NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives (Ex 6.3) Exercise 6.3

Students should use the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 to effortlessly go on with their preparation for Class 12 examinations. PDF of the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 has all the necessary data required to understand the chapter on the Application of Derivatives.

The National Council of Educational Research and Training (NCERT) is a self-governing body established by the Indian government in 1961 to assist and provide advice to the Central and State Governments on policies and programmes aimed at improving the quality of schooling. This is why they have developed and published the NCERT Textbooks for all students from Classes 1-12. Hence, NCERT books are the standard and go-to books for all students. Students should download the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 made available by Extramarks to ensure that their NCERT textbook revision does not lag.

Every student should have a dependable resource that is well-founded and at par with the NCERT Syllabus. The NCERT Solutions are very vital for students to include in their studying and learning process. NCERT Solutions prove to be a great guide and revision material for all students from Class 1-12. Extramarks provides premium study materials, prepared by subject-matter experts. The NCERT Solutions provided by Extramarks have simple and accurate solutions for every question in the NCERT books systematically. Students can check the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 provided on the Extramarks website to check the pattern and style of questions that can appear in the final exam. The answer writing style and pattern are well-adapted for the understanding level of each class of students.

Class Wise-List for NCERT Solutions:

Given below are the details of the NCERT Solutions provided for each class and their importance in the academic career of the students. These solutions would aid the preparation of students for them to achieve excellent results.

NCERT Solutions Class 12

Students might find challenges while practising for the pre-board and board examinations, but students also have to deal with the forthcoming competitive examinations and college entrances. For all these important examinations, NCERT is the standard reading material. This being the case, students should never skip out on solving NCERT Class 12 books. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 have a proper step-by-step explanation for every sum to help students through the process. Class 12 students need to prepare well for their examinations and choose resources that would benefit them in their preparation process and also in their final examination. The NCERT Solutions Class 12 provided by Extramarks contain all the necessary topics for all these examinations. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 are available to download for the students preparing for their Maths exam. Students can use the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 as a sample to know the pattern and technique to be followed by the students. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 will showcase the expertly curated Extramarks’ learning resources that the students can utilise for their preparation. Extramarks makes accessible NCERT Solutions Class 12 for all the subjects of Class 12, categorized chapter-wise for ease of access. The NCERT Solutions Class 12 has sufficient resource material for students to solve and practice all their doubts from their NCERT books. Students can download the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 for detailed study material that gives an in-depth knowledge of the Application of Derivatives. The download link for NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 is provided below.

NCERT Solutions Class 11

After the Class 10 board examination and just before the Class 12 board examination, students may wish to take a break before beginning the preparation process. During this timeline, Class 11 also becomes a very crucial stage. This is where students can start preparing and setting up a strong foundation for the forthcoming board examinations. It is in Class 11 that students have to choose their academic stream, which further helps in deciding their future professions. Arts or Humanities, Science and Commerce, all these streams have their own set of subjects. Most students select these streams based on their interests or past years’ scores in certain subjects.

Extramarks has NCERT Solutions Class 11 for all these subjects that any student of Class 11 would require for their examination. The syllabus for Class 11 & 12 are interlinked with each otherThis is the reason to give that extra attention to this exam. Concentration and sincerity are the keys to the preparation for any examination. The NCERT Solutions provided by Extramarks are simple and easy to understand, which does not confuse  students . It keeps the student engaged in the properly categorised document. This helps students enjoy their studies and also gain the best results in their examinations. Students can use these NCERT Solutions as they exist or use them to make personalised notes for themselves. All important points and key factors are mentioned in these NCERT Solutions.

NCERT Solutions Class 10

Class 10 Board Examination is a prominent point in the career path of every Class 10 student. Some students might even get fearful if not supported and guided well. A well-formed and well-researched study material and a fixed routine can go a long way. Subject-matter experts at Extramarks have done the hard task of curating reliable study material. This puts the load off of students by filtering what is essential and what is not. The NCERT Solutions are everything they would need to ace their upcoming examination. Using the NCERT Solutions and other resource material from Extramarks students can make this stressful preparation easier for them. Past years’ papers should be solved by the students to keep a constant check on their preparation level and strive for a better performance every time. Sample papers are a great way to understand the level of questions set by the board. This also helps to understand the idea of time allocation required for the examination.

NCERT Solutions Class 9

Class 9 is the foundation required for a well-formed and stable base for the forthcoming board examination. The syllabus of Class 9 is quite vast and detailed. It deals with all the topics and concepts that students would need for their further examination. Class 9 examinations are a test to understand the preparation level of students. All the chapters of NCERT are made to provide for the overall development of students. The NCERT Solutions Class 9 is made by proper research by a faculty of expert teachers who understand pedagogy and technology. The faculty understands the thought processes of students and their study patterns. The NCERT Solutions for Class 9 deals with all the chapters of all the subjects of Class 9. These materials are based on CBSE guidelines to provide concept-based learning at their convenience. It is difficult for students to get personalized attention at school, this is why with Extramarks, students get to do that on their own. Students can use these NCERT Solutions to learn by themselves or use them with other study materials for better preparation. Extramarks’ step-by-step solution to the NCERT textbooks can be easily downloaded from their website.

