NCERT Solutions Class 12 Mathematics Chapter 6 - Application of Derivatives
Mathematics is a subject requiring a strong conceptual understanding of its applications. Hence, students are advised to study the subject from good resources to gain in-depth knowledge about Mathematics chapters with examples, solutions, theorems as well as exercises and to score well in exams.
The Class 12 Mathematics Chapter 6 'Application of derivatives' is a part of the Calculus section of Mathematics. It covers the various applications of derivatives and their role while doing calculations. The vital topics covered in Chapter 6 Class 12 Mathematics include:
- Rate of Change of Quantities
- Increasing and Decreasing Functions
- Tangents and Normals
- Maxima and Minima
- Maximum and Minimum Values of a Function in a Closed Interval
The chapter's important formulas are listed in the NCERT Solutions Class 12 Mathematics Chapter 6. Also, one can find easy ways to carry out calculations after referring to them thoroughly and consistently.
Extramarks is a reliable and trustworthy source for all the NCERT-related study material. One can find NCERT textbooks, NCERT solutions, NCERT exemplars, NCERT revision notes, NCERT formulas, NCERT-based mock tests and the NCERT solutions Class 12 Mathematics Chapter 6 on the Extramarks official website.
Key Topics Covered in NCERT Solutions Class 12 Mathematics Chapter 6
NCERT Solutions Class 12 Mathematics Chapter 6 is about derivatives and their applications.
This chapter is a part of Calculus and requires the study of differential calculus. One should have a good command of it to excel in this chapter.
If you have a hold on calculus, this chapter will help students to learn to carry out calculations easily. As a result, they will be able to approach problems more logically. The entire chapter is covered thoroughly in the NCERT Solutions Class 12 Mathematics Chapter 6 and is available on the Extramarks website.
After completing Chapter 6 Mathematics Class 12, students will learn to analyse the problems with a better approach and be able to solve them easily.
A derivative is the rate of change or the amount on which a particular function changes at one given point.
In the above-given figure, the function is represented in black colour, and a tangent is represented in red colour.
Rate of Change of Quantities
The derivative ds/dt is used to show the rate of change of distance s to the time t.
Assume a particular quantity y varies with another quantity x which satisfies y = f(x), here dy/dx or f’(x) shows the rate of change of y w.r.t. x and [dy/dx]x = x0 or f’(x0) shows the rate of change of y w.r.t. x at x = x0.
Now, assume the two variables x and y are varying w.r.t. another variable t i.e. x = f(t) and y = g(t)
Thus, by chain rule
dy/dx = (dy/dt)/(dx/dt) here dx/dt ≠ 0
So, the rate of change of y to x could be calculated using the rate of change of y at a given time and that of x to t.
Increasing and Decreasing Functions
The function's derivative might determine if it is increasing or decreasing at any intervals in its domain.
Assume I will be an open interval contained in the domain of the real-valued function f. Then f is said to be
(i) increasing on I when x1 < x2 in I => f(x1) ≤ f(x2) for all x1, x2 Є I.
(ii) strictly increasing on I if x1 < x2 in I => f(x1) < f(x2) for all x1, x2 Є I.
(iii) decreasing on I if x1 < x2 in I => f(x1) ≥ f(x2) for all x1, x2 Є I.
(iv) strictly decreasing on I if x1 < x2 in I => f(x1) > f(x2) for all x1, x2 Є I.
However, some functions are neither increasing nor decreasing.
The graphical representations of all these functions are given below:
function f will be said to be increasing at x0 when there exists an interval I = (x0 – h, x0 + h), h > 0 such that for x1, x2 ∈ I,
x1 < x2 in I => f (x1) ≤ f (x2)
With the help of a theorem listed in the NCERT Solutions Class 12 Mathematics Chapter 6, you can test for increasing and decreasing functions for a given interval.
Tangents and Normals
For the given equation of a straight line that is passing through a given point (x0, y0) having a finite slope, m is represented as y – y0 = m(x – x0)
Assume the given curve y = f(x), and the tangent of the curve at that point (x0, y0) will be
[dy/dx](x0, y0) or f’(x0)
Now, the equation of the tangent of the curve y = f(x) at (x0, y0) will be
y – y0 = f’(x0)(x – x0)
The slope of the normal to curve y = f(x) at (x0, y0) is given by -1/ f’(x0) here f’(x0) ≠ 0
For the equation of normal to the curve, y = f(x) at (x0, y0) will be
Y – y0 = (-1/f’(x0))(x – x0)
(y – y0)* f’(x0) +(x – x0) = 0
(i) If the slope of the tangent line is zero, then tan θ = 0, i.e. θ = 0, which means a tangent line is parallel to the x-axis. For this case, the equation of the given tangent at the point (x0, y0) will be y = y0.
