NCERT Solutions for Class 12 Maths Chapter 5: Continuity and Differentiability - Exercise 5.6

Exercise 5.6  

1. Determine $\mathbf{\frac{dy}{dx}}$ from the equations x=2at2 , y=at4, without eliminating the  parameter t, where a,bare constants. 

Ans: The given equations are  

x=2at2 …… (1) 

and y=at4 …… (2) 

Then, differentiating both sides of the equation (1) with respect to t gives

$\frac{dx}{dy} = \frac{d}{dt}(2at^{2})= 2a \times \frac{d}{dt}(t^2)= 2a \times 2t = 4at$.......(3)

Also, differentiating both sides of the equation (2) with respect to t gives

$\frac{dy}{dt} = \frac{d}{dt}(2at^{4})= a \times 4 \times \frac{d}{dt}(t^4)= a \times 4 \times t^3 = 4at^3$.........(4)

Now, dividing the equations (4) by (3) gives 

$\frac{dy}{dx}= \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})}= \frac{4at^3}{4at} = t^2$

Hence $\frac{dy}{dx}=t^2$.

2. Determine $\mathbf{\frac{dy}{dx}}$ from the equations x=acosθ, y=bcosθ, without eliminating  the parameter θ, where a,bare constants. 

Ans: The given equations are  

x=acosθ …… (1) 

and y=bcosθ …… (2) 

Then, differentiating both sides of the equation (1) with respect to θ gives

$\frac{dx}{d\theta }= \frac{d}{d\theta }= (a\cos \theta ) =a(-\sin \theta ) = -a\sin \theta$.....(3)

Also, differentiating both sides of the equation (1) with respect to θgives

$\frac{dy}{d\theta }= \frac{d}{d\theta }= (b\cos \theta ) =b(-\sin \theta ) = -b\sin \theta$......(4)

Therefore, dividing the equation (4) by (3) gives 

$\frac{dy}{dx} = \frac{(\frac{dy}{d\theta })}{(\frac{dx}{d\theta })}= \frac{-b \sin\theta }{-a\sin\theta }= \frac{b}{a}$

Hence, $\frac{dy}{dx} =  \frac{b}{a}$

3. Determine $\mathbf{\frac{dy}{dx}}$  from the equations x=sint, y=cos2t without eliminating the  parameter t. 

Ans: The given equations are 

x=sint …… (1) 

and y=cos2t …… (2) 

Then, differentiating both sides of the equation (1) with respect to t gives 

$\frac{dx}{dt} = \frac{d}{dt}(\sin\theta ) = \cos\theta ......(3)$

Also, differentiating both sides of the equation (2) with respect to t gives

$\frac{dy}{dt} = \frac{d}{dt}(\cos2\theta ) = \sin 2\theta  \times \frac{d}{dt}(2t)= -2\sin2t......(4)$

Therefore,  by dividing the equation (4) by (3) gives 

$\frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})} = \frac{-2 \sin 2t}{cos t} = \frac{-2 \times 2 \sin t \cos t}{\cos t} = -4 \sin t$

Hence, $\frac{dy}{dx} =   -4 \sin t$

4. Determine $\mathbf{\frac{dy}{dx}}$   from the equations x=4t,  y = $\mathbf{\frac{4}{t}}$ without eliminating the  parameter t . 

Ans: The given equations are 

x=4t …… (1) 

 y = $\frac{4}{t}$ ……(2)

Now, differentiating both sides of the equation (1) with respect to t gives 

$\frac{dy}{dx} =\frac{d}{dt}(4t)=4......(3)$

Also, differentiating both sides of the equation (2) with respect to t gives 

$\frac{dy}{dt}= \frac{d}{dt}(\frac{4}{t}) = 4 \times \frac{d}{dt}\frac{1}{t}= 4 \times \frac{-1}{t^2}= \frac{-4}{t^2}.......(4)$

Therefore, dividing the equation (4) by (3) gives 

$\frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})} = \frac{(\frac{-4}{t^2})}{4} = \frac{-1}{t^2} $

Hence, $\frac{dy}{dx} = \frac{-1}{t^2} $

5. Determine $\mathbf{\frac{dy}{dx}}$  from the equations x=cosθ-cos2θ, y=sinθ-sin2θ, without  eliminating the parameter θ . 

