left(-\frac{3}{2},\infty\right) | (-)(+)\lt 0 | Strictly Decreasing |
Hence, the function f(x)=10-6x-2x^2 is
- Strictly increasing in \left(-\infty,-\frac{3}{2}\right).
- Strictly decreasing in \left(-\frac{3}{2},\infty\right).
(c) f(x)=-2x^3-9x^2-12x+1
Solution
Given
f(x)=-2x^3-9x^2-12x+1
Differentiating with respect to x,
We get, f'(x)=-6x^2-18x-12
\displaystyle =-6(x^2+3x+2)
\displaystyle =-6(x+1)(x+2)
To find the critical points, let f'(x)=0. Then
-6(x+1)(x+2)=0
x=-1,\,-2
Thus, the critical points x=-2 and x=-1 divide the number line into three disjoint intervals, namely (-\infty,-2), (-2,-1) and (-1,\infty).
| Interval | Sign of f'(x)=-6(x+1)(x+2) | Nature of Function |
|---|---|---|
| (-\infty,-2) | (-)(-)(-)\lt 0 | Strictly Decreasing |
| (-2,-1) | (-)(-)(+)\gt 0 | Strictly Increasing |
| (-1,\infty) | (-)(+)(+)\lt 0 | Strictly Decreasing |
Hence, the function f(x)=-2x^3-9x^2-12x+1 is
- Strictly increasing in (-2,-1).
- Strictly decreasing in (-\infty,-2)\cup(-1,\infty).
(d) f(x)=6-9x-x^2
Solution
Given
f(x)=6-9x-x^2
Differentiating with respect to x, we get
f'(x)=-9-2x
\displaystyle =-(2x+9)
To find the critical points, let f'(x)=0. Then
-(2x+9)=0
\Rightarrow x=-\frac{9}{2}
Thus, the critical point x=-\frac{9}{2} divides the number line into two disjoint intervals, namely \left(-\infty,-\frac{9}{2}\right) and \left(-\frac{9}{2},\infty\right).
| Interval | Sign of f'(x)=-(2x+9) | Nature of Function |
|---|---|---|
| \left(-\infty,-\frac{9}{2}\right) | (-)(-)\gt 0 | Strictly Increasing |
| \left(-\frac{9}{2},\infty\right) | (-)(+)\lt 0 | Strictly Decreasing |
Hence, the function f(x)=6-9x-x^2 is
- Strictly increasing in \left(-\infty,-\frac{9}{2}\right).
- Strictly decreasing in \left(-\frac{9}{2},\infty\right).
In other words, the function is increasing for x\lt -\frac{9}{2} and decreasing for x\gt -\frac{9}{2}.
(e) f(x)=(x+1)^3(x-3)^3
Solution
Given
f(x)=(x+1)^3(x-3)^3
Using the product rule,
We get, \displaystyle f'(x)=(x+1)^3\frac{d}{dx}(x-3)^3+(x-3)^3\frac{d}{dx}(x+1)^3
\displaystyle =(x+1)^3\cdot 3(x-3)^2+(x-3)^3\cdot 3(x+1)^2
\displaystyle =3(x+1)^2(x-3)^2[(x+1)+(x-3)]
So, \displaystyle f'(x)=6(x+1)^2(x-3)^2(x-1)
To find the critical points, let f'(x)=0. Then
6(x+1)^2(x-3)^2(x-1)=0
\Rightarrow x=-1,\;1,\;3
Since 6\gt 0, (x+1)^2\geq 0 and (x-3)^2\geq 0, the sign of f'(x) depends only on the factor (x-1).
Thus, the critical points x=-1, x=1 and x=3 divide the number line into four disjoint intervals, namely (-\infty,-1), (-1,1), (1,3) and (3,\infty).
| Interval | Sign of f'(x)=6(x+1)^2(x-3)^2(x-1) | Nature of Function |
|---|---|---|
| (-\infty,-1) | (+)(+)(-)\lt 0 | Strictly Decreasing |
| (-1,1) | (+)(+)(-)\lt 0 | Strictly Decreasing |
| (1,3) | (+)(+)(+)\gt 0 | Strictly Increasing |
| (3,\infty) | (+)(+)(+)\gt 0 | Strictly Increasing |
Hence, the function f(x)=(x+1)^3(x-3)^3 is
- Strictly increasing in (1,3)\cup(3,\infty).
- Strictly decreasing in (-\infty,-1)\cup(-1,1).
