NCERT Solutions for Class 12 Maths Chapter 5

NCERT Solutions for class 12 Maths chapter 5 Continuity and Differentiability all exercises (ex. 5.1, 5.2, 5.3, 5.4, 5.5, 5.6, 5.7, 5.8 and miscellaneous exercise) English Medium and Hindi Medium. All the NCERT sols are updated for academic session 2020-21 for all boards who are following NCERT Books for their course. CBSE Solutions are in the format of PDF file and Video. Videos related to Class 12 Mathematics chapter 5 all exercises are given below in Hindi and English Medium. UP Board Students also using the same NCERT textbooks as CBSE Students. So, these solutions are useful for those also in solving their doubts.

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NCERT Solutions for class 12 Maths chapter 5

Class: 12Maths (English and Hindi Medium)
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 and English Medium free to download. Download NCERT Books for class 12 all subjects for 2020-21 based on latest CBSE Syllabus. Join the discussion forum to ask the doubts related to NIOS and CBSE Board.

Class 12 Maths Exercise 5.1 Solutions in Videos

Class 12 Maths Exercise 5.1 Question 1, 2 Solution in English
Class 12 Maths Exercise 5.1 Question 1, 2 Solution in Hindi

Class 12 Maths Exercise 5.2 Solutions in Videos

Class 12 Maths Exercise 5.2 Question 1, 2, 3 Solution in English
Class 12 Maths Exercise 5.2 Question 1, 2, 3 Solution in Hindi

Class 12 Maths Exercise 5.3 Solutions in Videos

Class 12 Maths Exercise 5.3 Question 1, 2, 3 Solution in English
Class 12 Maths Exercise 5.2 Question 1, 2, 3 Solution in Hindi

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)].
ROLLE’S THEOREM

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

Find dy/dx in the following: 2x+3y = sin⁡x.
2x+3y=sin⁡x
Differentiating both sides with respect to x, we get
d/dx (2x) + d/dx (3y) = d/dx sin⁡x
⇒ 2+3 dy/dx = cos⁡x
⇒ dy/dx = (cos⁡x-2)/3
निम्नलिखित फलनों का अवकलन कीजिए: sin⁡〖(x^2+5)〗
माना y=sin⁡〖(x^2+5)〗
इसलिए,
dy/dx = cos⁡〖(x^2+5).d/dx〗 (x^2+5)
= cos(x^2+5).2x
Find the second order derivatives of x^2+3x+2.
Let y=x^2+3x+2, therefore,
dy/dx = d/dx (x^2+3x+2)
=2x+3
⇒(d^2 y)/(dx^2 )
= d/dx (2x+3)=2
If y=3 cos⁡〖(log⁡x )+4 sin⁡〖(log⁡x)〗 〗, show that x^2 y_2+xy_1+y=0
Given that: y=3 cos⁡〖(log⁡x )+4 sin⁡〖(log⁡x)〗 〗, therefore,
dy/dx=d/dx (3 cos⁡〖(log⁡x )+4 sin⁡〖(log⁡x)〗 〗 )
=-3 sin⁡〖(log⁡x ).1/x+4 cos⁡〖(log⁡x ).〗 〗 1/x
⇒x dy/dx=-3 sin⁡〖(log⁡x )+4 cos⁡(log⁡x ) 〗
⇒x (d^2 y)/(dx^2 )+dy/dx.d/dx x
=d/dx [-3 sin⁡〖(log⁡x )+4 cos⁡(log⁡x ) 〗 ]
=-3 cos⁡〖(log⁡x ).1/x-4 sin⁡(log⁡x ) 〗.1/x
=-1/x [3 cos⁡〖(log⁡x )+4 sin⁡(log⁡x ) 〗 ]=-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
⇒x^2 y_2+xy_1+y=0
फलन f(x)=x^2+2x-8,x∈[-4,2] के लिए रोले के प्रमेय को सत्यापित कीजिए।
दिया गया फलन f(x)=x^2+2x-8,x∈[-4,2]
(i) फलन f एक बहुपद है। अतः, यह संवृत अंतराल [-4,2] में संतत है।
(ii) f'(x)=2x+2
अतः, फलन f विवृत अंतराल (-4,2) में अवकलनीय है।
(iii) f(-4)=(-4)^2+2(-4)-8=16-8-8=0
तथा f(2)=(2)^2+2(2)-8=4+4-8=0
⇒f(-4)=f(2)
यहाँ, रोले की तीनों परिस्थितियाँ सत्य हैं। इसलिए, विवृत अंतराल (-4,2) में किसी ऐसे c का अस्तित्व है कि f'(c)=0 है।
⇒f^’ (c)=2c+2=0
⇒c=-1∈(-4,2)
अतः, फलन f(x)=x^2+2x-8,x∈[-4,2] के लिए रोले की प्रमेय सत्यापित हो जाती है।
Using mathematical induction prove that d/dx (x^n )=nx^(n-1) for all positive integers n.
Let, P(n):d/dx (x^n )=nx^(n-1)
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.
To prove:
P(k+1):d/dx (x^(k+1) )=(k+1)x^k is also true.
LHS
=d/dx (x^(k+1) )
=d/dx (x^k.x)
=x^k d/dx (x)+x d/dx x^k
=x^k.1+x.kx^(k-1)
=(1+k) x^k
=RHS
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.