Exercise 6.5 class 12

dy/dx = 4x 3 – 18x 2 + 26x – 10 dy/dx = 4(1) 3 – 18(1) 2 + 26(1) – 10 dy/dx = 4 – 18 + 26 – 10 = 2 dy/dx = 2 -dx/dy = -1/2 Using point slope form, equation of tangent is y – y 1 = dy/dx(x – x1) y – 3 = 2(x – 1) y – 2x = 1 is the equation of tangent. Using point slope form, equation of normal is y –y 1 = -dx/dy(x – 1) y – 3 = -1/2(x – 1) 2y – 6 = -x + 1 2y + x = 7 is the equation of normal. (iii) Given curve : y = x 3 Given point is (1, 1) dy/dx = 3x 2 dy/dx = 3(1) 2 = 3 dy/dx = 3 & -dx/dy = -1/3 Using point slope form, equation of tangent is y – y 1 = dy/dx(x – x 1) dy/dx = -1 & -dx/dy = 1 Now for t = Ï€/4, y 1 = sin t = sin( Ï€/4) = 1/√2 x 1 = cos t = cos( Ï€/4) = 1/√2 The point is(1/√2, 1/√2) y – y 1 = y – (1/√2) = -1(x – 1/√2) y – 1/√2 = -x + 1/√2 x + y = √2 is the equation of normal is y – 1/√2 = 1(x – 1/√2) x = y is the equation of normal. Question 15. Find the equation of the tangent line to the curve y = x 2 – 2x + 7 which is (i) Parallel to line 2x – y + 9 = 0 (ii) Perpendicular to the line 5y – 15x = 13 Solution: Given curve: y = x 2 – 2x + 7 On differentiating w.r.t. x, we get dy/dx = 2x – 2 -(1) (i) Tangent is parallel to 2x – y + 9 = 0 that means, Slope of tangent = slope of 2x – y + 9 = 0 y = 2x + 9 Slope = 2 -(Comparing with y = mx + e) dy/dx = slope = 2 2x – 2 = 2 x 1 = 2 Corresponding to x 1 = 2, y 1 = x 1 2– 2x 1 + 7 y 1 = (2) 2 – 2(2) + 7 y 1 = 7 The point of contact is (2, 7). Using point slope form, equation of tangent is y – y 1 =...

NCERT Solutions for Class 10 Maths Chapter 6 - Triangles Exercise 6.5 Chapter 6, Triangles, Exercise 6.5 begins on page number 150 of the NCERT textbook. The exercise covers the portions in which theorems related to right angles are present. In other words, the exercise-wise The subject experts at BYJU’S make sure that the questions present in the NCERT textbooks are solved in the easiest way possible. The Download PDF carouselExampleControls112 Previous Next Access other exercise solutions of Class 10 Maths Chapter 6 – Triangles Access Answers to Maths NCERT Class 10 Chapter 6 – Triangles Exercise 6.5 1.  Sides of 4 triangles are given below. Determine which of them are right triangles. In the case of a right triangle, write the length of its hypotenuse. (i) 7 cm, 24 cm, 25 cm (ii) 3 cm, 8 cm, 6 cm (iii) 50 cm, 80 cm, 100 cm (iv) 13 cm, 12 cm, 5 cm Solution: (i) Given, the sides of the triangle are 7 cm, 24 cm, and 25 cm. Squaring the lengths of the sides of the, we will get 49, 576, and 625. 49 + 576 = 625 (7) 2 + (24) 2 = (25) 2 Therefore, the above equation satisfies Pythagoras theorem. Hence, it is a right-angled triangle. Length of Hypotenuse = 25 cm (ii) Given, the sides of the triangle are 3 cm, 8 cm, and 6 cm. Squaring the lengths of these sides, we will get 9, 64, and 36. Clearly, 9 + 36 ≠64 Or, 3 2 + 6 2 ≠8 2 Therefore, the sum of the squares of the lengths of two sides is not equal to the square of the length of the hypotenuse. Hence, the given t...