Ans: The given equations are 

x=cosθ-cos2θ …… (1) 

and y=sinθ-sin2θ …… (2) 

Then, differentiating both sides of the equation (1) with respect to θ gives 

$\frac{dx}{d\theta } = \frac{d}{d\theta }(\cos \theta - \cos 2\theta ) = \frac{d}{dt}(\cos \theta )- \frac{d}{d\theta }(cos 2\theta )= -\sin \theta (-2 \sin 2\theta ) = 2 \sin 2\theta - \sin\theta ………..(3)$

Also, differentiating both sides of the equation (2) with respect to θ gives 

$\frac{dy}{d\theta } = \frac{d}{d\theta }(\sin \theta - \sin 2\theta ) = \frac{d}{d\theta }(\sin \theta )- \frac{d}{d\theta }(sin 2\theta )= \cos \theta - 2 \cos\theta .......(4)$

Therefore, dividing the equation (4) by (3) gives

$\frac{dy}{dx} = \frac{(\frac{dy}{d\theta })}{(\frac{dx}{d\theta })}= \frac{\cos\theta - 2 \cos\theta  }{2 \sin 2\theta - \sin\theta }$

Hence, $\frac{dy}{dx} =  \frac{\cos\theta - 2 \cos\theta  }{2 \sin 2\theta - \sin\theta }$

6. Determine $\mathbf{\frac{dy}{dx}}$  from the equations x=a(θ-sinθ), y=a(1+cosθ), without  eliminating the parameter θ, where a,bare constants. 

Ans: The given equations are 

x=a(θ-sinθ) …… (1) 

and y=a(1+cosθ) …… (2) 

Then, differentiating both sides of the equation (1) with respect to θ gives

$\frac{dx}{d\theta }= a[\frac{d}{d\theta }(\theta ) - \frac{d}{d\theta }(\sin \theta )] = a(1 - \cos \theta )$

Also, differentiating both sides of the equation (2) with respect to θ gives 

$\frac{dy}{d\theta }= a[\frac{d}{d\theta }(1) - \frac{d}{d\theta }(\cos \theta )] = a[0 + (- \sin \theta )] = -a \sin \theta ........(4)$

Therefore, by dividing the equation (4) by (3) gives 

$\frac{dy}{dx} = \frac{(\frac{dy}{d\theta })}{(\frac{dx}{d\theta })}= \frac{-a \sin \theta }{a(1- \cos\theta )}=\frac{-2 \sin\frac{\theta }{2} \cos \frac{\theta }{2}}{2 \sin ^2\frac{\theta }{2}}= \frac{-\cos\frac{\theta }{2}}{\sin \frac{\theta }{2}}= -\cos\frac{\theta }{2}$

Hence, $\frac{dy}{dx} =  -\cos\frac{\theta }{2}$

7. Determine  $\mathbf{\frac{dy}{dx}}$  from the equations  $-\frac{sin^3 t}{\sqrt{cos 2t}} , y = \frac{cos^3 t}{\sqrt{cos 2t}}$  without   eliminating the parameter t . 