(ii) If θ -> π/2 then tan θ → ∞, that means the tangent line will be perpendicular to the x-axis, that is parallel to the y-axis. For this case, the equation of the tangent at (x0, y0) will be x = x0.
An approximation is anything which is similar but not the same as something else.
Assume f : D => R, D Ì R, will be a given function and assume y = f (x). Assume ∆x denotes a small increment in terms of x.
Now, consider the increment in y corresponding to the increment in x will be ∆y = f (x + ∆x) – f (x).
(i) The differential of x is represented as dx = ∆x.
(ii) The differential of y is represented as dy = f’(x) dx or dy = (dy/dx) * ∆x
Maxima and Minima
In the section, you will learn the different methods of calculating a function's maximum and minimum values in a given domain. You will also learn about the absolute maximum and minimum of a function used to find the solution to many applied problems.
You will clearly understand this topic through the definitions and theorems included in the NCERT Solutions Class 12 Mathematics Chapter 6 available on the Extramarks official website.
Maximum and Minimum Values of a Function in a Closed Interval
You will learn about the two theorems to find out the absolute maximum and minimum values of a function upon a closed interval I.
Theorem: Let f be a continuous function on an interval I = [a, b]. Then f has the absolute maximum value, as well as f attains it at least once in I. Also, f has the absolute minimum value and attains it at least once in I.
Theorem: Let f be a differentiable function on a closed interval I and let c be any interior point of I. Then
(i) f′(c) = 0 when f attains its absolute maximum value at c.
(ii) f′(c) = 0 when f attains its absolute minimum value at c.
NCERT Solutions Class 12 Mathematics Chapter 6 Exercise & Solutions
Find all the NCERT solutions to the exercises covered in the chapter in the NCERT Solutions Class 12 Mathematics Chapter 6. It is prepared by the subject experts while adhering to the NCERT book and following the CBSE guidelines and curriculum.
All the vital topics and key concepts are covered in a point-wise manner. NCERT Solutions Class 12 Mathematics Chapter 6 is written end-to-end, highlighting everything in detail by the faculty experts. . Therefore, it is trusted by students and teachers across the private and government schools.
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- Class 12 Mathematics Chapter 6: Questions and Answers
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NCERT Exemplar Class 12 Mathematics
NCERT Exemplar Class 12 Mathematics book is an excellent source for students preparing for JEE Mains, NEET, MHT-CET etc. It has questions according to the topics and concepts covered in the NCERT textbook. This aids students in solving all types of questions confidently.
It has different questions with varying levels of difficulty, which helps students to improve their performance in various tests and exams and definitely builds their confidence level in the process. The questions are selected from different sources for students to prepare and improve their performance. It is also a complete source of information for CBSE students preparing for their 12th standard examinations. The students think logically about a problem after referring to the NCERT Solutions Class 12 Mathematics Chapter 6 and NCERT Exemplar.
The NCERT solutions Class 12 Mathematics Chapter 6 is prepared after analysing CBSE past years’ question papers. It contains extra questions from the NCERT Class 12 Mathematics textbook. Students can refer to NCERT Exemplar for Class 12 Mathematics for more practice and face their examinations with courage. They can assure themselves that nothing remains untouched in the chapter and will score very well in all their examinations. To get good grades in exams students must refer to multiple study resources, practice a lot of questions and stick to a study schedule and follow it religiously to come out with flying colours.
Key Features for NCERT Solutions Class 12 Mathematics Chapter 6
Regular practice is quite necessary for the students to excel. Hence, NCERT Solutions Class 12 Mathematics Chapter 6 helps students develop the habit of regular practice and to clarify their doubts then and there, take regular tests to assess their performance and get proper feedback to step up their learning. . The key features are as follows:
- You can find all the topics covered as well as in text to end text exercises from the chapter covered in the NCERT Solutions Class 12 Mathematics Chapter 6.
- It helps students analyse the easy and difficult topics in the chapter.
- After completing the NCERT Solutions Class 12 Mathematics Chapter 6, students will become experts in solving derivatives problems.
Q.1 Find the rate of change of the area of a circle with respect to its radius r when
(a) r = 3 cm (b) r = 4 cm
Q.2 The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm?
Q.3 The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.
Q.4 An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?
Q.5 A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?
Q.6 The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?
Q.7 The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.
Q.8 A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.
Q.9 A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.
Q.10 A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall?
Q.11 A particle moves along the curve 6y = x3 + 2. Find the points on the curve at which the y coordinate is changing 8 times as fast as the x-coordinate.
Q.12 The radius of an air bubble is increasing at the rate of ½ cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?