NCERT Solutions Class 8

The NCERT Solutions Class 8 provide solutions for all questions in the exercise of the textbook. The answers are explained in easy-to-comprehend language and proper format. By using Extramarks’ K12 study materials, students can study with the help of thorough and convenient study materials to help them succeed in their examinations. The in-depth performance reports provided by Extramarks help students in self-assessment and also help them track their academic progress. Live doubt-solving classes provide students access to live sessions with the expert educationalists of Extramarks.

The PDF made available by Extramarks for the NCERT Solutions Class 8 helps in building a strong conceptual base which will be helpful in students’ examinations. The solutions strictly adhere to the guidelines provided by NCERT. This helps students score well by helping them revise the key points. The NCERT Solutions will prove to be a morale booster for students using the resource diligently.

NCERT Solutions Class 7

It is important for students in Class 7 to decide on a preparation schedule for their examination. Accessing good learning resources is a necessity for students to acquaint themselves with the final exam pattern. The NCERT Solutions Class 7 have all the answers categorically with simple explanations. Extramarks’ NCERT Solutions become an aid for students who want to revise their subject or do  last-minute preparation. The NCERT Solutions are available in PDF format. Using NCERT Solutions made available through Extramarks provides insight into the level of preparation of students. This helps them improve upon their problem areas and enhance their preparation.

NCERT Solutions Class 6

The importance of primary education is such that many countries have made it compulsory and a basic right. The holistic development of a kid starts at an early age and this is where proper guidance and support for education are required. If a student has not been guided well in Class 6, their secondary education might be affected. The NCERT Solutions provided by Extramarks aid in the educational understanding of the topics in the NCERT textbooks. Understanding the fundamental concepts well at the primary level helps children grow into adults with confidence and sensibility.

NCERT Solutions Class 5

Using NCERT Solutions Class 5 is a great way to improve students’ learning methodologies. The NCERT Solutions are an interactive and efficient learning resource for students to quickly learn their syllabus. The NCERT Solutions Class 5 provided by Extramarks can be utilised by students by helping them finish their revisions quickly and easily. By practising with the help of these study materials, students can easily analyse their preparation levels and self-assess the areas they need to work on. This gives students a positive attitude towards studies and life in general.

NCERT Solutions Class 4

The learning methods and pace of each child is different. Class 4 is an important academic grade for students to excel in. When students get the learning support they need, it helps boost their confidence and morale. NCERT Solutions Class 4 helps students decide their own pace of learning. When students have a readily available bank of solutions while they are practising, it allows them to become familiar with the variety of questions that can appear in the final exam. The NCERT Solutions Class 4 complement the students’ revision schedules and their ways of understanding. The simple solutions and interactive material can prove helpful in boosting their understanding of particular subjects.

NCERT Solutions Class 3

The NCERT Solutions Class 3 made available by Extramarks are readily accessible online in downloadable PDF format for the support of Class 3 students. A team of in-house experts in the industry assembles the curriculum-based study material on the platform. They also provide students with the various types of questions that can appear in the examination, since the same kind of questions can repeatedly appear in the board examinations. Also, there is a certain way in which the answers should be written in the examination to achieve a good score in the subject. Reading engaging and informative content is a great way for young children to prepare for their exams.

NCERT Solutions Class 2

Parents and teachers need to provide students with the right tool for learning to facilitate their development. When it comes to efficient learning resources, Extramarks’ NCERT Solutions Class 2 allows students to prepare themselves before the examinations. By getting them familiar with the blueprint of the structure of answers for the NCERT questions, Extramarks aids students by making available a wide variety of tools for preparation.

NCERT Solutions Class 1

The NCERT Solutions Class 1 by Extramarks is an engaging and fun solution for little children. Instead of using NCERT Solutions as the sole preparation material, students should try to read the chapter from the NCERT textbook first. Then using the NCERT Solutions, students can develop a better understanding of the chapter and solve exercises with ease. Solutions for all the subjects like Hindi, English, Mathematics, Science and Social Science are available for students in proper categorisation to help organise their revision. Using NCERT Solutions would help students save time and focus better on every subject equally. This helps students avoid the worry and pressure of multiple subjects.

Advantages of NCERT Solutions for Students

There are numerous advantages of practising using NCERT Solutions. The solutions provided by Extramarks are categorized by exercises of every chapter, like NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3, so that students do not get confused. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 and other exercise solutions make learning easy for students of Class 12 who need to prepare well in order to develop a good foundational base for their higher education. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 and other such resources are created by special faculty members at Extramarks with utmost care and concentration, allowing students to learn from the best faculty and expert minds. Students must use the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 and other exercise solutions for their revision and doubt clearance. NCERT textbooks and NCERT Solutions are a significant resource for CBSE Board Examinations. It helps students practice for the real examination.

The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 are available in downloadable PDF form. Students can use NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 to understand the concepts and questions of Chapter 6.

NCERT Class 12 Maths Chapter 6 Exercise 6.3 – Free PDF Download

CBSE Class 12 Board Examination is held every year in which over 16 lakh students appear. It could be a stressful and daunting time for students. Students of different streams, that of Arts or Humanities, Science and Commerce appear for the examination. Every student has their strengths and their weaknesses. Mathematics has been a troublesome subject for a lot of students. Extramarks has a lot of learning resources such as the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 to help students with all the reading and learning resources. Mathematics need not be difficult if dealt with tactfully, by students. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 provided by Extramarks are available to be downloaded from the website, to be utilised at the convenience of the students.

Similar to the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3, Extramarks has made available the solutions for all the chapters of all the subjects of CBSE Class 12.