Ans: The given equations are, 

$x = - \frac{\sin^3 t}{\sqrt{\cos 2t}} ......... (1)$

$y = - \frac{\cos^3 t}{\sqrt{\cos 2t}} ......... (2)$

Then, differentiating both sides of the equation (1) with respect to t gives 

$\begin{aligned}&\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left[\frac{\sin ^{3} \mathrm{t}}{\sqrt{\cos 2 \mathrm{t}}}\right] \\ \\&=\frac{\sqrt{\cos 2 \mathrm{t}}-\frac{\mathrm{d}}{\mathrm{dt}}\left(\sin ^{3} \mathrm{t}\right)-\sin ^{3} \mathrm{t} \times \frac{\mathrm{d}}{\mathrm{dt}} \sqrt{\cos 2 \mathrm{t}}}{\cos 2 \mathrm{t}} \\  \\&=\frac{\sqrt{\cos 2 \mathrm{t}} \times 3 \sin ^{2} \mathrm{t} \times \frac{\mathrm{d}}{\mathrm{dt}}(\sin \mathrm{t})-\sin ^{3} \mathrm{t} \times \frac{1}{2 \sqrt{\cos 2 \mathrm{t}}} \times \frac{\mathrm{d}}{\mathrm{dt}}(\cos 2 \mathrm{t})}{\cos 2 \mathrm{t}} \\ \\&=\frac{3 \sqrt{\cos 2 \mathrm{t}} \times \sin ^{2} \mathrm{t} \cos \mathrm{t}-\frac{\sin ^{3} \mathrm{t}}{2 \sqrt{\cos 2 \mathrm{t}}} \times(-2 \sin 2 \mathrm{t})}{\cos 2 \mathrm{t} \sqrt{\cos 2 \mathrm{t}}}\end{aligned}$

Also, differentiating both sides of the equation (2) with respect to $t$ gives 

$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{3 \cos 2 \mathrm{t} \sin ^{2} t \cos t+\sin ^{2} t \sin 2 \mathrm{t}}{\cos 2 \mathrm{t} \sqrt{\cos 2 \mathrm{t}}} . \ldots \ldots$ (3)$

$\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left[\frac{\cos ^{3} \mathrm{t}}{\sqrt{\cos 2 \mathrm{t}}}\right]$

$\begin{aligned}&=\frac{\sqrt{\cos 2 \mathrm{t}} \times \frac{\mathrm{d}}{\mathrm{dt}}\left(\cos ^{3} \mathrm{t}\right)-\cos ^{3} \mathrm{t} \times \frac{\mathrm{d}}{\mathrm{dt}}(\sqrt{\cos 2 \mathrm{t}})}{\cos 2 \mathrm{t}} \\ \\&=\frac{3 \sqrt{\cos 2 \mathrm{t} \cos ^{2} \mathrm{t}(-\sin t)-\cos ^{3} t \times \frac{1}{2(\sqrt{\cos 2 t})}} \times \frac{\mathrm{d}}{\mathrm{dt}}(\cos 2 \mathrm{t})}{\cos 2 \mathrm{t}} \\&\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{-3 \cos 2 \mathrm{t} \times \cos ^{2} \mathrm{t} \times \sin t+\cos ^{3} \mathrm{t} \sin 2 \mathrm{t}}{\cos 2 \mathrm{t} \times \sqrt{\cos 2 \mathrm{t}}} \ldots \ldots \text { (4) }\end{aligned}$

Thus, dividing the equation (4) by the equation (3) gives

$\frac{d y}{d x}=\frac{\left(\frac{d x}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{-3 \cos 2 t \times \cos ^{2} t \times \sin t+\cos ^{3} t \sin 2 t}{3 \cos 2 t \cos t \sin ^{2} t+\sin ^{3} t \sin 2 t}$

$\begin{aligned}&=\frac{\sin t \cos t\left[-3 \cos 2 t \times \cos t+2 \cos ^{3} t\right]}{\sin t \cos t\left[3 \cos 2 t \sin t+2 \sin ^{3} t\right]}  \\ \\&=\frac{\left[-3\left(2 \cos ^{2} t-1\right) \cos t+2 \cos ^{3} t\right]}{\left[3\left(1-2 \sin ^{3} t\right) \sin t+2 \sin ^{3} t\right]} \quad\left[\begin{array}{l}\cos 2 t=\left(2 \cos ^{2} t-1\right) \\\cos 2 t=\left(1-2 \sin ^{2} t\right)\end{array}\right] \\ \\&=\frac{-4 \cos ^{3} t+3 \cos t}{3 \sin t-4 \sin ^{3} t} \\ \\&=\frac{-\cos 3 t}{\sin 3 t} \\  \\&\text { Hence, } \frac{d y}{d x}=-\cot 3 t .\end{aligned}$

8. Determine $\mathbf{\frac{d y}{d x}}$ from the parametric equations $\mathbf{x=a\left(\operatorname{cost}+\log \tan \frac{t}{2}\right), y=a \sin t}$ , without eliminating the parameter $\mathbf{t}$.

