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

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

Solution 12

Thus, from the above observation, it can be concluded that x = 1 is the only point of discontinuity.

Solution 20

The given function is f(x) = x2 - sinx + 5

It is evident that f is defined at x = ∏

Solution 22

Therefore, cosecant is continuous except at x = np, n Z
Therefore, cotangent is continuous except at x = np, n Z

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 =

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)

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

Continuity and Differentiability Exercise Misc. Ex.

Solution 9

where sin x > cosx

Hence, proved.

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 h0 as x1-]

Now,

RHD =

[Take x=1+h, h>0 such that h0 as x1+]

≠ LHD

Therefore, f is not differentiable at x=1.

At x=2

LHD =

[Take x=2-h, h>0 such that h0 as x2-]

Now,

RHD =

[Take x=2+h, h>0 such that h0 as x2+]

≠ LHD

Therefore, f is not differentiable at x=2.

Hence, f is not differentiable at exactly two points.