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Derivatives, a key concept in calculus, have practical applications in various fields like physics, business, and everyday life. They are used to analyse rates of change, predict future trends, and optimise processes. The "Application of Derivatives" chapter of the NCERT includes all these topics. After studying this chapter, you came across the concept of instantaneous change, increasing and decreasing functions, local maxima and minima, and absolute maxima and minima. In this miscellaneous exercise of Chapter 6, Class 12 Application of Derivatives, you will find various questions from each topic, which will help you understand the concept with clarity. These NCERT Solutions are prepared by the subject matter experts at Careers360, which will help you to learn the concept and approach better.
Question:1(a) Using differentials, find the approximate value of each of the following:
Answer:
Let
Now, we know that
So,
Now,
Hence,
Question:1 Using differentials, find the approximate value of each of the following:
Answer:
Let
Now, we know that
So,
Now,
Hence,
Question:2. Show that the function given by
Answer:
Given function is
Hence, x =e is the critical point
Now,
Hence, x = e is the point of maxima
Question:3(i) Find the intervals in which the function f given by
increasing
Answer:
Given function is
But
So,
Now three ranges are there
In interval
Hence, the given function
in interval
Question:3 (ii) Find the intervals in which the function f given by f x is equal to
decreasing
Answer:
Given function is
But
So,
Now three ranges are there
In interval
Hence, given function
in interval
Hence, given function
Question:4(i) Find the intervals in which the function f given by
Increasing
Answer:
Given function is
Hence, three intervals are their
In interval
Hence, given function
In interval (-1,1) ,
Hence, given function
Question:4(ii) Find the intervals in which the function f given by
decreasing
Answer:
Given function is
Hence, three intervals are their
In interval
Hence, given function
In interval (-1,1) ,
Hence, given function
Answer:
Given the equation of the ellipse
Now, we know that ellipse is symmetrical about x and y-axis. Therefore, let's assume coordinates of A = (-n,m) then,
Now,
Put(-n,m) in equation of ellipse
we will get
Therefore, Now
Coordinates of A =
Coordinates of B =
Now,
Length AB(base) =
And height of triangle ABC = (a+n)
Now,
Area of triangle =
Now,
Now,
but n cannot be zero
therefore,
Now, at
Therefore,
Now,
Now,
Therefore, Area (A)
Answer:
Let l , b and h are length , breath and height of tank
Then, volume of tank = l X b X h = 8
h = 2m (given)
lb = 4 =
Now,
area of base of tank = l X b = 4
area of 4 side walls of tank = hl + hl + hb + hb = 2h(l + b)
Total area of tank (A) = 4 + 2h(l + b)
Now,
Hence, b = 2 is the point of minima
So, l = 2 , b = 2 and h = 2 m
Area of base = l X B = 2 X 2 =
building of tank costs Rs 70 per sq metres for the base
Therefore, for
Area of 4 side walls = 2h(l + b)
= 2 X 2(2 + 2) =
building of tank costs Rs 45 per square metre for sides
Therefore, for
Therefore, total cost for making the tank is = 720 + 280 = 1000 Rs
Answer:
It is given that
the sum of the perimeter of a circle and square is k =
Let the sum of the area of a circle and square(A) =
Now,
Hence,
Hence proved that the sum of their areas is least when the side of the square is double the radius of the circle
Answer:
Let l and bare the length and breadth of rectangle respectively and r will be the radius of circle
The total perimeter of window = perimeter of rectangle + perimeter of the semicircle
=
Area of window id given by (A) =
Now,
Hence, b = 5/2 is the point of maxima
Hence, these are the dimensions of the window to admit maximum light through the whole opening
Question:9 A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is
Answer:
It is given that
A point on the hypotenuse of a triangle is at a distance a and b from the sides of the triangle
Let the angle between AC and BC is
So, the angle between AD and ED is also
Now,
CD =
And
AD =
AC = H = AD + CD
=
Now,
When
Hence,
AC =
Hence proved
Question:10 Find the points at which the function f given by
Answer:
Given function is
Now, for value x close to
Thus, point x =
Now, for value x close to 2 and to the Right of 2 ,
Thus, point x = 2 is the point of minima
There is no change in the sign when the value of x is -1
Thus x = -1 is the point of inflexion
Question:11 Find the absolute maximum and minimum values of the function f given by
Answer: Given function is
Now,
Hence, the point
And
Hence, the point
Answer:
The volume of a cone (V) =
The volume of the sphere with radius r =
By Pythagoras theorem in
V =
Now,
Hence, the point
Hence, the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is
Answer:
Let's do this question by taking an example
suppose
Now, also
Hence by this, we can say that f is an increasing function on (a, b)
Question:14 Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is
Answer:
The volume of the cylinder (V) =
By Pythagoras theorem in
h = 2OA
Now,
Hence, the point
Hence, the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is
and maximum volume is
Answer:
Let's take radius and height of cylinder = r and h ' respectively
Let's take radius and height of cone = R and h respectively
Volume of cylinder =
Volume of cone =
Now, we have
Now, since
Now,
Now,
Now,
at
Hence,
Hence proved
Now, Volume (V) at
hence proved
(A) 1 m/h
(B) 0.1 m/h
(C) 1.1 m/h
(D) 0.5 m/h
Answer:
It is given that
Volume of cylinder (V) =
Hence, (A) is correct answer
Also Read,
(a) increasing on an interval
(b) decreasing on
(c) constant in
(i) If
(ii) If
(iii) If
As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters
(i)
(ii)
(iii) The test fails if
Also, read,
These are links to other subjects' NCERT textbook solutions. Students can check and analyse these well-structured solutions for a deeper understanding.
Students can check these NCERT exemplar links for further practice purposes.
Questions related to the tropics rate of change of quantities, approximation, increasing and decreasing functions, tangents and normals and maxima and minima are covered in Class 12 Maths chapter 6 miscellaneous solutions.
24 questions are present in Class 12 Maths chapter 6 miscellaneous exercise solutions
10 miscellaneous questions are solved in Class 12 NCERT Mathematics book.
Including miscellaneous, there are 6 exercises. For more questions students can use NCERT exemplar.
51 solved examples are given in the chapter 6 application of derivatives
The questions regarding maximum and minimum of a function using derivatives are covered in the class 12 maths exercise 6.5
Exercise 6.4 and a few questions of Class 12 Maths chapter 6 miscellaneous solutions covers the concept of approximation.
The two tests discussed are the first derivative test and the second derivative test.
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