Ans: The given equations are

$x=a\left(\cos t+\log \tan \frac{t}{2}\right) \ldots \ldots$ (1)

and $y=a \sin t \quad \ldots \ldots(2)$

Then, differentiating both sides of the equation (1) with respect to $t$ gives $\frac{d x}{d t}=a \times\left[\frac{d}{d \theta}(\operatorname{cost})+\frac{d}{d \theta}\left(\log \tan \frac{t}{2}\right)\right]$

$\begin{aligned}&=\mathrm{a}\left[-\operatorname{sint}+\frac{1}{\tan\frac{\mathrm{t}}{2}} \times \frac{\mathrm{d}}{\mathrm{dt}}\left(\tan \frac{\mathrm{t}}{2}\right)\right] \\&=\mathrm{a}\left[-\sin \mathrm{t}+\cot \frac{\mathrm{t}}{2} \times \sec ^{2} \frac{\mathrm{t}}{2}\times\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{t}}{2}\right)\right]\\&=\mathrm{a}\left[-\operatorname{sint}+\frac{\cos \frac{\mathrm{t}}{2}}{\sin \frac{\mathrm{t}}{2}} \times \frac{1}{\cos ^{2} \frac{\mathrm{t}}{2}} \times \frac{1}{2}\right)\end{aligned}$

$\begin{aligned}&=a\left(-\sin t+\frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}}\right) \\&=a\left(-\sin t+\frac{1}{\sin t}\right) \\&=a\left(\frac{-\sin ^{2} t+1}{\sin t}\right) \\&\text { Therefore, } \frac{d x}{d t}=a \frac{\cos ^{2} t}{\sin t}\end{aligned}$

Also, differentiating both sides of the equation (2) with respect to $t$ gives $\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{a} \frac{\mathrm{d}}{\mathrm{dt}}(\sin t)=\mathrm{acost} \ldots \ldots(4)$

Thus, dividing the equation (4) by the equation (3) gives

$\frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{a \cos t}{\left(a \frac{\cos ^{2} t}{\sin t}\right)}=\frac{\sin t}{\cos t}=\operatorname{tant}$

Hence, $\frac{d y}{d x}=\tan t$.

9. Determine $\mathbf{\frac{d y}{d x}}$ from the parametric equations $\mathbf{x=a s e c \theta, y=b \tan \theta}$, without eliminating the parameter $\theta$, where $\mathbf{a}, \mathbf{b}$ are constants.

Ans: The given equations are

$\mathrm{x}=\operatorname{asec} \quad \ldots \ldots$ (1)

and $y=\tan \theta \quad \ldots \ldots$ (2)

Then, differentiating both sides of the equation (1) with respect to $\theta$ gives $\frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a} \times \frac{\mathrm{d}}{\mathrm{d} \theta}(\sec \theta)=\operatorname{asec} \theta \tan \theta \ldots \ldots$ (3)

Also, differentiating both sides of the equation (2) with respect to $\theta$ gives $\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{b} \times \frac{\mathrm{d}}{\mathrm{d} \theta}(\tan \theta)=\mathrm{b} \sec ^{2} \theta \ldots \ldots$ (4)

Thus, dividing the equation (4) by the equation (3) gives

$\frac{d y}{d x}=\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}=\frac{b \sec ^{2} \theta}{a \sec \theta \tan \theta}=\frac{b}{a} \sec \theta \tan \theta=-\frac{b \cos \theta}{a \cos \theta \sin \theta}=\frac{b}{a} \times \frac{1}{\sin \theta}=\frac{b}{a} \cos$

Hence, $\frac{d y}{d x}=\frac{b}{a} \operatorname{cosec} \theta$.

10. Determine $\mathbf{\frac{d y}{d x}}$ from the parametric equations $\mathbf{x=a(\cos \theta+\theta \sin \theta), y=a(\sin \theta-\theta \cos \theta)}$, without eliminating the parameter $\theta$, where a,b are constants.