Q.15 The total cost C (x) in Rupees associated with the production of x units of an item is given
by C(x) = 0.007x3 – 0.003 x2 + 15x + 4000 Find the marginal cost when 17 units are produced
Q.16 The total revenue in Rupees received from the sale of x units of a product is given by
R(x) = 13x2 + 26x + 15 Find the marginal revenue when x = 7.
Q.19 Show that the function given by f(x) = 3x + 17 is strictly increasing on R.
Q.20 Show that the function given by f(x) = e2x is strictly increasing on R
Q.22 Find the intervals in which the function f given by f(x) = 2x2 − 3x is
(a) strictly increasing (b) strictly decreasing
Q.23 Find the intervals in which the function f given by f(x) = 2x3 − 3x2 − 36x + 7 is
(a) strictly increasing (b) strictly decreasing
Q.24 Find the intervals in which the following functions are strictly increasing or decreasing:
(a) x2 + 2x − 5 (b)10 − 6x − 2x2
(c) −2x3 − 9x2 − 12x + 1 (d) 6 − 9x − x2
(e) (x + 1)3 (x − 3)3
Q.29 Prove that the function f given by f(x)= x2 − x + 1 is neither strictly increasing
nor strictly decreasing on (−1, 1).
Q.32 Find the least value of a such that the function f given f(x) =x2+ax+1 is strictly increasing on (1, 2).
Q.38 Find the slope of the tangent to the curve y = 3x4 − 4x at x = 4
Q.40 Find the slope of the tangent to curve y = x3−x+1 at the point whose x-coordinate is 2.
Q.41 Find the slope of the tangent to curve y = x3−3x+2 at the point whose x-coordinate
Q.44 Find points at which the tangent to the curve
y = x3 − 3x2 − 9x + 7 is parallel to the x axis.
Q.45 Find a point on the curve y = (x − 2)2 at which the tangent is parallel to the chord joining the
points (2, 0) and (4, 4).
Q.46 Find the point on the curve y = x3 − 11x + 5
at which the tangent is y = x − 11.
Q.51 Find the equations of the tangent and normal to the given curves at the indicated points:
(i) y = x4 − 6x3+13x2−10x + 5 at (0, 5)
(ii) y = x4 − 6x3+13x2−10x + 5 at (1, 3)
(iii) y = x3 at (1, 1)
(iv) y = x2 at (0, 0)
Q.52 Find the equation of the tangent line to the curve y = x2 − 2x + 7 which is
(a) parallel to the line 2x – y + 9 = 0
(b) perpendicular to the line 5y −15x = 13.
Q.53 Show that the tangents to the curve y = 7x3 + 11 at the points where x = 2 and x = −2 are parallel.
Q.54 Find the points on the curve y = x3 at which the slope of the tangent is equal to the y coordinate of the point.
Q.55 For the curve y = 4x3 − 2x5, find all the points at which the tangents
passes through the origin.
Q.56 Find the points on the curve x2 + y2 − 2x − 3 = 0 at which the tangents are
parallel to the x-axis.
Q.57 Find the equation of the normal at the point (am2, am3) for the curve ay2 = x3
Q.58 Find the equation of the normals to the curve y = x3 + 2x + 6 which are parallel to the line
x + 14y + 4 = 0.
Q.59 Find the equations of the tangent and normal to the parabola y2= 4ax at the point (at2, 2at).
Q.60 Prove that the curves x = y2 and xy = k cut at right angles if 8k2 = 1.
Q.63 The slope of the normal to the curve y = 2x2 + 3 sin x at x = 0 is
(C) – 3
Q.64 The line y = x + 1 is a tangent to the curve y2 = 4x at the point
(A) (1, 2) (B) (2, 1) (C) (1, −2) (D) (−1, 2)
Q.67 Find the approximate value of f (5.001), where f (x) = x3 – 7x2 + 15.
Q.68 Find the approximate change in the volume V of a cube of side x metres caused by increasing side by 1%.
Q.69 Find the approximate change in the surface area of a cube of side x metres caused by decreasing the side by 1%.
Q.70 If the radius of a sphere is measured as 7 m with an error of 0.02m, then find the approximate error in calculating its volume.
Q.71 If the radius of a sphere is measured as 9 m with an error of 0.03 m, then find the approximate error in calculating ‘its surface area’
Q.72 If f (x) = 3x2 + 15x + 5, then the approximate value of f (3.02) is
A. 47.66 B. 57.66 C. 67.66 D. 77.66
Q.73 The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is
A. 0.06 x3 m3 B. 0.6 x3 m3
C. 0.09 x3 m3 D. 0.9 x3 m3
Q.74 Find the maximum and minimum values, if any, of the following functions given by
(i) f(x) = (2x − 1)2 + 3 (ii) f(x)= 9x2+12x+2
(iii) f(x) = −(x − 1)2 + 10 (iv) g(x)= x3 + 1