Class 12-science NCERT Solutions Maths Chapter 5 - Continuity and Differentiability
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
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/9af4d0edef3d96458dc0dda4a36ba0f15d52a3dab87727.82984083continuity.png)
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
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/6b70be60fb2d40267d577aecc6e24cb05d53e0e76f92b8.85559924continuity3.png)
![begin mathsize 12px style element of end style](https://images.topperlearning.com/topper/tinymce/cache/0ca6d0224f106e0e0d3b944284e1787c.png)
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/d28071bb2181310bb38fedf7e982d4525d53e1e0b83cf6.47324824continuity4.png)
![element of](https://images.topperlearning.com/topper/tinymce/cache/bbd43e05551f5ee0bbf5c52658980071.png)
Solution 23
Solution 24
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/7bda54e071edb403edc37ef6a6eb22f15d53e58656ed24.44048424continuity5.png)
![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](https://images.topperlearning.com/topper/tinymce/cache/d296ddbec3c87879a856130abf68abd8.png)
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/ab90487f943a6d0e61919065918bfe505d53e5c4ce2494.00608612continuity6.png)
Solution 25
Solution 26
Solution 27
Solution 28
The given function f is continuous at x = , if f is defined at x =
and if the value
of f at x = equals the limit of f at x =
Solution 29
Solution 30
Solution 31
Solution 32
Solution 33
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/8ea376da7e155504cc3f32ed412898265d53f1ab2ccca5.55586757continuity10.png)
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/e620b4db09541859ee1bcb4cbecec2215d53f1c06615a8.45509587continuity11.png)
Solution 34
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
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/1dd9c8f9fe22d5e5e35686ca1ab685f65d54018ed8a4d4.81550047continuity14.png)
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
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
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
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
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
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
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Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
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 h→0 as x→1-]
Now,
RHD =
[Take x=1+h, h>0 such that h→0 as x→1+]
≠ LHD
Therefore, f is not differentiable at x=1.
At x=2
LHD =
[Take x=2-h, h>0 such that h→0 as x→2-]
Now,
RHD =
[Take x=2+h, h>0 such that h→0 as x→2+]
≠ LHD
Therefore, f is not differentiable at x=2.
Hence, f is not differentiable at exactly two points.