NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7

The Central Board of Secondary Education CBSE is one of the biggest and most reputed boards in the country. CBSE is a national-level board with schools affiliated with it all over the country. According to CBSE, education has an objective approach, and its curriculum requires students to have an in-depth understanding of the subject matter. CBSE is the ideal board for students who wish to take a variety of competitive exams and succeed in them. CBSE’s higher secondary curriculum is well known throughout the country for its level of difficulty. It is found that students who switch from another board to CBSE for their eleventh standard face many obstacles. This might seem challenging but there are effective ways that can be followed by the student to get over these struggles.

NCERT stands for National Council of Educational Research and Training and is a resource-based institution founded by the Indian Government. NCERT assists the Central Government and the State Government on matters related to education and academia. CBSE follows all the NCERT suggested textbooks. CBSE emphasises improving the cognitive development of its students. Regular revision is necessary for CBSE students, as well as hard work, excellent reference work, and a solid understanding of all subjects.

Mathematics is one of the core subjects for all the students who have opted for the subject after their Class 10 board examinations. The change from Class 10 to a 10 + 2 level can be challenging. Many students suffer from great anxiety when it comes to Mathematics, and the 10+2 level makes the subject even more difficult. The CBSE Class 12 Mathematics has a massive syllabus and finishing the whole syllabus can be challenging.

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability (Ex 5.7) Exercise 5.7

Chapter 5 of Class 12 CBSE is Continuity and Differentiation which introduces the students to concepts from Physics. This exercise primarily focuses on finding out and calculating the second order derivative of functions, calculating the value of d²y/dx² regarding finding its expression in question which asks the students to ‘proof’ something.

Mathematics is a core subject that has interrelations with every science subject. Exercise 5.7 Class 12th Maths NCERT Solutions teaches the students to calculate second-order derivatives and these techniques and concepts are extremely crucial for understanding Coulomb’s Law when a student is preparing for Electric Charges and Fields. The mathematical proof of the law itself implements the concepts of second-order derivatives and therefore the numerical-based questions of the same chapter use the same concepts. These areas of the syllabus which merge different subjects are extremely important, suggesting teachers and students must take these chapters seriously. It is suggested that these chapters should be given a lot of priority not only because they carry a lot of weightage but also because it makes sense in the context of a good studying strategy. When a student is learning Exercise 5.7 they are not just preparing for Mathematics but in turn, they are also making progress in their Physics Syllabus.

Extramarks compiles the notes provided by esteemed subject experts from all over the country. The solutions are based on the questions that students in CBSE Class 12 find difficult and therefore the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 assist these students greatly. Leading educators provide the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 in the field. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 provides not just solutions but explanations for every step taken to solve a mathematical problem. Highly qualified professionals provide the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 and therefore the students can heave a breath of relief. Therefore, the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 provide detailed solutions which are checked multiple times to ensure complete accuracy.

Teachers all over the country have repeatedly emphasised the importance of inculcating a habit of regularly practising mathematical problems to master them. For a student to combat both the vast Mathematics syllabus and the advanced level of difficulty that comes with being in CBSE Class 12. Teachers and Students unanimously have observed that these reasons hinder the smooth progression of the syllabus. Students often complain about how the doubts that arise for them while solving mathematical problems sometimes follow the same trend consecutively for many questions.

The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7, provides great and highly valued assistance to deal with the problems faced by a student. The solutions provided in NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 are accurate. Mathematical problems are often solved by using multiple approaches. It is in such a situation that the solution that seems the most acceptable and understandable is selected.

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Access NCERT Solutions for Class 12 Maths Chapter 5 – Continuity and Differentiability

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Class 12 students can often get intimidated by the vast syllabus of Class 12 Mathematics. If a student has a thorough understanding of all the concepts and topics of the Mathematics syllabus, it can be one of the most scoring subjects in Class 12. Students should have a thorough understanding of all the important concepts and topics of Mathematics. Understanding all the formulas will help the students solve the exercise questions with ease.

What Do You Learn from NCERT Maths 5.7 Exercise Class 12?

Chapter 5 of the CBSE is Continuity and Differentiability and the 5.7 Exercise is one of the most important exercises of the chapter. Concepts like Differentiation and Integration are introduced to students in Class 11 and Students who have taken up science as their stream know the importance of these concepts. These ideas constantly reiterate themselves in subjects like Chemistry and Physics all the time. As observed in the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 the implications of this exercise are huge. One just does not learn the mathematical concepts but also the interdisciplinary context of these mathematical; concepts, and their practical applications.

The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 addresses the various facets of this exercise very skilfully. Students are introduced to the concepts of Physics which have a direct link with this exercise so that a student can have an idea about the chapters of Physics.