Students can learn the Applications of Derivatives by using all the solutions of the exercises of the chapter, including the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 to understand the concepts properly. The chapter could be confusing for some students because it can be interpreted to be quite different from the topics students have studied before. Hence, the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 are an easy way to deal with this problem. All the solutions to the questions of Chapter 6 Exercise 6.3 are available in the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3. A total of 27 questions are solved in the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3.

Access NCERT Solutions for Class 12 Maths Chapter 6 – Application of Derivatives

Download the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 curated by Extramarks from below.

Chapter 6 Exercise 6.3 Class 12 Mathematics Solutions

Mathematics is a very significant subject in Class 12. It can also be very time-consuming. Some chapters might take more time and effort than others to solve. Every chapter holds significant scoring value and important concepts that should be revised properly to appear for the examination. To make sure time management is a skill that they can master, students can thoroughly practice the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3. NCERT syllabus is a standard for the education system in India and therefore, every competitive exam uses this as a benchmark for its questions. Being thorough with the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 gives a competitive edge to the students aspiring to appear for any of these examinations. They can keep learning from the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 for preparation for further exams. The Exercise in Chapter 6 Exercise 6.3 has a total of 27 questions. These are based on the concepts of Tangents and Normal. Most questions are asking students to calculate the value of the slope based on the tangents. Here, with the format of the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3, students will have step-by-step methods to solve each sum. The interactive examples incorporated in the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 are sufficient to gain a clear understanding of this topic.

The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 covers two parts which are Normals and Tangents, along with the exercises.

Application of Derivatives Exercise 6.3 is based on the following topics:

Tangents and Normals

In Chapter 6 the explanation, formulas and examples are given on the topic of Tangent Lines and Normal Lines. The chapter deals with theoretical and formula-based knowledge which is very crucial for students to understand, and it includes all the terms and figures to solve complex problems. They must understand all the theoretical concepts before solving Ex 6.3 Class 12. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 have all the important concepts mentioned along with the sums. In this branch of Mathematics, the concept of a straight line and slope is required. So, the theory of the slope, the equation, and the slope of the tangent to the curve are explained very well in the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3.

Tangents are defined as a line that touches the curve at a point and does not cross.

Normals are defined as perpendicular lines that contact the tangent at the point.

While solving Ex 6.3 Class 12, students come across many common problems like forgetting the formula, not recalling the core concept, and so on. To avoid these kinds of concerns, they should refer to the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 have simplified the exercise for the ease of students. Not only the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 but also the other exercise solutions made accessible through Extramarks, all make learning easier for the students by providing curated solutions that are detailed and clear.

Finding Tangents and Normals 

The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 have solutions for Exercise 6.3 from Chapter 6 dealing with Tangents and Normals. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 explain the details of the exercise and the chapter. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 also provide step-by-step solutions for all the questions in the exercise.

A line that touches the curve at a point and does not cross it is known as a tangent. When a perpendicular line contacts the tangent at the point, then it is called normal. To solve Class 12th Ex 6.3, one can refer to the NCERT Solutions For Class 12 Maths Chapter 6 Exercise 6.3. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 will make the task of preparation easier and save time as it has simplified solutions for the ease of the students.

Benefits of Chapter 6 Exercise 6.3 Class 12 Mathematics Solved Solutions

The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 are curated in a manner that students can use these even at the last moment of the preparation. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 are curated by the experts at Extramarks, which means they are a reliable source and provide a better learning and understanding experience. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 gives the step-by-step method to solve each problem thereby making it simple and quickly recognisable. Practising Class 12th Ex 6.3 along with the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 allows students to understand the topics of Tangents and Normals very well without any hesitation. By referring repeatedly to the NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3, they can retain theories and formulas for a longer period. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 are quite suitable for promoting Mathematics proficiency in students and building a deeper knowledge base.

Access Other Exercise Solutions of Class 12 Maths Chapter 6 – Application of Derivatives 

The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 are solved according to the updated CBSE Board syllabus and guidelines so that students do not have to verify and access different websites to get help in downloading Class 12 Math Exercise 6.3 Solution. The NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 are prepared by the Extramarks experts by referring to the relevant sources to make exercises easy to understand and interesting. It can be used to solve Ex 6.3 Class 12 during the revision schedule by the students.

Students can download NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.3 by clicking on this link.

Q.1 Find the slope of the tangent to the curve y = 3x4 − 4x at x = 4


The given curve is y=3x4 4x.Differentiating w.r.t. x, we get        dydx=ddx(3x4 4x)    =12x34Then, the slope of the tangent to the given curve at x=4 is given by,(dydx)x=4=(12x34)x=4    =12(4)34    =7684    =764Thus,​ the slope of tangent to the given curve is 764.

Q.2 Find the slope of the tangent to the curve y = x 1 x 2 , x 2 a t x = 10 .


The given curve is y=x1x2Differentiating w.r.t. x, we getdydx=ddx(x1x2)      =(x2)ddx(x1)(x1)ddx(x2)(x2)2      =(x2)(10)(x1)(10)(x2)2      =x2x+1(x2)2      =1(x2)2Thus, the slope of the tangent at x = 10 is given by,(dydx)x=10=(1(x2)2)x=10    =1(102)2    =164Thus,​ the slope of tangent to the given curve is 164.

Q.3 Find the slope of the tangent to curve y = x3−x+1 at the point whose x-coordinate is 2.