Ans: The given equations are

$\mathrm{x}=\mathrm{a}(\cos \theta+\theta \sin \theta) \quad \ldots \ldots$ (1)

and $\mathrm{y}=\mathrm{a}(\sin \theta-\theta \cos \theta) \ldots \ldots$ (2)

Then, differentiating both sides of the equation (1) with respect to $\theta$ gives $\frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}\left[\frac{\mathrm{d}}{\mathrm{d} \theta} \cos \theta+\frac{\mathrm{d}}{\mathrm{d} \theta}(\theta \sin \theta)\right]=\mathrm{a}\left[-\sin \theta+\theta \frac{\mathrm{d}}{\mathrm{d} \theta}(\sin \theta)+\sin \theta \frac{\mathrm{d}}{\mathrm{d} \theta}(\theta)\right]$ $=\mathrm{a}[-\sin \theta+\theta \cos \theta+\sin \theta]$.

Therefore, $\frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \theta}=\mathrm{a} \theta \cos \theta$(3)

Also, differentiating both sides of the equation (2) with respect to $\theta$ gives $\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a}\left[\frac{\mathrm{d}}{\mathrm{d} \theta}(\sin \theta)-\frac{\mathrm{d}}{\mathrm{d} \theta}(\theta \cos \theta)\right]=\mathrm{a}\left[\cos \theta-\left\{\theta \frac{\mathrm{d}}{\mathrm{d} \theta}(\cos \theta)+\cos \theta \times \frac{\mathrm{d}}{\mathrm{d} \theta}(\theta)\right\}\right]$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a}[\cos \theta+\theta \sin \theta-\cos \theta]$

Therefore, $\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a} \theta \sin \theta \quad \ldots \ldots$ (4)

Thus, dividing the equation (4) by the equation (3) gives

$\frac{d y}{d x}=\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}=\frac{a \theta \sin \theta}{a \theta \sin \theta}=\tan \theta \text {. }$

Hence, $\frac{d y}{d x}=\tan \theta$.

11. Prove that $\mathbf{\frac{d y}{d x}=-\frac{y}{x}}$, where it is provided that $\mathbf{x=\sqrt{a^{\sin ^{-1} t}}, y=\sqrt{a^{\cos ^{-1} t}}}$.

Ans: The given parametric equations are $x=\sqrt{a^{\sin ^{-1} t}}$ and $y=\sqrt{a^{\cos ^{-1} t}}$.

Now, $x=\sqrt{a^{\sin ^{-1} t}}$ and $y=\sqrt{a^{\cos ^{-1} t}}$

$\Rightarrow \mathrm{x}=\left(\mathrm{a}^{\sin ^{-1} \mathrm{t}}\right)$ and $\mathrm{y}=\left(\mathrm{a}^{\cos ^{-1} \mathrm{t}}\right)^{\frac{1}{2}}$

$\Rightarrow \mathrm{x}=\mathrm{a}^{\frac{1}{2} \sin ^{-1 \mathrm{t}}}$ and $\mathrm{y}=\mathrm{a}^{\frac{1}{2} \cos ^{-1} \mathrm{t}}$

Therefore, first consider $x=a^{\frac{1}{2} \sin ^{-1} t}$.

Take logarithms on both sides of the equation.

Then, we have

$\log x=\frac{1}{2} \sin ^{-1}$ tloga .

Then, differentiating both sides of the equation with respect to $t$ gives

$\begin{aligned}&\frac{1}{x} \times \frac{d x}{d t}=\frac{1}{2} \log a \times \frac{d}{d t}\left(\sin ^{-1} t\right) \\&\Rightarrow \frac{d x}{d t}=\frac{x}{2} \log a \times \frac{1}{\sqrt{1-t^{2}}} \\&\text { Therefore, } \frac{d x}{d t}=\frac{x \log a}{2 \sqrt{1-t^{2}}} . \ldots \ldots \text { (1) }\end{aligned}$

Again, consider the equation $y=a^{\frac{1}{2} \cos ^{-1} t}$.

Take logarithm both sides of the equation.

Then, we have

$\log y=\frac{1}{2} \cos ^{-1} \text { tloga }$

Differentiating both sides of the equation with respect to $\mathrm{t}$ gives $\frac{1}{y} \times \frac{d x}{d t}=\frac{1}{2} \log a \times \frac{d}{d t}\left(\cos ^{-1} t\right)$

$\begin{aligned}&\frac{1}{y} \times \frac{d x}{d t}=\frac{1}{2} \log a \times \frac{d}{d t}\left(\cos ^{-1} t\right) \\&\Rightarrow \frac{d x}{d t}=\frac{y \log a}{2} \times\left(\frac{1}{\sqrt{1-t^{2}}}\right)\end{aligned}$

Therefore, $\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{-\mathrm{yloga}}{2 \sqrt{1-\mathrm{t}^{2}}}$.