The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 provide an effective introduction and sets up the ground for Chapter 12 Physics CBSE Class 12 Chapter 12 – Coulombs Law. Starting from the derivation of the law to solving the numerical-based questions of these chapters a student has to have a very good understanding of Exercise 5.7 and the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 greatly helps students in that regard.

Extramarks Class 12 Physics Chapter 1 NCERT Solutions

Physics is one of the core subjects and therefore students from the Science stream have to give  a lot of importance to it. Physics is a subject which is completely analytical and students take a considerable amount of time understanding the concepts because these topics require the fundamental concepts of the chapter to be very clear. Physics and Mathematics as they are two sides of the same coin. To be good at Physics students have to be good at Mathematics and vice-versa.

Chapter 12 of Class 12 CBSE Physics explicates more on the concept of Electric Charges and Electric Fields. Students have learned about these concepts in their earlier classes but in Class 12 they learn about these concepts in full depth. Having a good hold over the Class 12 Maths NCERT Solutions Chapter 5 Exercise 5.7 is advised by teachers.

Chapter 12 discusses new concepts about conductors, insulators, and charging through induction. The chapter includes diagrams delineating the charge flow and it also helps visualise the force field around a source charge. The chapter discusses electric dipoles, while the theory of these concepts is important, the derivations of the formulas in these chapters are based on everything discussed in the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7.

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Benefits of Extramarks Chapter 1 Physics Class 12 NCERT Solutions

The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 that are provided are based on the NCERT books. Physics has both theoretical and mathematical aspects. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 are compiled and curated very carefully looking out for any errors. The NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 make sure that nothing is difficult for the students to understand. NCERT Solutions are often overlooked by a lot of students. The NCERT textbooks are the primary source of teaching and this has resulted in a higher demand for NCERT solutions. For senior classes like Class 12, NCERT solutions are extremely essential for the preparation for their board examinations.

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Q.1

Find the second order derivative of​ the function : x20

Ans

Let y=x20Differentiating both sides w.r.t. x, we get  dydx=ddxx20=20x19Differentiating again w.r.t. x, we get∴d2ydx2=ddx(20x19)   =20×19 x18   =380 x18Thus, second derivative of given function is 380 x18.

Q.2

Find the second order derivative of the function:x.cosx

Ans

Let y=xcosxDifferentiating both sides w.r.t. x, we get  dydx=ddx(xcosx)=xddx cosx+cosxddxx[By product rule]=x(−sinx)+cosx×1=−xsinx+cosxDifferentiating again w.r.t. x, we get∴d2ydx2=ddx(−xsinx+cosx)   =−(xddxsinx+sinxddxx)+ddxcosx   =−(xcosx+sinx×1)−sinx   =−xcosx−sinx−sinx   =−xcosx−2sinxThus, second derivative of given function is −(xcosx+2sinx).

Q.3

Find the second order derivative​ of ​the function: logx

Ans

Let y=logxDifferentiating both sides w.r.t. x, we get  dydx=ddxlogx=1x=x−1Differentiating again w.r.t. x, we get∴d2ydx2=ddx(x−1)   =−1 x−2   =−x−2Thus, second derivative of given function is −x−2.

Q.4 

Find the second order derivative of the function:x3 logx

Ans

Let y=x3logxDifferentiating both sides w.r.t. x, we get  dydx=ddx(x3logx)=x3ddxlogx+logxddxx3[By product rule]=x3×1x+logx×3x2=x2+3x2logxDifferentiating again w.r.t. x, we get∴d2ydx2=ddx(x2+3x2logx)   =ddxx2+3ddx(x2logx)   =2x+3(x2ddxlogx+logxddxx2)   =2x+3(x2×1x+logx.2x)   =2x+3x+6xlogx   =5x+6xlogxThus, second derivative of given function is x (5+6logx).

Q.5

Find the second order derivative of the function:ex sin5x

Ans

Let y=exsin5xDifferentiating both sides w.r.t. x, we get  dydx=ddxexsin5x=exddxsin5x+sin5xddxexBy product rule=ex5cos5x+sin5x.ex=ex5cos5x+sin5xDifferentiating again w.r.t. x, we get∴d2ydx2=ddxex5cos5x+sin5x   =exddx5cos5x+sin5x+5cos5x+sin5xddxexBy product rule   =ex−25sin5x+5cos5x+5cos5x+sin5xex   =ex−25sin5x+5cos5x+5cos5x+sin5x   =2ex−12sin5x+5cos5x