The given curve isy=x3x+1Differentiating w.r.t.x, we getdydx=ddx(x3x+1)= 3x21The slope of tangent to the curve is(dydx)x=2=3(2)21=121=11Thus, the slope of the tangent at the point where thexcoordinate is 2 is11.

Q.4 Find the slope of the tangent to curve y = x3−3x+2 at the point whose x-coordinate is 3.


The given curve is      y=x33x+2Differentiating w.r.t. x, we get  dydx=ddx(x33x+2)=3x23The slope of tangent to the curve is(dydx)x=3=3(3)23    =273=24Thus, the slope of the tangent at the point where the xcoordinate is 3 is 24.


Find the slope of the normal to the curve x =acos3θ, y=asin3θat θ=π4.


Given that x = acos3θ and y = asin3θDifferentiating w.r.t. θ, we getdx=adcos3θ        =3acos2θdcosθ=3acos2θsinθdy=adsin3θ        =3asin2θdsinθ        =3asin2θcosθdydx=(dy)(dx)        =3asin2θcosθ3acos2θsinθ=tanθTherefore, the slope of the tangent at  θ=π4,is(dydx)(θ=π4)=(tanθ)(θ=π4)        =tanπ4        =1Hence, the slope of the normal atθ=π4,isslope of normal(M)=1slope of tangent(m)=11=1

Q.6 Find the slope of the normal to the curve x = 1 a s i n θ , y = b c o s 2 θ a t θ = π 2


Given that:x=1asinθ and y=bcos2θDifferentiating w.r.t. θ, we get    dx=d(1asinθ)=0acosθ=acosθ  dy=d(bcos2θ)=2bcosθdcosθ=2bcosθsinθdydx=(dy)(dx)=2bcosθsinθacosθ=2basinθSlope of tangent at θ=π2,is given by    m=(dydx)θ=π2=2basinπ2=2baslope of normal    (M)=1m=1(2ba)=a2bThus, slope of normal is a2b.

Q.7 Find points at which the tangent to the curve
y = x3 − 3x2 − 9x + 7 is parallel to the x axis.


The equation of the given curve is y= x 3 3x 2 9x + 7 Differentiating w.r.t. x, we get dy dx = d dx ( x 3 3x 2 9x + 7 ) =3 x 2 6x9 Now, the tangent is parallel to the x-axis if the slope of the tangent is zero. 3 x 2 6x9=03( x 2 2x3 )=0 3( x3 )( x+1 )=0 x=3,1 When x=3, then y= ( 3 ) 3 3 ( 3 ) 2 9( 3 ) + 7 =272727+7 =20 When x=1, then y= ( 1 ) 3 3 ( 1 ) 2 9( 1 ) + 7 =13+9+7 =12 Hence, the points at which the tangent is parallel to the x-axis are ( 3,20 ) and ( 1, 12 ). MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@2D49@

Q.8 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).


Given curve is y= ( x2 ) 2 Differentiating both sides, w.r.t. x, we get dy dx = d dx ( x2 ) 2 =2( x2 ) d dx ( x2 ) =2( x2 ) The slope of the chord = 40 42 [ m= y 2 y 1 x 2 x 1 ] =2 Since, the slope of the tangent = slope of the chord, we have: 2( x2 )=2 x=3 When x=3, y= ( 32 ) 2 =1 Hence, the required point is ( 3, 1 ). MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6F0D@

Q.9 Find the point on the curve y = x3 − 11x + 5
at which the tangent is y = x − 11.


The equation of the given curve is y= x 3 11x+5 The equation of the tangent to the given curve is y = x11 comparing it with y = mx + c, we get m=1 slope of the tangent( m )=1 Now, the slope of the tangent to the given curve at the point (x, y) is, dy dx = d dx ( x 3 11x+5 ) =3 x 2 11 3 x 2 11=1 [ Given ] x 2 = 12 3 =4 x=±2 When x=2, y= (2) 3 11(2) + 5 =822 + 5 =9. When x=2, y=(2)311 (2) + 5 =8 + 22 + 5 =19. Hence, the required points are ( 2,9 ) and ( 2, 19 ). MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@44E6@


Find the equation of all lines having slope1 that aretangents to the curvey=1x1,x1.


Equation of given curve is y=1x1,x1Differentiating w.r.t. x, we getdydx=ddx1x1      =1(x1)2ddx(x1)      =1(x1)2So,m=(dydx)(x,y)      =1(x1)2According​ to question,1(x1)2=1(x1)2=1x1=±1x1=1  or  x1=1x=2,0When x=0, y=1Whenx=2,  y=1Thus, there are two tangents to the given curve having slope1.These are passing through the points (0, -1) and (2, 1).

And equation of two tangents are y 1 = 1 ( x 2 ) y 1 = x + 2 x + y 3 = 0 A n d , y + 1 = 1 ( x 0 ) y + 1 = x x + y + 1 = 0

Q.11 Find the equation of all lines having slope 2 which are tangents to the curve y = 1 x 3 , x 3 .


Equation of given curve is y=1x3,x3Differentiating w.r.t. x, we getdydx=ddx(1x3)      =1(x3)2ddx(x3)    =1(x3)2So,m=(dydx)(x,y)      =1(x3)2According​ to question,1(x3)2=2(x3)2=2This is impossible that square of a number is negative.Hence, there is no tangent to the given curve having slope 2.

