NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 Continuity and Differentiability

Class 12 Maths Chapter 5 Exercise 5.7 Continuity and Differentiability:- Class 12 Maths Chapter 5, "Continuity and Differentiability," covers essential concepts that are foundational in calculus. Exercise 5.7 focuses on applying these principles to various functions, examining when and how a function is continuous and differentiable. This exercise challenges students to check functions for points of continuity, use differentiation techniques such as the chain rule, and apply methods like implicit and logarithmic differentiation. 

Understanding these concepts is crucial for higher-level calculus and helps students gain a deeper insight into the behaviour of functions, enhancing their problem-solving skills. Get the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 Continuity and Differentiability below. 

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NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 Continuity and Differentiability

Check out the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 Continuity and Differentiability below:-

Find the second order derivatives of the functions given in Exercises 1 to 5.

Question 1. x2 + 3x + 2

Solution :
Let y = x2 + 3x + 2

Question 2. x20

Solution :
Let  x20

Question 3. x.cos x

Solution :
Let x. cos x

Check out: NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.1

Question 4. log x

Solution :
Let log x

Check out: NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.2

Question 5. x3 log x

Solution :
Let x3 log x

Find the second order derivatives of the functions given in Exercises 6 to 10.

Question 6. ex sin 5x

Solution :
Let ex sin 5x

Check out: NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.5

Question 7. e6x cos x

Solution :
Let e6x cos x

Check out: NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.6

Question8. tan-1 x

Solution :
Let tan-1 x

Question9. log (log x)

Solution :

Let log (log x)

Question10. sin(log x)

Solution :
Let sin (log x)

Question11. If y = 5 cos x –  3 sin x prove that

Solution :
Let y = 5 cos x –  3 sin x

Question12. If y = cos-1 x Find in terms of y alone.

Solution :
Given: y = cos-1 x

Question13. If y = 3 cos (log x) + 4 sin (log x), show that x2y2 + xy1 + y = 0

Solution :
Given: y = 3 cos (log x) + 4 sin (log x)

Hence proved.

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Question14. If y = Aemx + Benx show that

Solution :
Given: y = Aemx + Benx

Hence proved.

Question15. If  500e7x  + 600e-7x show that  

Solution :
Given: 500e7x  + 600e-7x

Hence proved.

Question16. If  ey (x + 1) = 1, show that

Solution :
Given: ey (x + 1) = 1

Taking log on both sides,

Differentiating this relationship with respect to x, we obtain

Hence proved.

Question17. If y = (tan-1 x)2 show that (x2 + 1)2y2 + 2(x2 + 1)y1 = 2

Solution :
Given:  y = (tan-1 x)2

Hence proved.

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Class 12 Maths Chapter 5 Exercise 5.7 Continuity and Differentiability Summary 

In Exercise 5.7, the focus is on advanced concepts of continuity and differentiability. Here’s a summary to help you understand the key points:

Key Concepts:

1. Continuity at a Point:

• A function f(x)f(x)f(x) is continuous at x=cx = cx=c if lim⁡x→cf(x)=f(c)\lim_{{x \to c}} f(x) = f(c)limx→c​f(x)=f(c).

• This implies there are no breaks, jumps, or holes in the graph of the function at x=cx = cx=c.

2. Differentiability at a Point:

• A function f(x)f(x)f(x) is differentiable at x=cx = cx=c if lim⁡h→0f(c+h)−f(c)h\lim_{{h \to 0}} \frac{f(c+h) - f(c)}{h}limh→0​hf(c+h)−f(c)​ exists.

• Differentiability implies continuity, but the reverse is not always true.

3. Chain Rule:

• If y=f(u)y = f(u)y=f(u) and u=g(x)u = g(x)u=g(x), then dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}dxdy​=dudy​⋅dxdu​.

• This rule is essential for finding the derivative of composite functions.

4. Implicit Differentiation:

• Used when the function is not explicitly solved for one variable in terms of the other.

• For example, if xxx and yyy are related by an equation, differentiate both sides with respect to xxx and solve for dydx\frac{dy}{dx}dxdy​.

5. Logarithmic Differentiation:

• Useful for functions of the form y=[f(x)]g(x)y = [f(x)]^{g(x)}y=[f(x)]g(x).

• Take the natural logarithm of both sides, then differentiate.

Class 12 Maths Chapter 5 Exercise 5.7 Continuity and Differentiability FAQs

1. What is the Intermediate Value Theorem?

    Ans. The Intermediate Value Theorem states that if a function f(x)f(x)f(x) is continuous on the interval [a,b][a, b][a,b] and NNN is any            number between f(a)f(a)f(a) and f(b)f(b)f(b), then there exists a point ccc in [a,b][a, b][a,b] such that f(c)=Nf(c) = Nf(c)=N.

2. What is Rolle’s Theorem?

    Ans. Rolle’s Theorem states that if a function f(x)f(x)f(x) is continuous on [a,b][a, b][a,b], differentiable on (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.

3. 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 [a,b][a, b][a,b] and differentiable on (a,b)(a, b)(a,b),       then there exists at least one point ccc in (a,b)(a, b)(a,b) such that f′(c)=f(b)−f(a)b−af'(c) = \frac{f(b) - f(a)}{b - a}f′(c)=b−af(b)−f(a)​.

4. How do you prove that a function is continuous at a point?

    Ans. To prove that a function is continuous at a point x=cx = cx=c, show that f(c)f(c)f(c) is defined, lim⁡x→cf(x)\lim_{x \to c} f(x)limx→c​f(x)  exists, and lim⁡x→cf(x)=f(c)\lim_{x \to c} f(x) = f(c)limx→c​f(x)=f(c).

5. How do you determine points of discontinuity?

    Ans. To determine points of discontinuity, identify where the function fails to meet the criteria for continuity. Check for points where the        function is not defined, where the limit does not exist, or where the limit does not equal the function’s value at that point.