Continuity and differentiability class 12 miscellaneous exercise

In the Class 12 Maths chapter 5 miscellaneous exercise solutions, you will get a mixture of questions from all the previous exercises of this Class 12 Maths NCERT textbook chapter. You will get questions related to first-order derivatives of different types of functions, second-order derivatives, mean-value theorem, Rolle's theorem in the miscellaneous exercise chapter 5 Class 12. This Class 12 NCERT syllabus exercise is a bit difficult as compared to previous exercises, so you may not be able to solve NCERT problems from this exercise at first. You can take help from NCERT solutions for Class 12 Maths chapter 5 miscellaneous exercise to get clarity. There are not many questions asked in the board exams from this exercise, but Class 12 Maths chapter 5 miscellaneous solutions are important for the students who are preparing for competitive exams like Also, see • • • • • • • •

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8 Get Free NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Continuity and Differentiability Exercise 5.8 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 5 Class 12 Continuity and Differentiability Ex 5.8 provided in NCERT Textbook. Free download NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20. The topics and sub-topics included in the Continuity and Differentiability chapter are the following: • Continuity and Differentiability • Introduction • Algebra of continuous functions • Differentiability • Derivatives of composite functions • Derivatives of implicit functions • Derivatives of inverse trigonometric functions • Exponential and Logarithmic Functions • Logarithmic Differentiation • Derivatives of Functions in Parametric Forms • Second Order Derivative • Mean Value Theorem • Summary There are total eight exercises and one misc exercise( 144 Questions fully solved) in the class 12th maths chapter ...

du/dx = x x(1 + log x) ………(2) Now we take v = 2 sin x On taking log on both sides, we get log v = log (2 sinx) log v = sin x log2 Now, on differentiating w.r.t x, we get dv/dx = v(log2cos x) dv/dx = 2 sin xcos xlog2 ………(3) Now put all the values from eq(2) and (3) into eq(1) dy/dx = x x(1 + log x) – 2 sin xcos xlog2 Question 5. (x + 3) 2.(x + 4) 3.(x + 5) 4 Solution: Let us considered y = (x + 3) 2.(x + 4) 3.(x + 5) 4 Now taking log on both sides, we get log y = log[(x + 3) 3.(x + 4) 3.(x + 5) 4] log y = 2 log(x + 3) + 3 log(x + 4) + 4 log(x + 5) Now, on differentiating w.r.t x, we get Given: y = (sin x) x + sin –1√x Let us considered y = u + v Where u = (sin x) x and v = sin –1√x so, dy/dx = du/dx + dv/dx ………(1) Now first we takeu = (sin x) x On taking log on both sides, we get log u = log(sin x) x log u = xlog(sin x) Now, on differentiating w.r.t x, we get ………(2) Now we takev = sin –1√x On taking log on both sides, we get log v = log sin –1√x Now, on differentiating w.r.t x, we get ………(3) Now put all the values from eq(2) and (3) into eq(1) Question 9. x sin x + (sin x) cos x Solution: Given: y = x sin x + (sin x) cos x Let us considered y = u + v Where u = x sin x and v = (sin x) cos x so, dy/dx = du/dx + dv/dx ………(1) Now first we takeu = x sin x On taking log on both sides, we get log u = log x sin x log u = sin x(log x) Now, on differentiating w.r.t x, we get ………(2) Now we takev =(sin x) cos x On taking log on both sides, we get log v = log(sin x) cos x log v = cosx l...

Transcript Ex 5.1, 1 Prove that the function 𝑓 (𝑥) = 5𝑥 – 3 is continuous at 𝑥 = 0, at 𝑥 = –3 and at 𝑥 = 5 Given 𝑓(𝑥)= 5𝑥 –3 At 𝒙=𝟎 f(x) is continuous at x = 0 if (𝐥𝐢𝐦)┬(𝐱→𝟎) 𝒇(𝒙) = 𝒇(𝟎) (𝐥𝐢𝐦)┬(𝐱→𝟎) 𝒇(𝒙) "= " lim┬(x→0) ""(5𝑥−3) Putting x = 0 = 5(0) − 3 = −3 𝒇(𝟎) = 5(0) − 3 = 0 − 3 = −3 Since L.H.S = R.H.S Hence, f is continuous at 𝒙 = 𝟎 At x = −3 f(x) is continuous at x = −3 if ( lim)┬(x→−3) 𝑓(𝑥)= 𝑓(−3) Since, L.H.S = R.H.S Hence, f is continuous at 𝒙 =−3 (𝐥𝐢𝐦)┬(𝐱→𝟑) 𝒇(𝒙) "= " lim┬(x→3) ""(5𝑥−3) Putting x = −3 = 5(−3) − 3 = −18 𝒇(−𝟑) = 5(−3) − 3 = −15 − 3 = −18 At 𝒙 =𝟓 f(x) is continuous at x = 5 if ( lim)┬(x→5) 𝑓(𝑥)= 𝑓(5) Since, L.H.S = R.H.S Hence, f is continuous at x = 5 Thus, the function is continuous at x = 0, x = −3, x = 5 (𝒍𝒊𝒎)┬(𝒙→𝟓) 𝒇(𝒙) "= " (𝑙𝑖𝑚)┬(𝑥→5) ""(5𝑥−3) Putting x = 5 = 5(5) − 3 = 22 𝒇(𝟓) = 5(5) − 3 = 25 − 3 = 22 Show More