Q.12 Find the equations of all lines having slope 0 which are tangent to the curve y = 1 x 2 2 x + 3


The equation of the given curve isy= 1 x 2 2x+3 The slope of the tangent to the given curve at any point ( x, y ) ( dy dx ) ( x,y ) = d dx 1 x 2 2x+3 = 1 ( x 2 2x+3 ) 2 d dx ( x 2 2x+3 ) = ( 2x2 ) ( x 2 2x+3 ) 2 Since, the slope of the tangent is 0, then we have: ( 2x2 ) ( x 2 2x+3 ) 2 =0( 2x2 )=0 x=1 Whenx=1, y= 1 12+3 = 1 2 The point on the curve is ( 1, 1 2 ). The equation of the tangent through( 1, 1 2 )is y 1 2 =0( x1 ) y= 1 2 Hence, the equation of the required line isy= 1 2 . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@C6CF@


Findpointsonthecurvex29+y216 = 1atwhichthetangentsare (i) paralleltoxaxis(ii)paralleltoyaxis.


The equation of the given curve isx29+y216=1(i)On differentiating both sides with respect to x, we have:    2x9+2y16dydx=0dydx=2x9×162y=169xy(i)When the tangent is parallel to xaxis, thenslope of tangent(m)=0dydx=0169xy=0x=0Putting​ value of x in equation (i),we ​have(0)29+y216=1y2=16  y=±4Hence, the points at which the tangents are parallel to the xaxis are (0, 4) and (0,4).(b)The tangent is parallel to the yaxis if the slope of the normal(M) is 0, i.e., M=0 So,  1169xy=0  [M=1m]yx=0y=0Putting​ value of y in equation (i),we ​havex29+(0)216=1x2=9x=±3Hence, the points at which the tangents are parallel to the yaxis are (3, 0) and (- 3, 0).

Q.14 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)

(v) x = cos t, y = sin t at t= π 4 . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeqaamaaeaqbaaGcbaacbeGaa8hEaiaa=bcacqGH9aqpcaWFGaGaa83yaiaa=9gacaWFZbGaa8hiaiaa=rhacaWFSaGaa8hiaiaa=LhacaWFGaGaeyypa0Jaa8hiaiaa=nhacaWFPbGaa8NBaiaa=bcacaWF0bGaa8hiaiaa=fgacaWF0bGaaGPaVlaaykW7caWF0bGaeyypa0ZaaSaaaeaarmWu51MyVXgaiyaacqGFapaCaeaacaWH0aaaaiaac6caaaa@5B2A@


(i) The equation of the curve is y = x 4 6 x 3 + 13 x 2 10 x + 5 . On differentiating with respect to x, we get: d y d x = d d x ( x 4 6 x 3 + 13 x 2 10 x + 5 )

=4x318x2+26x10(dydx)(0,5)=4(0)318(0)2+26(0)10=10Thus, the slope of the tangent at (0,5)is10.The equation of the tangent is given as:y5=10(x0)y=10x+510x+y=5The slope of the normal (M)at(0,5)=110=110There fore, thee quation of the normal at (0,5) is given as:y5=110(x0)[yy1=M(xx1)]10y50=xx10y+50=0(ii) The equation of the curve is y=x46x3+13x210x+5.On differentiating with respect to x, we get:dydx=ddx(x46x3+13x210x+5)=4x318x2+26x10(dydx)(1,3)=4(1)318(1)2+26(1)10=2Thus, the slope of the tangent at(1,3)is2.The equation of the tangent is given as:

y 3 = 2 ( x 1 ) y 3 = 2 x 2 y = 2 x + 1 T h e s l o p e o f t h e n o r m a l ( M ) a t ( 1 , 3 ) = 1 2 T h e r e f o r e , t h e e q u a t i o n o f t h e n o r m a l a t ( 1 , 3 ) i s g i v e n a s : y 3 = 1 2 ( x 1 ) [ y y 1 = M ( x x 1 ) ] x + 2 y 7 = 0 ( i i ) T h e e q u a t i o n o f t h e c u r v e i s y = x 3 O n d i f f e r e n t i a t i n g w i t h r e s p e c t t o x , w e g e t : d y d x = d d x ( x 3 ) = 3 x 2 ( d y d x ) ( 1 , 1 ) = 3 ( 1 ) 2 = 3 T h u s , t h e s l o p e o f t h e t a n g e n t a t ( 1 , 1 ) i s 3 . E q u a t i o n o f t a n g e n t a t ( 1 , 1 ) o f t h e t a n g e n t i s : y 1 = 3 ( x 1 ) y = 3 x 2 T h e s l o p e o f t h e n o r m a l a t ( 1 , 1 ) = 1 3 T h e r e f o r e , t h e e q u a t i o n o f t h e n o r m a l a t ( 1 , 1 ) i s : y 1 = 1 3 ( x 1 ) x + 3 y 4 = 0 ( i v ) T h e e q u a t i o n o f t h e c u r v e i s y = x 2 O n d i f f e r e n t i a t i n g w i t h r e s p e c t t o x , w e g e t :