Q.6

Find the second order derivative of the function:e6x cos3x

Ans

Let y=e6xcos3xDifferentiating both sides w.r.t. x, we get  dydx=ddx(e6xcos3x)=e6xddxcos3x+cos3xddxe6x[By product rule]=e6x(−3sin3x)+cos3x.6e6x=e6x(6cos3x−3sin3x)Differentiating again w.r.t. x, we get∴d2ydx2=ddx{e6x(6cos3x−3sin3x)}   =e6xddx(6cos3x−3sin3x)+(6cos3x−3sin3x)ddxe6x[By product rule]   =ex(−18sin3x−9cos3x)+(6cos3x−3sin3x)6e6x   =e6x(−18sin3x−9cos3x+36cos3x−18sin3x)   =e6x(−36sin3x+27cos3x)   =9e6x(−4sin3x+3cos3x)Thus, second derivative of given function is 9e6x(3cos3x−4sin3x).

Q.7

Find the second order derivative of the function:tan−1x

Ans

Let y=tan−1xDifferentiating both sides w.r.t. x, we get  dydx=ddxtan−1x=11+x2Differentiating again w.r.t. x, we get∴d2ydx2=ddx(11+x2)   =ddx(1+x2)−1   =−(1+x2)−2ddx(1+x2)=−(1+x2)−2(0+2x)=−2x(1+x2)2Thus, second derivative of given function is −2x(1+x2)2.

Q.8

Find the scond orderderivative of the function:log(logx)

Ans

Let y=log(logx)Differentiating both sides w.r.t. x, we get  dydx=ddxlog(logx)=1logxddxlogx[By  chain  rule]=1logx(1x)=1xlogxDifferentiating again w.r.t. x, we get∴d2ydx2=ddx(1xlogx)   =ddx(xlogx)−1   =−(xlogx)−2ddx(xlogx)[By  chain  rule]   =−1(xlogx)2(xddxlogx+logxddxx)   =−1(xlogx)2(x×1x+logx×1)   =−1(xlogx)2(1+logx)   =−(1+logx)(xlogx)2Thus, second derivative of given function is −(1+logx)(xlogx)2.

Q.9 Find the second order derivative of the function:
sin (log x)

Ans

Let y=sin(logx)Differentiating both sides w.r.t. x, we get  dydx=ddxsin(logx)=cos(logx)ddxlogx[By  chain  rule]=cos(logx)(1x)=cos(logx)xDifferentiating again w.r.t. x, we get∴d2ydx2=ddx(cos(logx)x)   =xddxcos(logx)−cos(logx)ddxxx2   =x×−sin(logx)ddxlogx−cos(logx)×1x2[By  chain  rule]   =−xsin(logx)×1x−cos(logx)x2   =−{sin(logx)+cos(logx)}x2Thus, second derivative of given function is −{sin(logx)+cos(logx)}x2.

Q.10

If y=5cosx−3sinx, prove that d2ydx2+y=0

Ans

Given:    y=5cosx−3sinxDifferentiating w.r.t. x, we get   dydx=ddx(5cosx−3sinx)=5ddxcosx−3ddxsinx=−5sinx−3cosxAgain differentiating w.r.t. x, we get   d2ydx2=−ddx(5sinx+3cosx)  =−(5ddxsinx+3ddxcosx)  =−(5cosx−3sinx)  =−y∴ d2ydx2+y=0Hence Proved.

Q.11

If y=cos−1x. Findd2ydx2 in term so fy alone.

Ans

Given:    y=cos−1x      cosy=xDifferentiating w.r.t. y, we get ddycosy=ddyx⇒       dxdy=−siny⇒       dydx=−1siny⇒       dydx=−cosecy  ...(i)Again differentiating w.r.t. x, we get   d2ydx2=ddx(−cosecy)  =cosecycotydydx  =cosecycoty×−cosecy  =−cosec2ycoty [From equation (i)]