The Application of Derivatives is covered in NCERT Maths Class 12 Exercise 6.5 Solutions. According to the CBSE, it is one of the most significant chapters. The following article includes a detailed examination of exercise 6.5 class 12 maths and a few other important points. The post also has examples of the types of problems found in the Class 12 Maths Chapter 6 Exercise 6.5 What Is The Significance Of Ex 6.5 Class 12 NCERT Solutions? Derivatives are used in a variety of ways. Exercise 6.5 is a crucial chapter regarding the CBSE board exam’s weighting of marks. The Application of Derivatives exercises 6.5 is used to apply all of the Differential Calculus. Infinity Learn’s The following are the implications of class 12 Maths NCERT solutions chapter 6 exercise 6.5: • Regularly practicing exercise 6.5 class 12 Maths NCERT solutions will undoubtedly improve your preparedness. • According to CBSE norms, the questions in exercise 6.5 Maths class 12 are completely prepared. • By practicing class 12 Maths ex 6.5 solutions, you can recognize your flaws. • Calculus is the most difficult section of class 12 maths, and you received a perfect grade in class 12 maths ch 6 ex 6.5. • NCERT solutions class 12 mathematics chapter 6 exercise 6.5 are the greatest materials to improve your level of preparation after completing the typical textbook questions. • You won’t be able to solve all of the issues on your own. However, it would help if you attempted to solve them on your own first. The...

Download the CBSE NCERT Solutions for Class 12 Maths Chapter-6 PDF from the official website of Vedantu. These solutions are prepared by subject matter experts who give lucid explanations for each of the topics discussed in the Application of Derivatives Class 12. The solution providers have a lot of experience in the education domain and they understand how to design a solution that caters to the understanding level of students of a particular class. In AOD Class 12 Solutions, even complex solutions are made easy by these experienced teachers by breaking them up into smaller chunks and giving you tips to remember major formulas and functions. You will also receive timely help from our experts and teachers if you face any doubts while going through the Application of Derivatives Class 12 NCERT Solutions. Topics Covered in NCERT Class 12 Maths Chapter 6 - Application of Derivatives The topics covered in Class 12 Maths Chapter 6 Application of Derivatives are given below. Exercise 6.1: This exercise deals with finding the rate of change of a quantity using derivatives. It covers problems related to finding the instantaneous rate of change, average rate of change, and marginal cost and revenue. Exercise 6.2: In this exercise, you will learn about increasing and decreasing functions, and how to find the intervals of increase or decrease of a function. Exercise 6.3: This exercise is about finding the maximum and minimum values of a function using derivatives. It covers problems...

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 are part of • • • • Board CBSE Textbook NCERT Class Class 12 Subject Maths Chapter Chapter 6 Chapter Name Application of Derivatives Exercise Ex 6.5 Number of Questions Solved 29 Category NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Ex 6.5 Class 12 Maths Question 1. 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) = x 3 + 1 Solution: (i) Minimum value of (2x – 1)² is zero. Minimum value of (2x – 1)² + 3 is 3 Clearly it does not have maximum value, (ii) f(x) = 9x² + 12x + 2 ⇒ f(x) = (3x + 2)² – 2 Minimum value of (3 + 2)² is zero. ∴ Min.value of (3x + 2)² – 2 = 9x² + 12x + 2 is – 2 f (x) does not have finite maximum value (iii) f(x) = – (x – 1)² + 10 Maximum value of – (x – 1)² is zero Maximum valuer f f(x) = – (x – 1)² + 10 is 10 f (x) does not have finite minimum value. (iv) As x—»∞,g(x)—»∞;Also x—»-∞,g(x)—»-∞ Thus there is no maximum or minimum value of f(x) Ex 6.5 Class 12 Maths Question 2. Find the maximum and minimum values, if any, of the following functions given by (i) f(x) = |x + 2| – 1 (ii) g(x) = -|x + 1| + 3 (iii) h (x) = sin 2x + 5 (iv) f(x) = |sin(4x + 3)| (v) h(x) = x + 1,x∈(-1,1) Solution: (i) We have :f(x) = |x + 2 |-1 ∀x∈R Now |x + 2|≥0∀x∈R |x + 2| – 1 ≥ – 1 ∀x∈R , So -1 is the min. value of f(x) now f(x) = -1 ⇒ |x + 2|-1 ⇒ ...

