NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.5 Continuity and Differentiability

Author at PW
April 23, 2025

Class 12 Maths Chapter 5 Exercise 5.5 Continuity and Differentiability:- In calculus, continuity and differentiability refer to the core of how functions behave and change. Continuity describes how smooth a function is at every point in the domain; it does not have breaks or jumps in its graph. If a function has a limit at a point that equates to its value, then that will be considered continuous.

On the other hand, differentiability extends the concept of continuity and includes the rate of change of the function. A function is itself described as differentiable at a point if it contains a derivative at that point—it means the tangent line, or equivalently, the rate of change at that particular point, is well-defined. Get NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.5 Continuity and Differentiability from the below article.

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NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.5 Continuity and Differentiability

Here are NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.5 Continuity and Differentiability:-

Question 1. Differentiate the function with respect to x.

cos x.cos 2x.cos3x

Solution :
Let y = cos x.cos 2x.cos3x

Taking logarithm on both the sides, we obtain

chapter 5-Continuity & Differentiability Exercise 5.5

Question 2. Differentiate the function with respect to x.

chapter 5-Continuity & Differentiability Exercise 5.5

Solution :
Let y = chapter 5-Continuity & Differentiability Exercise 5.5

Taking logarithm on both the sides, we obtain

NCERT Solutions class 12 Continuity & Differentiability/d328c07.gif

Question 3. Differentiate the function with respect to x.

/chapter 5-Continuity & Differentiability Exercise 5.5/783d3a39.gif

Solution :
Let, y = /chapter 5-Continuity & Differentiability Exercise 5.5/783d3a39.gif

Taking logarithm on both the sides, we obtain

log y = cos x .log(log x)

Differentiating both sides with respect to x, we obtain

/NCERT Solutions class 12 Continuity & Differentiability/7df01e.gif

Question 4. Differentiate the function with respect to x.

chapter 5-Continuity & Differentiability Exercise 5.5/71de10.gif

Solution :
chapter 5-Continuity & Differentiability Exercise 5.5/m2a70b6e1.gif

Question 5. Differentiate the function with respect to x.

NCERT Solutions class 12 Continuity & Differentiability/73da24a.gif

Solution :
chapter 5-Continuity & Differentiability Exercise 5.5/m23f8f68.gif

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Question 6. Differentiate the function with respect to x.

chapter 5-Continuity & Differentiability Exercise 5.5/6db13fca.gif

Solution :
chapter 5-Continuity & Differentiability Exercise 5.5/5ce4fc8c.gif

Differentiating both sides with respect to x, we obtain

chapter 5-Continuity & Differentiability Exercise 5.5/m67161841.gif

Differentiating both sides with respect to x, we obtain

chapter 5-Continuity & Differentiability Exercise 5.5/m291a6d50.gif

Question 7. Differentiate the function with respect to x.

NCERT Solutions class 12 Continuity & Differentiability/b6d35e3.gif

Solution :
NCERT Solutions class 12 Continuity & Differentiability/1ae62d8.gif

Differentiating both sides with respect to x, we obtain

NCERT Solutions class 12 Continuity & Differentiability/bd5e52e.gif

Differentiating both sides with respect to x, we obtain

chapter 5-Continuity & Differentiability Exercise 5.5/5324cf14.gif

Question 8. Differentiate the function with respect to x.

chapter 5-Continuity & Differentiability Exercise 5.5/1a8d3394.gif

Solution :
chapter 5-Continuity & Differentiability Exercise 5.5/1596bc50.gif

Differentiating both sides with respect to x, we obtain

chapter 5-Continuity & Differentiability Exercise 5.5/m220de476.gif

chapter 5-Continuity & Differentiability Exercise 5.5/66e67501.gif

Question 9. Differentiate the function with respect to x.

chapter 5-Continuity & Differentiability Exercise 5.5/263eebdd.gif

Solution :
Let, y = chapter 5-Continuity & Differentiability Exercise 5.5/263eebdd.gif

chapter 5-Continuity & Differentiability Exercise 5.5/555a94f9.gif

Differentiating both sides with respect to x, we obtain

chapter 5-Continuity & Differentiability Exercise 5.5/me8cba15.gif

Differentiating both sides with respect to x, we obtain

chapter 5-Continuity & Differentiability Exercise 5.5/m61700c16.gif

Question 10. Differentiate the function with respect to x.

NCERT Solutions class 12 Continuity & Differentiability/7b64539.gif

Solution :
Let, y = NCERT Solutions class 12 Continuity & Differentiability/7b64539.gif

chapter 5-Continuity & Differentiability Exercise 5.5/193f9bb9.gif

Differentiating both sides with respect to x, we obtain

chapter 5-Continuity & Differentiability Exercise 5.5/771ab082.gif

Differentiating both sides with respect to x, we obtain

NCERT Solutions class 12 Continuity & Differentiability/adb796c.gif

Question 11. Differentiate the function with respect to x.

chapter 5-Continuity & Differentiability Exercise 5.5/3e605c11.gif

Solution :
Let, y = chapter 5-Continuity & Differentiability Exercise 5.5/3e605c11.gif

chapter 5-Continuity & Differentiability Exercise 5.5/m59e0dc54.gif

Differentiating both sides with respect to x, we obtain

chapter 5-Continuity & Differentiability Exercise 5.5/33ac2b2.gif

Differentiating both sides with respect to x, we obtain

chapter 5-Continuity & Differentiability Exercise 5.5/760c0f18.gif

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Question 12. Find dy/dx of function.

xy+ yx = 1

Solution :
The given function is xy + yx = 1

Let xy = u and yx = v

Then, the function becomes u + v = 1

∴du/dx + dv/dx = 1

Differentiating both sides with respect to x, we obtain

chapter 5-Continuity & Differentiability Exercise 5.5/m2c850875.gif

Differentiating both sides with respect to x, we obtain

NCERT Solutions class 12 Continuity & Differentiability/4966692.gif

Question 13. Find dy/dx of function.

yx = xy

Solution :
The given function is yx = xy

Taking logarithm on both the sides, we obtain

x log y = y log x

Differentiating both sides with respect to x, we obtain

chapter 5-Continuity & Differentiability Exercise 5.5/1c72ea1e.gif

Question 14. Find dy/dx of function.

