# NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

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 2019-20 in PDF form. NCERT Solutions and corresponding Offline Apps are also available to free download.

 Class: 12 Subject: Maths – गणित Chapter 5: Continuity and Differentiability ## NCERT Solutions for class 12 Maths chapter 5

### NCERT Solutions of Continuity and Differentiability

• Class 12 Maths Chapter 5 Exercise 5.1 Solutions
• Class 12 Maths Chapter 5 Exercise 5.2 Solutions
• Class 12 Maths Chapter 5 Exercise 5.3 Solutions
• Class 12 Maths Chapter 5 Exercise 5.4 Solutions
• Class 12 Maths Chapter 5 Exercise 5.5 Solutions
• Class 12 Maths Chapter 5 Exercise 5.6 Solutions
• Class 12 Maths Chapter 5 Exercise 5.7 Solutions
• Class 12 Maths Chapter 5 Exercise 5.8 Solutions
• Class 12 Maths Chapter 5 Miscellaneous Exercise 5 Solutions

### Assignments for practice

#### Mixed Chapter Tests

Chapter 5 & 6

##### Level 3 Test 1 Test 2
• 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.
• 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.
• 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]

#### 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.

## 3 thoughts on “NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability”

1. ram sharma says:

nice sir

2. ram sharma says:

3. max says: