NCERT Solutions for Class 12 Maths Chapter 5
NCERT Solutions for class 12 Maths chapter 5 Continuity and Differentiability exercise 5.1, 5.2, 5.3, 5.4, 5.5, 5.6, 5.7, 5.8 and miscellaneous exercise English Medium and Hindi Medium for 2020-21 in PDF form.NCERT Solutions and corresponding Offline Apps are also available to free download. All the contents on this website are free to use.
NCERT Solutions for class 12 Maths chapter 5
|Chapter 5:||Continuity and Differentiability|
12th Maths Chapter 5 Solutions
NCERT Solutions for class 12 Maths chapter 5 Continuity and Differentiability all exercises with miscellaneous exercises in Hindi & English Medium free to download. Download NCERT Books 2020-21 based on latest CBSE Syllabus. Join the discussion forum to ask the doubts related to NIOS and CBSE Board.
Chapter 5 Solutions in PDF
- Download Exercise 5.1 in PDF
- Download Exercise 5.2 in PDF
- Download Exercise 5.3 in PDF
- Download Exercise 5.4 in PDF
- Download Exercise 5.5 in PDF
- Download Exercise 5.6 in PDF
- Download Exercise 5.7 in PDF
- Download Exercise 5.8 in PDF
- Download Miscellaneous Exercise 5
- NCERT Book Chapter 5
- NCERT Book Answers
- Revision Book Chapter 5
- Revision Book Answers
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Important Terms related to Chapter 5
1. The main points of the chapter are continuous functions, algebra of continuous functions, differentiation and continuity, chain rule, rules for derivative of inverse functions, derivative of implicit function, parametric and logarithmic functions. Second order derivatives, mean value theorems (LMV) and Rolle’s Theorem.
2. Calculus (differentiability) deals with related variables and constants. In differential calculus we investigate the way in which ‘one quantity varies when the other related quantity is made to vary’. We find the rate of change (in Application of derivatives) one variable quantity relative to another variable. The relation between a function and its derivative is same as between displacement of a particle and its velocity.
3. The derivative of f(x) at x = a is represented by the slope (gradient) of the tangent to the curve y = f(x) at the point P[a, f(a)].
Let f be a real function defined in the closed interval [a, b] such that
(i) f is continuous in the closed interval [ a, b ]
(ii) f is differentiable in the open interval ( a, b )
(iii) f (a) = f (b)
Then f'(c) = 0, where c lies in (a, b).
MEAN VALUE THEOREM
Let f be a real valued function defined on the closed interval [a, b] such that
(a) f is continuous on [a, b], and
(b) f is differentiable in (a, b)
Then there exists a point c in the open interval (a, b) such that f ‘(c) = [f (b) – f (a)]/[b – a]
Important Questions on 12th Maths Chapter 5
Differentiating both sides with respect to x, we get
d/dx (2x) + d/dx (3y) = d/dx sinx
⇒ 2+3 dy/dx = cosx
⇒ dy/dx = (cosx-2)/3
dy/dx = cos〖(x^2+5).d/dx〗 (x^2+5)
dy/dx = d/dx (x^2+3x+2)
⇒(d^2 y)/(dx^2 )
= d/dx (2x+3)=2
dy/dx=d/dx (3 cos〖(logx )+4 sin〖(logx)〗 〗 )
=-3 sin〖(logx ).1/x+4 cos〖(logx ).〗 〗 1/x
⇒x dy/dx=-3 sin〖(logx )+4 cos(logx ) 〗
⇒x (d^2 y)/(dx^2 )+dy/dx.d/dx x
=d/dx [-3 sin〖(logx )+4 cos(logx ) 〗 ]
=-3 cos〖(logx ).1/x-4 sin(logx ) 〗.1/x
=-1/x [3 cos〖(logx )+4 sin(logx ) 〗 ]=-1/x.y
⇒x (d^2 y)/(dx^2 )+dy/dx=-1/x y
⇒x^2 (d^2 y)/(dx^2 )+x dy/dx=-y
⇒x^2 (d^2 y)/(dx^2 )+x dy/dx+y=0
(i) फलन f एक बहुपद है। अतः, यह संवृत अंतराल [-4,2] में संतत है।
अतः, फलन f विवृत अंतराल (-4,2) में अवकलनीय है।
यहाँ, रोले की तीनों परिस्थितियाँ सत्य हैं। इसलिए, विवृत अंतराल (-4,2) में किसी ऐसे c का अस्तित्व है कि f'(c)=0 है।
अतः, फलन f(x)=x^2+2x-8,x∈[-4,2] के लिए रोले की प्रमेय सत्यापित हो जाती है।
Putting n=1, we have LHS=d/dx (x^1 )=1 and RHS=1x^(1-1)=x^0=1
Hence, P(n) is true for n=1.
Let, P(k):d/dx (x^k )=kx^(k-1) is true.
P(k+1):d/dx (x^(k+1) )=(k+1)x^k is also true.
=d/dx (x^(k+1) )
=x^k d/dx (x)+x d/dx x^k
Hence, P(n) is true for n=k+1.
Therefore, by the principle of Mathematical Induction P(n) is true for all natural numbers n.