Question 7: Exercise 6.2
7. Show that y=\log(1+x)-\frac{2x}{2+x},\;x\gt -1 is an increasing function of x throughout its domain.
Solution
Given
y=\log(1+x)-\frac{2x}{2+x},\qquad x\gt -1
Differentiating with respect to x, we get
\displaystyle \frac{dy}{dx}=\frac{1}{1+x}-\frac{(x+2)(2)-2x(1)}{(x+2)^2}
\displaystyle =\frac{1}{x+1}-\frac{4}{(x+2)^2}
Taking the LCM,
We get, \displaystyle \frac{dy}{dx}=\frac{(x+2)^2-4(x+1)}{(x+1)(x+2)^2}
\displaystyle =\frac{x^2+4x+4-4x-4}{(x+1)(x+2)^2}
\displaystyle =\frac{x^2}{(x+1)(x+2)^2}
Now, x^2\geq 0, x+1\gt 0 for x\gt -1 and (x+2)^2\gt 0.
Therefore,
\displaystyle \frac{dy}{dx}=\frac{x^2}{(x+1)(x+2)^2}\geq 0
for all x\gt -1.
Hence, the function y=\log(1+x)-\frac{2x}{2+x} is an increasing function of x throughout its domain.
Factors with even powers such as (x-a)^2 do not change sign when crossing x=a. Therefore, they may not affect the increasing or decreasing nature of a function.
Question 8: Application of Derivatives 6.2
8. Find the values of x for which y=[x(x-2)]^2 is an increasing function.
Solution
Given
y=[x(x-2)]^2
Using the chain rule,
We get, \displaystyle \frac{dy}{dx}=2[x(x-2)]\frac{d}{dx}[x(x-2)]
\displaystyle =2[x(x-2)](2x-2)
\displaystyle =4x(x-2)(x-1)
To find the critical points, let \frac{dy}{dx}=0. Then
4x(x-2)(x-1)=0
\Rightarrow x=0,\;1,\;2
Thus, the critical points x=0, x=1 and x=2 divide the number line into four disjoint intervals, namely (-\infty,0), (0,1), (1,2) and (2,\infty).
| Interval | Sign of \displaystyle \frac{dy}{dx}=4x(x-2)(x-1) | Nature of Function |
|---|---|---|
| (-\infty,0) | (-)(-)(-)\lt 0 | Decreasing |
| (0,1) | (+)(-)(-)\gt 0 | Increasing |
| (1,2) | (+)(-)(+)\lt 0 | Decreasing |
| (2,\infty) | (+)(+)(+)\gt 0 | Increasing |
Hence, the function y=[x(x-2)]^2 is increasing in (0,1) and (2,\infty).
In other words, the function is increasing for 0\lt x\lt 1 and x\gt 2.
NCERT Class 12 Maths includes several important exercises across both Part 1 and Part 2, and I’ll cover them one by one with clear explanations and step-by-step solutions. Many of these questions are also available in video format on my YouTube Channel, @MathsBetter, to help you understand the concepts more visually.
Now, let’s proceed to the next question.
Question 9: Application of Derivatives 6.2
9. Prove that y=\frac{4\sin\theta}{2+\cos\theta}-\theta is an increasing function of \theta in \left[0,\frac{\pi}{2}\right].
Solution
Given
y=\frac{4\sin\theta}{2+\cos\theta}-\theta
Differentiating with respect to \theta,
We get, \displaystyle \frac{dy}{d\theta}=\frac{(2+\cos\theta)(4\cos\theta)-4\sin\theta(-\sin\theta)}{(2+\cos\theta)^2}-1
\displaystyle =\frac{8\cos\theta+4\cos^2\theta+4\sin^2\theta}{(2+\cos\theta)^2}-1
\displaystyle =\frac{8\cos\theta+4(\cos^2\theta+\sin^2\theta)}{(2+\cos\theta)^2}-1
or
\displaystyle =\frac{8\cos\theta+4}{(2+\cos\theta)^2}-1
\displaystyle =\frac{8\cos\theta+4-(2+\cos\theta)^2}{(2+\cos\theta)^2}
i.e.
\displaystyle =\frac{8\cos\theta+4-(4+4\cos\theta+\cos^2\theta)}{(2+\cos\theta)^2}
\displaystyle =\frac{4\cos\theta-\cos^2\theta}{(2+\cos\theta)^2}
So, \displaystyle \frac{dy}{d\theta}=\frac{\cos\theta(4-\cos\theta)}{(2+\cos\theta)^2}
Now, for \theta\in\left[0,\frac{\pi}{2}\right], \theta lies in the first quadrant. Therefore,
- 0\leq\cos\theta\leq 1.
