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

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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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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, nbegin mathsize 12px style element of end styleZ

Therefore, cotangent is continuous except at x = np, nelement ofZ

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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

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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 celement of (-4, -2) such that f'(c) = 0

Solution 2

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

Let n be an integer such that nbegin 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 cbegin mathsize 12px style element of end style (-5, 5) such that

Solution 4

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

Solution 5

Chapter 5 - Continuity and Differentiability Exercise Misc. Ex.

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where sin x > cosx

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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

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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.

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