NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.6 Continuity and Differentiability

Class 12 Maths Chapter 5 Exercise 5.6 Continuity and Differentiability:- Chapter 5 of Class 12 Maths focuses on Continuity and Differentiability. This chapter builds upon the concepts of calculus introduced in previous classes and explores them in greater depth. Exercise 5.6 is an integral part of this chapter, where you will work on various problems to understand the practical applications of continuity and differentiability. Go through NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.6 Continuity and Differentiability in the article below. 

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NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.6 Continuity and Differentiability

Get the NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.6 Continuity and Differentiability below:- 

If x  and y are connected parametrically by the equations given in Exercise 1 to 5, without eliminating the parameter, find dy/dx

Question 1.  

Solution :

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

Question 2. x = a cosθ , y = b cosθ

Solution :
Given: x = a cosθ , y = b cos

Check out: NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.3

Question 3. x = sin t, y = cos 2t

Solution :
Given: x = sin t, y = cos 2t

Question 4. x = 4t, y = 4/t

Solution :
Given:x = 4t, y = 4/t

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

Question 5. x = cosθ – cos2θ, y = sinθ – sin2θ

Solution :
Given: x = cosθ – cos2θ, y = sinθ – sin2θ

If x and y are connected parametrically by the equations given in Exercises 6 to 10, without eliminating the parameter, find dy/dx 

Question 6. x = a (θ – sinθ), y = a(1+ cosθ)

Solution :
Given: x = a (θ – sinθ), y = a(1+ cosθ)

Question 7.

Solution :
Given:

Question 8.

Solution :
Given:

Question 9. x = a sec θ, y = b tan θ

Solution :
Given: x = a sec θ and y = b tan θ

Question 10. x = a (cos θ + θ sin θ), y = b (sin θ – θ  cosθ)

Solution :
Given: x = a (cos θ + θ sin θ) and y = b (sin θ – θ  cosθ)

Check out: Class 12th Sample Papers

Question 11. If

Solution :
Given:

 

Check out: Class 12th Question Banks

Class 12 Maths Chapter 5 Exercise 5.6 Continuity and Differentiability Summary 

Continuity

A function is said to be continuous if there are no breaks, jumps, or holes in its graph. Mathematically, a function f(x)f(x)f(x) is continuous at a point x=ax = ax=a if:

1. f(a)f(a)f(a) is defined.

2. The limit of f(x)f(x)f(x) as xxx approaches aaa exists.

3. The limit of f(x)f(x)f(x) as xxx approaches aaa is equal to f(a)f(a)f(a).

Differentiability

A function is differentiable at a point if it has a defined derivative at that point. Differentiability implies continuity, but a function can be continuous without being differentiable. Mathematically, a function f(x)f(x)f(x) is differentiable at x=ax = ax=a if:

1. The derivative f′(a)f'(a)f′(a) exists.

2. The derivative is defined as the limit: f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{{h \to 0}} \frac{{f(a+h) - f(a)}}{h}f′(a)=h→0lim​hf(a+h)−f(a)​

Class 12 Maths Chapter 5 Exercise 5.6 Continuity and Differentiability FAQs

Q1. Can a function be continuous but not differentiable?

Ans. Yes, a function can be continuous at a point but not differentiable there. An example is f(x)=∣x∣f(x) = |x|f(x)=∣x∣ at x=0x = 0x=0, which is continuous but not differentiable at x=0x = 0x=0.

Q2. What is the relationship between continuity and differentiability?

Ans. Differentiability implies continuity. If a function is differentiable at a point, it must be continuous there. However, continuity does not imply differentiability.

Q3. How do you check the continuity of a function at a point?

Ans. To check the continuity at x=ax = ax=a:

1. Ensure f(a)f(a)f(a) is defined.

2. Compute the limit of f(x)f(x)f(x) as xxx approaches aaa.

3. Verify that the limit equals f(a)f(a)f(a).

Q4. How do you determine if a function is differentiable at a point?

Ans. To determine differentiability at x=ax = ax=a:

1. Check that the function is continuous at aaa.

2. Compute the derivative f′(a)f'(a)f′(a) using the limit definition: f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a) = \lim_{{h \to 0}} \frac{{f(a+h) - f(a)}}{h}f′(a)=h→0lim​hf(a+h)−f(a)​

Q5. What are some common techniques for finding derivatives?

Ans. Common techniques include:

• Power rule: ddx[xn]=nxn−1\frac{d}{dx}[x^n] = nx^{n-1}dxd​[xn]=nxn−1

• Product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'(uv)′=u′v+uv′

• Quotient rule: (uv)′=u′v−uv′v2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}(vu​)′=v2u′v−uv′​

• Chain rule: (f(g(x)))′=f′(g(x))⋅g′(x)(f(g(x)))' = f'(g(x)) \cdot g'(x)(f(g(x)))′=f′(g(x))⋅g′(x)