NCERT Solutions for Class 12-science Maths Chapter 5 - Continuity and Differentiability

Updated NCERT Textbook Solutions Coming Soon!

Chapter 5 - Continuity and Differentiability Exercise Ex. 5.1

Solution 1

The given function is f(x) = 5x - 3

At x = 0, f(0) = 5 × 0 - 3 = -3

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

 
Thus, from the above observation, it can be concluded that x = 1 is the only point of discontinuity.
 
Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

The given function is f(x) = x2 - sinx + 5

It is evident that f is defined at x = ∏

Solution 21

Solution 22

Therefore, cosecant is continuous except at x = np, n begin mathsize 12px style element of end styleZ
Therefore, cotangent is continuous except at x = np, n element ofZ
Solution 23

Solution 24

 
begin mathsize 12px style negative straight x squared less or equal than straight x squared sin 1 over straight x less or equal than straight x squared end style
Solution 25

Solution 26

Solution 27

Solution 28

 

The given function f is continuous at x = begin mathsize 12px style straight pi end style, if f is defined at x = begin mathsize 12px style straight pi end style and if the value

of f at x = begin mathsize 12px style straight pi end style equals the limit of f at x = begin mathsize 12px style straight pi end style

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

 
Solution 34

Chapter 5 - Continuity and Differentiability Exercise Ex. 5.2

Solution 1

Then, (v ο u)(x) = v(u(x)) = v(x2 + 5) = sin (x2 + 5) = f(x)

 

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Chapter 5 - Continuity and Differentiability Exercise Ex. 5.3

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Chapter 5 - Continuity and Differentiability Exercise Ex. 5.4

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Chapter 5 - Continuity and Differentiability Exercise Ex. 5.5

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Chapter 5 - Continuity and Differentiability Exercise Ex. 5.6

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Chapter 5 - Continuity and Differentiability Exercise Ex. 5.7

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Chapter 5 - Continuity and Differentiability Exercise Ex. 5.8

Solution 1

 

Rolle's theorem states that there is a point c element of (-4, -2) such that f'(c) = 0

Solution 2

 

then there exists some c begin mathsize 12px style element of end style(a, b) such that f'(c) = 0

Let n be an integer such that n begin mathsize 12px style element of end style (5, 9).

Let n be an integer such that n begin mathsize 12px style element of end style (-2, 2).

Solution 3

Therefore, by the Mean Value Theorem, there exists c begin mathsize 12px style element of end style (-5, 5) such that

 

Solution 4

 

Mean Value Theorem states that there is a point c begin mathsize 12px style element of end style(1, 4) such that f'(c) = 1

Solution 5

Solution 6


Chapter 5 - Continuity and Differentiability Exercise Misc. Ex.

Solution 1

Solution 2

Solution 3

 

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

 

where sin x > cosx 

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

begin mathsize 12px style fraction numerator open square brackets 1 plus open parentheses begin display style dy over dx end style close parentheses squared close square brackets to the power of begin display style 3 over 2 end style end exponent over denominator fraction numerator straight d squared straight y over denominator dx squared end fraction end fraction equals fraction numerator begin display style open square brackets 1 plus begin display style open parentheses straight x minus straight a close parentheses squared over open parentheses straight y minus straight b close parentheses squared end style close square brackets to the power of 3 over 2 end exponent end style over denominator negative open square brackets begin display style fraction numerator open parentheses straight y minus straight b close parentheses squared plus open parentheses straight x minus straight a close parentheses squared over denominator open parentheses straight y minus straight b close parentheses cubed end fraction end style close square brackets end fraction equals fraction numerator begin display style open square brackets fraction numerator begin display style open parentheses straight y minus straight b close parentheses squared plus open parentheses straight x minus straight a close parentheses squared end style over denominator begin display style open parentheses straight y minus straight b close parentheses squared end style end fraction close square brackets to the power of 3 over 2 end exponent end style over denominator negative open square brackets fraction numerator open parentheses straight y minus straight b close parentheses squared plus open parentheses straight x minus straight a close parentheses squared over denominator open parentheses straight y minus straight b close parentheses cubed end fraction close square brackets end fraction end style
begin mathsize 12px style equals fraction numerator open square brackets begin display style straight c squared over open parentheses straight y minus straight b close parentheses squared end style close square brackets over denominator begin display style fraction numerator negative straight c squared over denominator open parentheses straight y minus straight b close parentheses cubed end fraction end style end fraction equals fraction numerator begin display style straight c cubed over open parentheses straight y minus straight b close parentheses cubed end style over denominator begin display style fraction numerator negative straight c squared over denominator open parentheses straight y minus straight b close parentheses cubed end fraction end style end fraction equals negative straight c space which space is space constant space and space is space independent space of space straight a space and space straight b end style
 
Hence, proved.
Solution 16

begin mathsize 12px style rightwards double arrow sin open parentheses straight a plus straight y minus straight y close parentheses dy over dx equals cos squared open parentheses straight a plus straight y close parentheses
rightwards double arrow dy over dx equals fraction numerator cos squared open parentheses straight a plus straight y close parentheses over denominator sina end fraction end style

 

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

 Yes.

Consider the function f(x)=|x-1|+|x-2|

Since we know that the modulus function is continuous everywhere, so there sum is also continuous

Therefore, function f is continuous everywhere

Now, let us check the differentiability of f(x) at x=1,2

At x=1

LHD =

 [Take x=1-h, h>0 such that h0 as x1-]

  

Now,

RHD =

 [Take x=1+h, h>0 such that h0 as x1+]

≠ LHD

Therefore, f is not differentiable at x=1.

 

At x=2

LHD =

 [Take x=2-h, h>0 such that h0 as x2-]

  

Now,

RHD =

 [Take x=2+h, h>0 such that h0 as x2+]

≠ LHD

Therefore, f is not differentiable at x=2.

Hence, f is not differentiable at exactly two points.

Solution 22

Solution 23