d y d x = d d x ( x 2 ) = 2 x ( d y d x ) ( 0 , 0 ) = 2 ( 0 ) = 0 E q u a t i o n o f t a n g e n t a t ( 0 , 0 ) : y 0 = 0 ( x 0 ) y = 0 T h e s l o p e o f t h e n o r m a l a t ( 0 , 0 ) = 1 0 = 1 0 T h e r e f o r e , t h e e q u a t i o n o f t h e n o r m a l a t ( 0 , 0 ) i s : y 0 = 1 0 ( x 0 ) x = 0 ( v ) T h e e q u a t i o n o f t h e c u r v e i s x = c o s t , y = s i n t . D i f f e r e n t i a t i n g w . r . t . t , w e g e t d x d t = sin t a n d d y d t = cos t d y d x = d y d t d x d t = cos t sin t = cot t ( d y d x ) x = π 4 = cot ( π 4 ) = 1 E q u a t i o n o f t a n g e n t a t ( cos π 4 , sin π 4 ) i . e . , ( 1 2 , 1 2 ) :

y 1 2 = 1 ( x 1 2 ) x + y = 2 T h e s l o p e o f t h e n o r m a l a t t = π 4 : M = 1 ( d y d x ) x = π 4 = 1 1 = 1 T h e r e f o r e , t h e e q u a t i o n o f t h e n o r m a l a t ( 1 2 , 1 2 ) : y 1 2 = 1 ( x 1 2 ) y = x

Q.15 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.


The equation of the given curve is       y=x22x+7 ...(i)On differentiating with respect to x, we get:  dydx=2x2slope of tangent to curve=(dydx)(x,y)=2x2(a)The equation of the line is 2xy + 9 = 0y=2x+9comparing with y=mx+c, we get    m=2When tangent is parallel to given line than,(dydx)(x,y)=m2x2=2x=2Putting x=2 in equation(i), we gety=(2)22(2)+7=7Thus, the equation of the tangent passing through (2, 7) is:  y7=2(x2)y7=2x4  y2x3=0Thus, the equation of the tangent line to the given curve (which is parallel to line 2xy + 9 = 0) is y2x3=0.(b) The equation of the line is 5y15x=13.This equation can be written asy=3x+135 which in the form of y=mx+c,So, m=3If a tangent is perpendicular to the line 5y15=13,then the slope of the tangent is1(dydx)(x,y)=132x2=13x=56Whenx=56,y=(56)22(56) + 7=21736Thus, the equation of the tangent passing through(56,21736) is    y21736=13(x56)12x+36y227=0Hence, the equation of the required tangent line is      12x+36y227=0.

Q.16 Show that the tangents to the curve y = 7x3 + 11 at the points where x = 2 and x = −2 are parallel.


We have, y= 7x 3 + 11 Differentiating w.r.t. x, we get dy dx =21 x 2 Slope of tangent at x=2, m 1 = ( dy dx ) x=2 =21 ( 2 ) 2 =84 Slope of tangent at x=2, m 2 = ( dy dx ) x=2 =21 ( 2 ) 2 =84 Since, m 1 = m 2 Hence, the two tangents are parallel. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@FB3A@

Q.17 Find the points on the curve y = x3 at which the slope of the tangent is equal to the y coordinate of the point.


The equation of the given curve is y = x3.Diffferentiating both sides, w.r.t. x, we get            dydx=3x2Slope of tangent at point (x, y) is given by,    (dydx)(x,y)=3x2(dydx)(x,y)=y[Given]      3x2=yBut y=x3[Given]  x3=3x2x33x2=0    x2(x3)=0x=0 and 3When x = 0, then y = 0 and when x = 3, then y = 3(3)2= 27.Hence, the required points are (0, 0) and (3, 27).

Q.18 For the curve y = 4x3 − 2x5, find all the points at which the tangents passes through the origin.


The equation of the given curve is     y=4x32x5...(i)Differentiating with respect to x, we getdydx=12x210x4slope of tangent at the point (x1,y1) is(dydx)(x1,y1)=12x1210x14The equation of the tangent at (x1, y1) is given byyy1=(dydx)(x1,y1)(xx1)yy1=(12x1210x14)(xx1)When the tangent passes through the origin (0, 0), then x=y=0.0y1=(12x1210x14)(0x1)  y1=(12x1210x14)x1  y1=12x1310x15​​​   ...(ii)Point (x1,y1) lies on curve (i),then      y1=4x132x15    .  (iii)Fromequation (ii) and equation(iii),wehave    12x1310x15=4x132x15      8x138x15=08x13(1x12)=0x1=0,±1When x=0, y=4 (0)32 (0)5= 0.When x=1, y=4 (1)32(1)5= 2.When x=1, y=4 (1)32(1)5=2.Hence, the required points are (0, 0), (1, 2) and (1,2).

Q.19 Find the points on the curve x2 + y2 − 2x − 3 = 0 at which the tangents are
parallel to the x-axis.


The equation of the given curve is x 2 + y 2 2x3=0 ( i ) Differentiating both sides, w.r.t. x, we get 2x+2y dy dx 2=0 dy dx = 22x 2y = 1x y Slope of tangent of the given curve at point ( x 1 , y 1 )is ( dy dx ) ( x 1 , y 1 ) = 1 x 1 y 1 Itis given that ( dy dx ) ( x 1 , y 1 ) =0 1 x 1 y 1 =01 x 1 =0 x 1 =1 Since, point ( x 1 , y 1 ) lies on curve( i ), we get x 1 2 + y 1 2 2x 1 3=0 ( 1 ) 2 + y 1 2 2( 1 )3=0 y 1 2 =4 y 1 =±2 Hence, the points at which the tangents are parallel to the x-axis are ( 1, 2 ) and ( 1,2 ). 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Q.20 Find the equation of the normal at the point (am2, am3) for the curve ay2 = x3


The equation of the given curve is ay2=x3.On differentiating with respect to x, we have:2aydydx=3x2        dydx=3x22aySlope at the point (am2, am3)=(dydx)(am2, am3)=3(am2)22a(am3)=3a2m42a2m3=32m    Slope of normal=1(32m)=23mHence, the equation of the normal at (am2, am3) is given by,  yam3=23m(xam2)  3my3am4=2x+2am22x+3myam2(23m2)=0

Q.21 Find the equation of the normals to the curve y = x3 + 2x + 6 which are parallel to the line
x + 14y + 4 = 0.


