Chapter 5 : Continuity and Differentiability - Ncert Solutions for Class 12-science Maths CBSE

Your CBSE Class 12 syllabus for Maths consists of topics such as linear programming, vector quantities, determinants, etc. which lay the foundation for further education in science, engineering, management, etc. Studying differential equations will be useful for exploring subjects such as Physics, Biology, Chemistry, etc. where your knowledge can be applied for scientific investigations.

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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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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
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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
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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 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

 

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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

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Chapter 5 - Continuity and Differentiability Exercise Misc. Ex.

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where sin x > cosx 

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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

 

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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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