Q.12

Ify=3cos(logx)+ 4sin(logx), show that x2y2+xy1+y=0

Ans

Given:    y=3cos(logx) + 4 sin(logx)Differentiating w.r.t. x, we getdydx(=y1)=ddx{3cos(logx) +4 sin(logx)}=3ddxcos(logx) + 4ddxsin(logx)=−3sin(logx)ddxlogx+4cos(logx)ddxlogx[By chain rule]=−3sin(logx)1x+4cos(logx)1x=−3sin(logx)+4cos(logx)xAgain differentiating w.r.t. x, we get  d2ydx2(y2)=ddx(4cos(logx)−3sin(logx)x)=xddx{4cos(logx)−3sin(logx)}−{4cos(logx)−3sin(logx)}ddxxx2=x{−4sin(logx)ddxlogx−3cos(logx)ddxlogx}−{4cos(logx)−3sin(logx)}×1x2=x{−4sin(logx)1x−3cos(logx)1x}−{4cos(logx)−3sin(logx)}x2=−4sin(logx)−3cos(logx)−4cos(logx)+3sin(logx)x2=−sin(logx)−7cos(logx)x2Thus,d2ydx2=y2=−sin(logx)−7cos(logx)x2L.H.S.=x2y2+xy1+y=x2(−sin(logx)−7cos(logx)x2)+x(−3sin(logx)+4cos(logx)x)+3cos(logx)+4sin(logx)=−sin(logx)−7cos(logx)−3sin(logx)+4cos(logx)+3cos(logx)+ 4sin(logx)=−4sin(logx)+ 4sin(logx)−7cos(logx)+7cos(logx)=0=R.H.S.Hence proved.

Q.13

Ify=Aemx+Benx, show that d2ydx2−(m+n)dydx+mny=0

Ans

Given that y=A emx + B enxDifferentiating w.r.t. x, we get   dydx=Addxemx + Bddx enx=Amemx + Bn enxAgain differentiating w.r.t. x, we get d2ydx2=Amddxemx + Bn ddxenx=Am2emx + Bn2enxL.H.S.=d2ydx2−(m+n)dydx+mny=(Am2emx+Bn2enx)−(m+n)(Amemx+Bn enx)+mn(A emx + B enx)=Am2emx+Bn2enx−Am2emx−Bmn enx−Amnemx−Bn2 enx+Amnemx+Bmn enx=0=R.H.S.

Q.14

Ify=500e7x+600e−7x, show that d2ydx2= 49y

Ans

Given that y=500 e7x + 600 e–7xDifferentiating w.r.t. x, we get       dydx=500ddxe7x +600ddx e–7x    =500×7e7x+600×(−7) e–7x    =7(500e7x−600 e–7x)Again differentiating w.r.t. x, we get     d2ydx2=7(500×7e7x−600×(−7) e–7x)⇒ d2ydx2=49(500e7x+600 e–7x)⇒ d2ydx2=49yHence proved.

Q.15

If ey(x+1)=1, show that d2ydx2=(dydx)2

Ans

Given that e y ( x+1 )=1     e y = 1 x+1     y=log( 1 x+1 )     y=−log( x+1 ) Differentiating w.r.t. x, we get         dy dx = d dx ( −log( x+1 ) )     =− d dx log( x+1 )     =− 1 x+1 d dx ( x+1 )     =− 1 x+1 ×( 1+0 )     =− 1 x+1 Again differentiating w.r.t. x, we get       d 2 y d x 2 = d dx ( − 1 x+1 ) ⇒  d 2 y d x 2 =− d dx ( x+1 ) −1 ⇒  d 2 y d x 2 =−( −1 ) ( x+1 ) −2 d dx ( x+1 )     = 1 ( x+1 ) 2     = ( − 1 x+1 ) 2 ⇒  d 2 y d x 2 = ( dy dx ) 2 Hence proved. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@A490@

Q.16

If y=(tan−1x)2, show that (x2+1)y2+2x(x2+1)y1=2

Ans

The given function is  y=(tan−1x)2Differentiating both sides w.r.t. x, we get    dydx=ddx(tan−1x)2=2tan−1xddxtan−1x               y1=2tan−1x(11+x2)(1+x2)y1=2tan−1xDifferentiating again w.r.t. x, we get        ddx{(1+x2)y1}=2ddxtan−1x(1+x2)ddxy1+y1ddx(1+x2)=2×11+x2           (1+x2)y2+y1(0+2x)=2×11+x2   (1+x2)y2+2xy1=21+x2    (1+x2)2y2+2x(1+x2)y1=2⇒(x2+1)2y2+2x(x2+1)y1=2Hence proved.

Q.17

VerifyRollesTheoremforthefunctionf(x)=x2+2x−8,x∈[−4,2].

Ans

The given function, f(x)=x2+2x−8, is a polynomial. So, it is continuous in [−4, 2] and is differentiable in (−4,2).f(−4)=(−4)2+2(−4)−8   =16−8−8=0  f(2)=(2)2+2(2)−8   =4−4−8=0∴f(−4)=f(2)=0⇒The value of f(x) at −4 and 2 coincide.Rolle‘s Theorem states that there is a point c∈(−4, 2) such that f‘(c)=0f(x)=x2+2x−8Differentiating f(x) w.r.t. x, we getf‘(x)=ddx(x2+2x−8)         =2x+2f‘(c)=2c+2∵f‘(c)=0⇒2c+2=0⇒c=−1∈(−4, 2)Hence, Rolle‘s Theorem is verified for the given function.

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