The equation of the given curve is y = x3 + 2x + 6.Differentiating w.r.t. x, we getdydx=3x2+2The slope of the tangent to the given curve at any point (x1, y1)                 (dydx)(x1, y1)=3x12+2Slope of normal to the given curve at any point (x1, y1)        =1(dydx)(x1, y1)        =13x12+2The equation of the given line is           x + 14y + 4=0Differentiating w.r.t. x, we get  dydx=114Slope of lineat any point (x1, y1)        =(dydx)(x1, y1)        =114Since normal is parallel to line, so   slope of normal=slope of line        13x2+2=114  14=3x2+2    x2=123=4      x=±2When x=2, y=8+4+6=18.When x=2,y=84+6=6.Therefore, the equations of normals passing through the points (2,18) and (2,6).                          y18=114(x2)x+14y254=0and                      y+6=114(x+2)  x+14y+86=0Hence, the equations of the normals to the given curve (which are parallel to the given line) arex+14y254=0,x+14y+86=0.  

Q.22 Find the equations of the tangent and normal to the parabola y2= 4ax at the point (at2, 2at).


The equation of parabola is y2=4axDifferentiating both sides w.r.t. x, we get    2ydydx=4adydx=4a2y=2aySlope of tangent to parabola at point (at2, 2at)  (dydx)(at2, 2at)=2a2at=1tSlope of normal to parabola at point (at2, 2at)=1(dydx)(at2, 2at)=1(1t)=tThen, the equation of the tangent at(at2, 2at)    y2at=1t(xat2)    ty=x+at2And, the equation of the normal at(at2, 2at)    y2at=t(xat2)      tx+  y=2at+at3Thus, the required equations of tangent and normal arety=x+at2 and tx+  y=2at+at3.

Q.23 Prove that the curves x = y2 and xy = k cut at right angles if 8k2 = 1.


Given equations of the curves are x=y2 and xy=k.Putting x=y2 in xy=k, we get          y2.y=ky=k13Putting value of y in xy=k, we get  k13x=kx=k23Thus, the point of intersection of the given curves is(k23,k13).Differentiating the curves x=y2 and xy=k w.r.t. x, we get      ddxx=ddxy21=2ydydxdydx=12ySlope of tangent to the curvesx=y2at point (k23,k13)          m1=(dydx)(k23,k13)=12k13and ddx(xy)=ddxkxdydx+yddxx=0dydx=yxSlope of tangent to the curve xy=k at point (k23,k13)          m2=(dydx)(k23,k13)=k13k23=1k13Since,two curves intersect at right angles if     m1.m2=1      12k13×1k13=1      1=2k23Cubing both sides, we get      1=8k2.Hence, the given two curves cut at right angels if 8k2=1.

Q.24 Find the equations of the tangent and normal to the hyperbola x 2 a 2 y 2 b 2 =1 at the point x 0 ,y 0 .


Equation of hyperbola is x 2 a 2 y 2 b 2 =1 Differentiating w.r.t. x, we get 2x a 2 2y b 2 dy dx =0 dy dx = 2x a 2 × b 2 2y = b 2 a 2 . x y Therefore, the slope of the tangent at( x 0 , y 0 ) ( dy dx ) ( x 0 , y 0 ) = b 2 a 2 . x 0 y 0 Then, the equation of the tangent at( x 0 , y 0 )is given by, y y 0 = b 2 a 2 . x 0 y 0 ( x x 0 ) a 2 y 0 y a 2 y 0 2 = b 2 x 0 x b 2 x 0 2 y 0 y b 2 y 0 2 b 2 = x 0 x a 2 x 0 2 a 2 x 0 x a 2 y 0 y b 2 = x 0 2 a 2 y 0 2 b 2 x 0 x a 2 y 0 y b 2 =1 { Point( x 0 , y 0 )lies on hyperbola x 2 a 2 y 2 b 2 =1 } Now, the slope of the normal at( x 0 , y 0 ), = 1 ( dy dx ) ( x 0 , y 0 ) = 1 b 2 a 2 . x 0 y 0 = a 2 y 0 b 2 x 0 Hence, the equation of the normal at( x 0 , y 0 ) is given by y y 0 = a 2 y 0 b 2 x 0 ( x x 0 ) y y 0 a 2 y 0 + x x 0 b 2 x 0 =0 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Q.25 Find the equation of the tangent to the curve y= 3x2 which is parallel to the line 4x2y + 5 = 0.


