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NCERT Solutions for miscellaneous exercise chapter 6 class 12 Application of Derivatives are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Students can get the NCERT solutions for Class 12 Maths chapter 6 miscellaneous exercise on this page. The Class 12 Maths chapter 6 miscellaneous exercise solutions are designed to make the students understand the applications of concepts studies in the NCERT book. Class 12 Maths chapter 6 miscellaneous solutions covers all the topics discussed in the previous exercises. Miscellaneous exercise chapter 6 Class 12 presents questions to practice the complete chapter.
After completing all the topics of NCERT syllabus Class 12 Maths chapter 6 and solved and unsolved questions of all other exercises the miscellaneous exercise can be attempted. The miscellaneous exercise combines questions from the complete chapter and the level of questions compared to exercises will be a bit higher. Miscellaneous exercise class 12 chapter 5 are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise enumerated in NCERT Book together using the link provided below.
Question:1(a) Using differentials, find the approximate value of each of the following:
Answer:
Let and
Now, we know that is approximate equals to dy
So,
Now,
Hence, is approximately equal to 0.677
Question:1(b) Using differentials, find the approximate value of each of the following:
Answer:
Let and
Now, we know that is approximately equals to dy
So,
Now,
Hence, is approximately equals to 0.497
Question:2. Show that the function given by has maximum at x = e.
Answer:
Given function is
Hence, x =e is the critical point
Now,
Hence, x = e is the point of maxima
Answer:
It is given that the base of the triangle is b
and let the side of the triangle be x cm ,
We know that the area of the triangle(A) =
now,
Now at x = b
Hence, the area decreasing when the two equal sides are equal to the base is
Question:4 Find the equation of the normal to curve which passes through the point (1, 2).
Answer:
Given the equation of the curve
We know that the slope of the tangent at a point on the given curve is given by
We know that
At point (a,b)
Now, the equation of normal with point (a,b) and
It is given that it also passes through the point (1,2)
Therefore,
-(i)
It also satisfies equation -(ii)
By comparing equation (i) and (ii)
Now, equation of normal with point (2,1) and slope = -1
Hence, equation of normal is x + y - 3 = 0
Question:5 . Show that the normal at any point to the curve is at a constant distance from the origin.
Answer:
We know that the slope of tangent at any point is given by
Given equations are
We know that
equation of normal with given points and slope
Hence, the equation of normal is
Now perpendicular distance of normal from the origin (0,0) is
Hence, by this, we can say that
the normal at any point to the curve
is at a constant distance from the origin
Question:6(i) Find the intervals in which the function f given by is
increasing
Answer:
Given function is
But
So,
Now three ranges are there
In interval ,
Hence, the given function is increasing in the interval
in interval so function is decreasing in this inter
Question:6(ii) Find the intervals in which the function f given by f x is equal to
is
decreasing
Answer:
Given function is
But
So,
Now three ranges are there
In interval ,
Hence, given function is increasing in interval
in interval
Hence, given function is decreasing in interval
Question:7(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 is increasing in interval
In interval (-1,1) ,
Hence, given function is decreasing in interval (-1,1)
Question:7(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 is increasing in interval
In interval (-1,1) ,
Hence, given function is decreasing in interval (-1,1)
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, is the point of maxima
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 Rs = 4 X 70 = 280 Rs
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 Rs = 16 X 45 = 720 Rs
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, is the point of minima
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:12 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, is the point of minima
and
AC = =
Hence proved
Question:13 Find the points at which the function f given by has (i) local maxima (ii) local minima (iii) point of inflexion
Answer:
Given function is
Now, for value x close to and to the left of , ,and for value close to and to the right of
Thus, point x = is the point of maxima
Now, for value x close to 2 and to the Right of 2 , ,and for value close to 2 and to the left 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:14 Find the absolute maximum and minimum values of the function f given by
Answer: Given function is
Now,
Hence, the point is the point of maxima and the maximum value is
And
Hence, the point is the point of minima and the minimum value is
Answer:
The volume of a cone (V) =
The volume of the sphere with radius r =
By Pythagoras theorem in we ca say that
V =
Now,
Hence, the point is the point of maxima
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:17 Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is . Also, find the maximum volume.
