NCERT Solutions for Class 12-science Maths Chapter 5 - Continuity and Differentiability
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
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Thus, from the above observation, it can be concluded that x = 1 is the only point of discontinuity.
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Solution 20
The given function is f(x) = x2 - sinx + 5
It is evident that f is defined at x = ∏
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Solution 22
Therefore, cosecant is continuous except at x = np, n 
Z
Z
Therefore, cotangent is continuous except at x = np, n 
Z
ZSolution 23
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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 =
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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)
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Chapter 5 - Continuity and Differentiability Exercise Ex. 5.3
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Chapter 5 - Continuity and Differentiability Exercise Ex. 5.4
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Chapter 5 - Continuity and Differentiability Exercise Ex. 5.5
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Chapter 5 - Continuity and Differentiability Exercise Ex. 5.6
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Chapter 5 - Continuity and Differentiability Exercise Ex. 5.7
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Chapter 5 - Continuity and Differentiability Exercise Ex. 5.8
Solution 1
Rolle's theorem states that there is a point c 
(-4, -2) such that f'(c) = 0
Solution 2
then there exists some c (a, b) such that f'(c) = 0
Let n be an integer such that n (5, 9).
Let n be an integer such that n (-2, 2).
Solution 3
Therefore, by the Mean Value Theorem, there exists c (-5, 5) such that
Solution 4
Mean Value Theorem states that there is a point c (1, 4) such that f'(c) = 1
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Solution 6
Chapter 5 - Continuity and Differentiability Exercise Misc. Ex.
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where sin x > cosx
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Hence, proved.
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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 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.
Solution 22
Solution 23
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