The equation of the given curve is y=3x2Differentiating w.r.t. x, we get  dydx=ddx(3x2)12=12(3x2)12ddx(3x2)[By chain rule]=123x2×3=323x2The slope of the tangent to the given curve at any point (x1, y1),  m1=(dydx)(x1, y1)=323x12The equation of the given line is 4x2y + 5 = 0.Differentiating w.r.t. x, we get42dydx=0dydx=2The slope of the tangent to the given line at any point (x1, y1),  m2=(dydx)(x1, y1)=2Since,​ tangent is parallel to given line. So,    m1=m2323x12=2    3x12=34      3x12=9163x1=916+23x1=4116x1=4148Since, point (x1,y1) lies on given curve  y=3x2so,    y1=3x12=34Equation of the tangent passing through the point(4148,34)is                y34=2(x4148)  48x24y=23Hence, the equation of the required tangent is  48x24y=23.

Q.26 The slope of the normal to the curve y = 2x2 + 3 sin x at x = 0 is
(A) 3
(B) 1/3
(C) – 3
(D) –1/3


The equation of given curve is y= 2x 2 + 3sinx Differentiating w.r.t. x, we get dy dx = d dx ( 2x 2 + 3sinx ) =4x+3cosx Putting x= 0, we get m= ( dy dx ) x= 0 =4( 0 )+3cos0 =3 Slope of normal to the given curve is M= 1 m = 1 3 The correct answer is D. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabeqaamaaeaqbaaGceaqabeaacaWGubGaamiAaiaadwgacaqGGaGaaeyzaiaabghacaqG1bGaaeyyaiaabshacaqGPbGaae4Baiaab6gacaqGGaGaae4BaiaabAgacaqGGaGaae4zaiaabMgacaqG2bGaaeyzaiaab6gacaqGGaGaae4yaiaabwhacaqGYbGaaeODaiaabwgacaqGGaGaaeyAaiaabohacaqGGaGaaeiiaiaabMhacqGH9aqpcaqGYaGaaeiEamaaCaaaleqabaGaaeOmaaaakiaabUcacaqGGaGaae4maiaabohacaqGPbGaaeOBaiaabIhaaeaacaqGebGaaeyAaiaabAgacaqGMbGaaeyzaiaabkhacaqGLbGaaeOBaiaabshacaqGPbGaaeyyaiaabshacaqGPbGaaeOBaiaabEgacaqGGaGaae4Daiaab6cacaqGYbGaaeOlaiaabshacaqGUaGaaeiiaiaabIhacaqGSaGaaeiiaiaabEhacaqGLbGaaeiiaiaabEgacaqGLbGaaeiDaaqaaiaaykW7daWcaaqaaiaadsgacaWG5baabaGaamizaiaadIhaaaGaeyypa0ZaaSaaaeaacaWGKbaabaGaamizaiaadIhaaaWaaeWaaeaacaqGYaGaaeiEamaaCaaaleqabaGaaeOmaaaakiaabUcacaqGGaGaae4maiaabohacaqGPbGaaeOBaiaabIhaaiaawIcacaGLPaaaaeaacaWLjaGaeyypa0JaaGinaiaadIhacqGHRaWkcaaIZaGaci4yaiaac+gacaGGZbGaamiEaaqaaiaadcfacaWG1bGaamiDaiaadshacaWGPbGaamOBaiaadEgacaqGGaGaiGgGbIhacGaAagypaiacObyGGaGaiGgGbcdacaqGSaGaaeiiaiaabEhacaqGLbGaaeiiaiaabEgacaqGLbGaaeiDaaqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaeyBaiabg2da9maabmaabaWaaSaaaeaacaWGKbGaamyEaaqaaiaadsgacaWG4baaaaGaayjkaiaawMcaamaaBaaaleaacaqG4bGaaeypaiaabccacaqGWaaabeaaaOqaaiaaykW7caaMc8UaaCzcaiabg2da9iaaisdadaqadaqaaiaaicdaaiaawIcacaGLPaaacqGHRaWkcaaIZaGaci4yaiaac+gacaGGZbGaaGimaaqaaiaaxMaacqGH9aqpcaaIZaaabaGaam4uaiaadYgacaWGVbGaamiCaiaadwgacaqGGaGaae4BaiaabAgacaqGGaGaaeOBaiaab+gacaqGYbGaaeyBaiaabggacaqGSbGaaeiiaiaabshacaqGVbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGJbGaaeyDaiaabkhacaqG2bGaaeyzaiaabccacaqGPbGaae4CaiaabccaaeaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaab2eacqGH9aqpcqGHsisldaWcaaqaaiaaigdaaeaacaWGTbaaaaqaaiaaxMaacqGH9aqpcqGHsisldaWcaaqaaiaaigdaaeaacaaIZaaaaaqaaiaabsfacaqGObGaaeyzaiaabccacaqGJbGaae4BaiaabkhacaqGYbGaaeyzaiaabogacaqG0bGaaeiiaiaabggacaqGUbGaae4CaiaabEhacaqGLbGaaeOCaiaabccacaqGPbGaae4CaiaabccacaqGebGaaeOlaaaaaa@179E@

Q.27 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)


The equation of the given curve is  y2= 4xDifferentiating w.r.t. x, we get2ydydx=4dydx=42y=2yTherefore, the slope of the tangent to the given curve at any point (x1, y1)=(dydx)(x1, y1)=2y1The given line is y = x + 1 which is in the form of y=mx+c,      m=1Since, tangent and line are parallel.So(dydx)(x1, y1)=m2y1=1    y1=2Since,y=x+1,so   x=y1        =21=1Hence, the line y = x + 1 is a tangent to the given curve at the point (1, 2).The correct answer is A.

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