- Multiplying by -1 throughout, we get -1\leq-\cos\theta\leq 0.
- Adding 4 throughout, we obtain 3\leq 4-\cos\theta\leq 4. Hence, 4-\cos\theta\gt 0.
- Also, adding 2 throughout to 0\leq\cos\theta\leq 1, we get 2\leq 2+\cos\theta\leq 3. Hence, (2+\cos\theta)^2\gt 0.
Therefore,
\displaystyle \frac{dy}{d\theta}=\frac{\cos\theta(4-\cos\theta)}{(2+\cos\theta)^2}\geq 0
for all \theta\in\left[0,\frac{\pi}{2}\right].
Hence, the function y=\frac{4\sin\theta}{2+\cos\theta}-\theta is an increasing function of \theta in \left[0,\frac{\pi}{2}\right].
Question 10: Increasing and Decreasing Functions
10. Prove that the logarithmic function is increasing on (0,\infty).
Solution
Let
f(x)=\log x
The domain of \log x is (0,\infty).
Differentiating with respect to x, we get
\displaystyle f'(x)=\frac{d}{dx}(\log x)=\frac{1}{x}
Now, for every x\in(0,\infty), we have x\gt 0. Therefore,
\displaystyle \frac{1}{x}\gt 0
That is,
f'(x)\gt 0 for all x\in(0,\infty).
Hence, the logarithmic function f(x)=\log x is increasing on (0,\infty).
While applying the First Derivative Test, it is often easier to factorise f'(x) completely and analyse the sign of each factor separately.
Question 11: Application of Derivatives
11. Prove that the function f(x)=x^2-x+1 is neither strictly increasing nor decreasing on (-1,1).
Solution
Given
f(x)=x^2-x+1
Differentiating with respect to x, we get
f'(x)=2x-1
To find the critical points, let f'(x)=0. Then
2x-1=0
\Rightarrow x=\frac{1}{2}
Since \frac{1}{2}\in(-1,1), the critical point x=\frac{1}{2} divides the interval (-1,1) into two disjoint intervals, namely \left(-1,\frac{1}{2}\right) and \left(\frac{1}{2},1\right).
| Interval | Sign of f'(x)=2x-1 | Nature of Function |
|---|---|---|
| \left(-1,\frac{1}{2}\right) | (-)\lt 0 | Strictly Decreasing |
| \left(\frac{1}{2},1\right) | (+)\gt 0 | Strictly Increasing |
Thus, the function is strictly decreasing in \left(-1,\frac{1}{2}\right) and strictly increasing in \left(\frac{1}{2},1\right).
Hence, the function f(x)=x^2-x+1 is neither strictly increasing nor strictly decreasing on (-1,1).
Question 12: Application of Derivatives 6.2
12. Which of the following functions are decreasing on \left(0,\frac{\pi}{2}\right)?
- (A) \cos x
- (B) \cos 2x
- (C) \cos 3x
- (D) \tan x
Solution
(A) f(x)=\cos x
f'(x)=-\sin x
For x\in\left(0,\frac{\pi}{2}\right), x lies in the first quadrant. Therefore, \sin x\gt 0.
Hence, f'(x)=-\sin x\lt 0. Therefore, \cos x is decreasing on \left(0,\frac{\pi}{2}\right).
(B) f(x)=\cos 2x
f'(x)=-2\sin 2x
For x\in\left(0,\frac{\pi}{2}\right), multiplying throughout by 2, we get
0\lt 2x\lt\pi
Since 2x lies in the first or second quadrant, \sin 2x\gt 0. Therefore,
f'(x)=-2\sin 2x\lt 0
Therefore, \cos 2x is decreasing on \left(0,\frac{\pi}{2}\right).
(C) f(x)=\cos 3x
f'(x)=-3\sin 3x
For x\in\left(0,\frac{\pi}{2}\right), multiplying throughout by 3, we get
0\lt 3x\lt\frac{3\pi}{2}
Now, \sin 3x\gt 0 when 0\lt 3x\lt\pi, whereas \sin 3x\lt 0 when \pi\lt 3x\lt\frac{3\pi}{2}.
Therefore, \sin 3x changes sign in the interval \left(0,\frac{\pi}{2}\right).
Hence, f'(x)=-3\sin 3x also changes sign in this interval.