• English Grammar • Samacheer Kalvi 12th Books Solutions Menu Toggle • Tamil Nadu 12th Model Question Papers • Samacheer Kalvi 11th Books Solutions Menu Toggle • Tamil Nadu 11th Model Question Papers • Samacheer Kalvi 10th Books Solutions Menu Toggle • Samacheer Kalvi 10th Model Question Papers • Samacheer Kalvi 9th Books Solutions • Samacheer Kalvi 8th Books Solutions Menu Toggle • Samacheer Kalvi 7th Books Solutions • Samacheer Kalvi 6th Books Solutions • Samacheer Kalvi 5th Books Solutions • Samacheer Kalvi 4th Books Solutions You can Download Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 6 Applications of Vector Algebra Ex 6.5 Question 1. Find the parametric form of vector equation and Cartesian equations of a straight line passing through (5, 2, 8) and is perpendicular to the straight lines Solution: ∴ This vector is perpendicular to both the given straight lines. ∴ The required straight line is \(\vec\) Question 5. Find the shortest distance between the parallel lines. Solution:

• The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. Find the rate at which its area increases, when side is 10 cm long. [CBSE Sample Paper 2017] • The volume of a sphere is increasing at the rate of 3 cubic centimeter per second. Find the rate of increase of its surface area, when the radius is 2 cm. [Delhi 2017] • The side of an equilateral triangle is increasing at the rate of 2 cm/s. At what rate is its area increasing when the side of the triangle is 20 cm? [Delhi 2015] • Determine for what values of x, the function f(x) = x³ + 1/x³, where x ≠0, is strictly increasing or strictly decreasing. [CBSE Sample Paper 2017] • Show that the function f(x) = 4x³ – 18x² + 27x – 7 is always increasing on R. [Delhi 2017] • Find the interval in which f(x) = sin 3x – cos 3x, 0 < x < Ï€, is strictly increasing or strictly decreasing. [Delhi 2016] • Find the point on the curve y = x³ – 11x + 5 at which the tangent is y = x – 11. [CBSE Sample Paper 2017] 1. Find the equation of tangents to the curve y = cos(x + y), where x lies in [- 2Ï€, 2Ï€], that are parallel to the line x + 2y = 0. [Foreign 2016] 2. Find the shortest distance between the line x – y + 1 = 0 and the curve y² = x. [CBSE Sample Paper 2017] 3. If the sum of lengths of the hypotenuse and a side of a right angled triangle is given, show that the area of the triangle is maximum, when the angle between them is Ï€/3. [Delhi 2017] 4. Show that the altitude of the right circular con...

Transcript Ex 6.3, 5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (ii) f (𝑥) = sin⁡𝑥 + cos⁡𝑥 , 𝑥 ∈ [0, 𝜋 ] Finding f’(𝒙) f’(𝑥)=𝑑(𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥)/𝑑𝑥 f’(𝑥)=cos⁡〖𝑥 −sin⁡𝑥 〗 Putting f’(𝒙) cos⁡〖𝑥 −sin⁡𝑥 〗= 0 cos⁡〖𝑥=sin⁡𝑥 〗 1 = sin⁡𝑥/(cos⁡ 𝑥) 1 = tan 𝑥 tan 𝑥 = 1 We know that know tan θ = 1 at θ = 𝜋/4 ∴ 𝑥 = 𝜋/4 Since given interval 𝑥 ∈ [0 , 𝜋] Hence calculating f(𝑥) at 𝑥=0 , 𝜋/4 ,𝜋 Absolute Maximum value of f(𝑥) is √𝟐 at 𝒙 = 𝝅/𝟒 & Absolute Minimum value of f(𝑥) is –1 at 𝒙 = π Show More