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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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).

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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.
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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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