Therefore, \cos 3x is not decreasing on the entire interval \left(0,\frac{\pi}{2}\right).
(D) f(x)=\tan x
f'(x)=\sec^2x
Since \sec^2x\gt 0 for all x\in\left(0,\frac{\pi}{2}\right), \tan x is increasing on \left(0,\frac{\pi}{2}\right).
Hence, the functions which are decreasing on \left(0,\frac{\pi}{2}\right) are (A) and (B).
Important: The sign of the first derivative determines the behaviour of a function. If f'(x)\gt 0, the function is increasing, while if f'(x)\lt 0, the function is decreasing.
Question 13: Application of Derivatives Exercise 6.2
13. On which of the following intervals is the function f(x)=x^{100}+\sin x-1 decreasing?
- (A) (0,1)
- (B) \left(\frac{\pi}{2},\pi\right)
- (C) \left(0,\frac{\pi}{2}\right)
- (D) None of these
Solution
Given
f(x)=x^{100}+\sin x-1
Differentiating with respect to x, we get
f'(x)=100x^{99}+\cos x
We now examine the given intervals.
(A) (0,1)
For x\in(0,1), we have x^{99}\gt 0. Also, \cos x\gt 0 since 0\lt x\lt 1\lt\frac{\pi}{2}.
Therefore, f'(x)=100x^{99}+\cos x\gt 0. Hence, f(x) is increasing on (0,1).
(B) \left(\frac{\pi}{2},\pi\right)
Since x\in\left(\frac{\pi}{2},\pi\right), we have
x\gt\frac{\pi}{2}\gt1
Raising both sides to the power 99, we get
x^{99}\gt1
Multiplying by 100 throughout,
100x^{99}\gt100
Also, for any real number x,
-1\leq\cos x\leq1
Adding these inequalities, we obtain
100x^{99}+\cos x\gt100-1=99\gt0
Thus, f(x) is increasing on \left(\frac{\pi}{2},\pi\right).
(C) \left(0,\frac{\pi}{2}\right)
For x\in\left(0,\frac{\pi}{2}\right), both 100x^{99} and \cos x are positive.
Therefore, f'(x)=100x^{99}+\cos x\gt 0. Hence, f(x) is increasing on \left(0,\frac{\pi}{2}\right).
Thus, the function is not decreasing on any of the given intervals.
✅️ Hence, the correct answer is (D) None of these.
Question 14: Application of Derivatives 6.2
14. For what values of a the function f(x)=x^2+ax+1 is increasing on [1,2]?
Solution
Given
f(x)=x^2+ax+1
Differentiating with respect to x, we get
f'(x)=2x+a
For the function to be increasing on [1,2], we must have
f'(x)\geq 0
for every x\in[1,2].
Now, f'(x)=2x+a is a linear function of x with positive coefficient of x. Therefore, f'(x) increases as x increases.
Hence, the minimum value of f'(x) on [1,2] occurs at x=1.
Therefore, it is sufficient to ensure that
f'(1)\geq 0
2+a\geq 0
i.e. a\geq -2
Hence, the function f(x)=x^2+ax+1 is increasing on [1,2] if
\boxed{a\geq -2}
Question 15: Application of Derivatives 6.2
15. Let I be any interval disjoint from [-1,1]. Prove that the function f(x)=x+\frac{1}{x} is increasing on I.
Solution
Given
f(x)=x+\frac{1}{x}
Differentiating with respect to x,
We get, \displaystyle f'(x)=1-\frac{1}{x^2}
\displaystyle =\frac{x^2-1}{x^2}
\displaystyle =\frac{(x-1)(x+1)}{x^2}
Since I is disjoint from [-1,1], every point of I satisfies either
- x\lt -1, or
- x\gt 1.
Case I: x\lt -1
Since x\lt -1, we have x-1\lt 0 and x+1\lt 0. Also, x^2\gt 0.
Therefore,
\displaystyle f'(x)=\frac{(-)(-)}{(+)}\gt 0
Case II: x\gt 1
Since x\gt 1, we have x-1\gt 0 and x+1\gt 0. Also, x^2\gt 0.
Therefore,
\displaystyle f'(x)=\frac{(+)(+)}{(+)}\gt 0
Hence, f'(x)\gt 0 for every x\in I.
Therefore, the function f(x)=x+\frac{1}{x} is increasing on I.