(cos x)y = (cos y)x

Solution :
The given function is (cos x)y = (cos y)x

Taking logarithm on both the sides, we obtain

y log cos x = x log cos y

Differentiating both sides, we obtain

chapter 5-Continuity & Differentiability Exercise 5.5

Question 15. Find dy/dx of function.

xy = e(x-y)

Solution :
The given function is xy = e(x-y)

Taking logarithm on both the sides, we obtain

log(xy) = log(e(x-y))

Differentiating both sides with respect to x, we obtain

chapter 5-Continuity & Differentiability Exercise 5.5

Question 16. Find the derivative of the function given by NCERT Solutions class 12 Continuity & Differentiability/709a18f.gif and hence find f'(1)

Solution :
The given relationship is NCERT Solutions class 12 Continuity & Differentiability/709a18f.gif

Taking logarithm on both the sides, we obtain

chapter 5-Continuity & Differentiability Exercise 5.5

Differentiating both sides with respect to x, we obtain

chapter 5-Continuity & Differentiability Exercise 5.5

Question 17. Differentiate chapter 5-Continuity & Differentiability Exercise 5.5 in three ways mentioned below

(i) By using product rule.

(ii) By expanding the product to obtain a single polynomial.

(iii By logarithmic differentiation.

Do they all give the same answer?

Solution :
Let, y = chapter 5-Continuity & Differentiability Exercise 5.5

(i)

NCERT Solutions class 12 Continuity & Differentiability/3e84036.gif

(ii)

chapter 5-Continuity & Differentiability Exercise 5.5

(iii) y = chapter 5-Continuity & Differentiability Exercise 5.5

Taking logarithm on both the sides, we obtain

chapter 5-Continuity & Differentiability Exercise 5.5

Differentiating both sides with respect to x, we obtain

chapter 5-Continuity & Differentiability Exercise 5.5

From the above three observations, it can be concluded that all the results of dy/dx are same.

Question 18. If u, v and w are functions of x, then show that

chapter 5-Continuity & Differentiability Exercise 5.5

in two ways-first by repeated application of product rule, second by logarithmic differentiation. 

Solution :
Let y = u.v.w = u(v.w)

By applying product rule, we obtain

chapter 5-Continuity & Differentiability Exercise 5.5

By taking logarithm on both sides of the equation y = u.v.w, we obtain

log y = log u + log v + log w

Differentiating both sides with respect to x, we obtain

chapter 5-Continuity & Differentiability Exercise 5.5

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Class 12 Maths Chapter 5 Exercise 5.5 Continuity and Differentiability Summary

Continuity of a function f(x) at a point 'a' in its domain:

  • A function is considered continuous at 'a' if three conditions hold:

    1. f(a) exists (i.e., the function has a finite output at 'a').

    2. The limit of f(x) as x approaches 'a' exists (i.e., the function approaches a specific value as x gets arbitrarily close to 'a' from either side).

    3. The two limits are equal. In other words, lim f(x) = f(a).

Differentiability of a function f(x) at a point 'a' in its domain:

  • A function is differentiable at 'a' only if it's also continuous at 'a'.

  • The additional (shart) or condition for differentiability is that the limit of the difference quotient exists as x approaches 'a'. The difference quotient is: lim [f(x) - f(a)] / (x - a)

  • If this limit exists, it represents the derivative of the function at 'a' and is denoted by f'(a).

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Class 12 Maths Chapter 5 Exercise 5.5 Continuity and Differentiability FAQs

Q1. What is continuity of a function?

Ans. Continuity refers to the absence of breaks, jumps, or holes in the graph of a function. Formally, a function f(x)f(x)f(x) is continuous at a point x=ax = ax=a if lim⁡x→af(x)=f(a)\lim_{x \to a} f(x) = f(a)limx→a​f(x)=f(a).

Q2. How do you determine if a function is continuous over an interval?

Ans. Check if the function is continuous at each point within the interval and if the limits from both sides approach the same value at the endpoints of the interval.

Q3. What does it mean for a function to be differentiable at a point?

Ans. A function f(x)f(x)f(x) is differentiable at a point x=ax = ax=a if its derivative f′(a)f'(a)f′(a) exists. Geometrically, this means the function has a well-defined tangent line at x=ax = ax=a.

Q4. What is Rolle's Theorem?

Ans. Rolle's Theorem states that if a function f(x)f(x)f(x) is continuous on the closed interval [a,b][a, b][a,b], differentiable on the open interval (a,b)(a, b)(a,b), and f(a)=f(b)f(a) = f(b)f(a)=f(b), then there exists at least one point ccc in (a,b)(a, b)(a,b) where f′(c)=0f'(c) = 0f′(c)=0.

Q5. What is the Mean Value Theorem?

Ans. The Mean Value Theorem states that if a function f(x)f(x)f(x) is continuous on the closed interval [a,b][a, b][a,b] and differentiable on the open interval (a,b)(a, b)(a,b), then there exists at least one point ccc in (a,b)(a, b)(a,b) where f′(c)=f(b)−f(a)b−af'(c) = \frac{f(b) - f(a)}{b - a}f′(c)=b−af(b)−f(a)​.

 

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