Answer:
The volume of the cylinder (V) =
By Pythagoras theorem in
h = 2OA
Now,
Hence, the point is the point of maxima
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 are similar
Now,
Now,
Now,
at
Hence, is the point of maxima
Hence proved
Now, Volume (V) at and is
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
Question:20 The slope of the tangent to the curve at the point
(2,– 1) is
A ) 22/7
B ) 6/7
C ) 7/6
D ) -6 /7
Answer:
Given curves are
At point (2,-1)
Similarly,
The common value between two is t = 2
Hence, we find the slope of the tangent at t = 2
We know that the slope of the tangent at a given point is given by
Hence, (B) is the correct answer
Question:21 The line y is equal to is a tangent to the curve if the value of m is
(A) 1
(B) 2
(C) 3
(D)1/2
Answer:
Standard equation of the straight line
y = mx + c
Where m is lope and c is constant
By comparing it with equation , y = mx + 1
We find that m is the slope
Now,
we know that the slope of the tangent at a given point on the curve is given by
Given the equation of the curve is
Put this value of m in the given equation
Hence, value of m is 1
Hence, (A) is correct answer
Question:22 The normal at the point (1,1) on the curve is
(A) x + y = 0
(B) x – y = 0
(C) x + y +1 = 0
(D) x – y = 1
Answer:
Given the equation of the curve
We know that the slope of the tangent at a point on the given curve is given by
We know that
At point (1,1)
Now, the equation of normal with point (1,1) and slope = 1
Hence, the correct answer is (B)
Question:23 The normal to the curve passing (1,2) is
(A) x + y = 3
(B) x – y = 3
(C) x + y = 1
(D) x – y = 1
Answer:
Given the equation of the curve
We know that the slope of the tangent at a point on the given curve is given by
We know that
At point (a,b)
Now, the equation of normal with point (a,b) and
?
It is given that it also passes through the point (1,2)
Therefore,
-(i)
It also satisfies equation -(ii)
By comparing equation (i) and (ii)
Now, equation of normal with point (2,1) and slope = -1
Hence, correct answer is (A)
Question:24 The points on the curve , where the normal to the curve makes equal intercepts with the axes are
Answer:
Given the equation of the curve
We know that the slope of the tangent at a point on a given curve is given by
We know that
At point (a,b)
Now, the equation of normal with point (a,b) and
It is given that normal to the curve makes equal intercepts with the axes
Therefore,
point(a,b) also satisfy the given equation of the curve
Hence, The points on the curve , where the normal to the curve makes equal intercepts with the axes are
Hence, the correct answer is (A)
Practice questions related to all the 5 main topics covered in the Class 12 NCERT Mathematics chapter application of derivatives are covered in Class 12 Maths chapter 6 miscellaneous exercise solutions. All these solutions of miscellaneous exercise are detailed in this page and are solved by Mathematics experts. The NCERT solutions for Class 12 Maths chapter 6 miscellaneous exercise are given in detail and step by step manner.
Also Read| Application of Derivatives Class 12 Notes
Benefits of NCERT Solutions for Class 12 Maths Chapter 6 Miscellaneous Exercise
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.
Application Date:09 September,2024 - 14 November,2024
Application Date:09 September,2024 - 14 November,2024
Admit Card Date:04 October,2024 - 29 November,2024
Admit Card Date:04 October,2024 - 29 November,2024
Hello there! Thanks for reaching out to us at Careers360.
Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.
Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!
Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.
If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.
Let me know if you need any other tips for your math prep. Good luck with your studies!
It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.
Possible steps:
Re-evaluate Your Study Strategies:
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Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.
I hope this information helps you.
Hi,
Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
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Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
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Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.
Scholarship Details:
Type A: For candidates scoring 60% or above in the exam.
Type B: For candidates scoring between 50% and 60%.
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Cash Scholarship:
Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).
Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.
Hope you find this useful!
hello mahima,
If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.
hope this helps.
Hello Akash,
If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.
You can get the Previous Year Questions (PYQs) on the official website of the respective board.
I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.
Thank you and wishing you all the best for your bright future.
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