Alternative Method
We have
\displaystyle f'(x)=\frac{(x-1)(x+1)}{x^2}
To find the critical points, let f'(x)=0. Then
\frac{(x-1)(x+1)}{x^2}=0
\Rightarrow x=-1,\;1
Thus, the critical points x=-1 and x=1 divide the number line into three disjoint intervals, namely (-\infty,-1), (-1,1) and (1,\infty).
| Interval | Sign of \displaystyle f'(x)=\frac{(x-1)(x+1)}{x^2} | Nature of Function |
|---|---|---|
| (-\infty,-1) | \frac{(-)(-)}{(+)}\gt0 | Increasing |
| (-1,1) | \frac{(-)(+)}{(+)}\lt0 | Decreasing |
| (1,\infty) | \frac{(+)(+)}{(+)}\gt0 | Increasing |
Hence, f(x)=x+\frac1x is increasing on (-\infty,-1) and (1,\infty).
Since any interval I disjoint from [-1,1] lies entirely in either (-\infty,-1) or (1,\infty), the function is increasing on I.
Question 16: Application of Derivatives 6.2
16. Prove that the function f(x)=\log(\sin x) is increasing on \left(0,\frac{\pi}{2}\right) and decreasing on \left(\frac{\pi}{2},\pi\right).
Solution
Given
f(x)=\log(\sin x)
Differentiating with respect to x,
We get, \displaystyle f'(x)=\frac{1}{\sin x}\cdot\frac{d}{dx}(\sin x)
\displaystyle =\frac{\cos x}{\sin x}
\displaystyle =\cot x
For x\in\left(0,\frac{\pi}{2}\right)
For x\in\left(0,\frac{\pi}{2}\right), the angle x lies in the first quadrant. Since \cot x is positive in the first quadrant, we have
f'(x)=\cot x\gt 0
Therefore, f(x)=\log(\sin x) is increasing on \left(0,\frac{\pi}{2}\right).
For x\in\left(\frac{\pi}{2},\pi\right), the angle x lies in the second quadrant. Since \cot x is negative in the second quadrant, we have
f'(x)=\cot x\lt 0
Therefore, f(x)=\log(\sin x) is decreasing on \left(\frac{\pi}{2},\pi\right).
Question 17: Increasing and Decreasing Functions
17. Prove that the function f(x)=\log|\cos x| is decreasing on \left(0,\frac{\pi}{2}\right) and increasing on \left(\frac{3\pi}{2},2\pi\right).
Solution
Given
f(x)=\log|\cos x|
Since x\in\left(0,\frac{\pi}{2}\right)\cup\left(\frac{3\pi}{2},2\pi\right), we have \cos x\gt 0. Therefore,
|\cos x|=\cos x
Hence,
f(x)=\log(\cos x)
Differentiating with respect to x,
We get, \displaystyle f'(x)=\frac{1}{\cos x}\cdot\frac{d}{dx}(\cos x)
\displaystyle =\frac{-\sin x}{\cos x}
\displaystyle =-\tan x
For x\in\left(0,\frac{\pi}{2}\right)
The angle x lies in the first quadrant. Since \tan x is positive in the first quadrant, we have
f'(x)=-\tan x\lt 0
Therefore, f(x)=\log|\cos x| is decreasing on \left(0,\frac{\pi}{2}\right).
For x\in\left(\frac{3\pi}{2},2\pi\right)
The angle x lies in the fourth quadrant. Since \tan x is negative in the fourth quadrant, we have
f'(x)=-\tan x\gt 0
Therefore, f(x)=\log|\cos x| is increasing on \left(\frac{3\pi}{2},2\pi\right).
Question 18: Application of Derivatives 6.2
18. Prove that the function f(x)=x^3-3x^2+3x-100 is increasing in \mathbb{R}.
Solution
Given
f(x)=x^3-3x^2+3x-100
Differentiating with respect to x,
We get, f'(x)=3x^2-6x+3
\displaystyle =3(x^2-2x+1)
\displaystyle =3(x-1)^2
Now, 3\gt 0 and (x-1)^2\geq 0 for every real number x.
Therefore,
f'(x)=3(x-1)^2\geq 0
for all x\in\mathbb{R}.
Also, f'(x)=0 only when x=1.
Hence, the derivative is non-negative throughout \mathbb{R}. Therefore, the function f(x)=x^3-3x^2+3x-100 is increasing in \mathbb{R}.
Important: Expressions such as e^x, x^2 (for x\neq 0) and (x^2+1) are always positive and can simplify sign analysis considerably.
