# Class 12-commerce NCERT Solutions Maths Chapter 5 - Continuity and Differentiability

## Continuity and Differentiability Exercise Ex. 5.1

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

At x = 0, f(0) = 5 × 0 - 3 = -3

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The given function is f(x) = x^{2} - sinx + 5

It is evident that f is defined at x = ∏

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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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## Continuity and Differentiability Exercise Ex. 5.2

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Then, (v ο u)(x) = v(u(x)) = v(x^{2} + 5) = sin (x^{2} + 5) = f(x)

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## Continuity and Differentiability Exercise Ex. 5.3

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## Continuity and Differentiability Exercise Ex. 5.4

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## Continuity and Differentiability Exercise Ex. 5.5

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## Continuity and Differentiability Exercise Ex. 5.6

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## Continuity and Differentiability Exercise Ex. 5.7

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## Continuity and Differentiability Exercise Ex. 5.8

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

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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).

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Therefore, by the Mean Value Theorem, there exists c (-5, 5) such that

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Mean Value Theorem states that there is a point c (1, 4) such that f'(c) = 1

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## Continuity and Differentiability Exercise Misc. Ex.

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where sin x > cosx

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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.

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