Question 19: Application of Derivatives 6.2 – MCQ
19. The interval in which y=x^2e^{-x} is increasing is
- (A) (-\infty,\infty)
- (B) (-2,0)
- (C) (2,\infty)
- (D) (0,2)
Solution
Given
y=x^2e^{-x}
Differentiating using the product rule,
We get, \displaystyle \frac{dy}{dx}=x^2\frac{d}{dx}(e^{-x})+e^{-x}\frac{d}{dx}(x^2)
\displaystyle =x^2(-e^{-x})+e^{-x}(2x)
\displaystyle =e^{-x}(2x-x^2)
or \displaystyle \frac{dy}{dx}=e^{-x}x(2-x)
To find the critical points, let \frac{dy}{dx}=0. Then
e^{-x}x(2-x)=0
Since e^{-x}\gt 0 for all real values of x,
x(2-x)=0
\Rightarrow x=0,\;2
Thus, the critical points x=0 and x=2 divide the number line into three disjoint intervals, namely (-\infty,0), (0,2) and (2,\infty).
| Interval | Sign of \displaystyle \frac{dy}{dx}=e^{-x}x(2-x) | Nature of Function |
|---|---|---|
| (-\infty,0) | (+)(-)(+)\lt 0 | Decreasing |
| (0,2) | (+)(+)(+)\gt 0 | Increasing |
| (2,\infty) | (+)(+)(-)\lt 0 | Decreasing |
Hence, the function y=x^2e^{-x} is increasing in (0,2).
✅️ Therefore, the correct answer is (D).
Common Mistakes to Avoid
- Incorrect differentiation: Errors in finding f'(x) lead to incorrect intervals of increase and decrease. Differentiate carefully before proceeding to sign analysis.
- Ignoring the domain of the function: Always identify the domain first. Points where the function is undefined must not be included while determining intervals of increase or decrease.
- Missing critical points: Critical points occur where f'(x)=0 or where f'(x) is undefined. Missing any such point may lead to incorrect intervals.
- Not dividing the number line correctly: After finding the critical points, divide the number line into disjoint intervals and analyse the sign of f'(x) separately in each interval.
- Ignoring the sign of individual factors: When f'(x) is factorised, determine the sign of each factor carefully before finding the sign of the derivative.
- Forgetting that positive factors remain positive: Expressions such as x^2, (x-1)^2, e^x and (x^2+1) are always positive and simplify sign analysis considerably.
- Confusing increasing with strictly increasing: If f'(x)\gt 0 throughout an interval, the function is strictly increasing. If f'(x)\geq 0 throughout an interval, the function is increasing.
- Ignoring trigonometric signs: For trigonometric functions, determine the quadrant first and then use the signs of \sin x, \cos x, \tan x or \cot x accordingly.
- Drawing conclusions without sign analysis: The values of the critical points alone do not determine whether a function is increasing or decreasing. The sign of f'(x) in each interval must be checked.
- Incorrect final conclusion: Always state the required intervals clearly using interval notation and ensure they agree with the sign table obtained from f'(x).
Continue Learning
After completing Exercise 6.2 of Application of Derivatives, you should now be familiar with using derivatives to determine whether a function is increasing or decreasing on a given interval. You have also seen how the sign of the first derivative helps in identifying intervals of increase and decrease and in proving monotonic behaviour of functions.
To strengthen your understanding further, make sure that you revise:
- The definitions of increasing, decreasing, strictly increasing and strictly decreasing functions
- The relationship between the sign of f'(x) and the behaviour of a function
- The First Derivative Test for determining increasing and decreasing intervals
- How to find critical points by solving f'(x)=0 or identifying points where f'(x) is undefined
- Sign analysis of factorised derivatives using interval tables
- Functions that are neither increasing nor decreasing on a given interval
- The signs of trigonometric functions in different quadrants
- Monotonicity of logarithmic, exponential and trigonometric functions
- How to determine intervals of increase and decrease from a sign table
- Proper interpretation of interval notation and domain restrictions while applying the First Derivative Test
Explore More
Try solving a variety of questions on increasing and decreasing functions on your own. Differentiate carefully, find the critical points and examine the sign of the derivative in each interval. Practise constructing sign tables and interpreting them correctly to determine where a function is increasing or decreasing. Also revise the signs of trigonometric functions in different quadrants and pay attention to domain restrictions. Regular practice will strengthen your understanding of the First Derivative Test and prepare you for applications such as maxima and minima.
All the best and keep learning 👍