$\frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})} = \frac{(\frac{-y \log a}{2\sqrt{1- t^2}})}{(\frac{x \log }{2 \sqrt{1 - t^2}})} = \frac{y}{x}$

 Hence, $\frac{dy}{dx} = \frac{y}{x}$

Class 12 Maths Chapter 5 Exercise 5.6 NCERT Solutions: A Brief Introduction

As mentioned earlier, the Class 12 Maths syllabus is a huge one. The new chapters have innumerable sections. Each section teaches new concepts to the students. These concepts are used to solve the problems given in the exercises. If you take the example of Class 12 Maths Exercise 5.6, you will discover that all the concepts you have learned in this chapter are required to solve the problems.

To utilize your time well, you need to work hard and smart. You will find this chapter easier to comprehend when you have the best solution in your hand. Solving problems might be challenging. There will be some doubts left to be solved. You can use Exercise 5.6 Class 12 Maths NCERT Solutions to resolve your doubts and proceed to the next chapters. There is no need to wait for an entire class to complete the chapter when you can enjoy studying at your own pace.

The expert teachers are aware of the common doubts that arise in the minds of Class 12 students. They know how to tackle these doubts without even explaining them verbally. This is where the solution for Class 12 Maths Ch 5 Ex 5.6 comes into the picture. The experience of these teachers has helped them formulate the right solution for all the problems in this exercise. 

All these answers can be easily understood due to the simple explanation provided by the mentors. Follow the answering format and practice these problems. You can be assured that your answering skill for integral problems will become much better in due course of time. All the answers here are framed following the CBSE rules. You will also get accustomed to follow the CBSE answering guidelines and score the highest marks in the exam.

NCERT Solutions for Class 12 Maths PDF Download

• Chapter 1 - Relations and Functions

• Chapter 2 - Inverse Trigonometric Functions

• Chapter 3 - Matrices

• Chapter 4 - Determinants

• Chapter 5 - Continuity and Differentiability

• Chapter 6 - Application of Derivatives

• Chapter 7 - Integrals

• Chapter 8 - Application of Integrals

• Chapter 9 - Differential Equations

• Chapter 10 - Vector Algebra

• Chapter 11 - Three Dimensional Geometry

• Chapter 12 - Linear Programming

• Chapter 13 - Probability

Students can also download the following additional study material -

• Chapter 5: Important Questions

• Chapter 5: Revision Notes

• Chapter 5: Formulas

• Chapter 5: NCERT Exemplar Solutions

• Chapter 5: RD Sharma Solutions

• Chapter 5: RS Aggarwal Solutions

Chapter 5 - Continuity and Differentiability Exercises in PDF Format

• Exercise 5.1 - 34 Questions & Solutions (10 Short Answers, 24 Long Answers)

• Exercise 5.2 - 10 Questions & Solutions (2 Short Answers, 8 Long Answers)

• Exercise 5.3 - 15 Questions & Solutions (9 Short Answers, 6 Long Answers)

• Exercise 5.4 - 10 Questions & Solutions (5 Short Answers, 5 Long Answers)

• Exercise 5.5 - 18 Questions & Solutions (4 Short Answers, 14 Long Answers)

• Exercise 5.6 - 11 Questions & Solutions (7 Short Answers, 4 Long Answers)

• Exercise 5.7 - 17 Questions & Solutions (10 Short Answers, 7 Long Answers)

• Exercise 5.8 - 6 Questions & Solutions (2 Short Answers, 4 Long Answers)

Why Should You Prefer Using Solutions for Continuity and Differentiability Exercise 5.6?

There are a few reasons why students use the solution for the chapter on continuity and differentiability in the Class 12 Maths syllabus.

Completing Preparing the Chapter Perfectly

The reason for using a solution for Class 12 Maths Chapter 5 Exercise 5.6 is to ensure the perfect completion of the chapter. There will be no problem un-attempted when you use the solution as a reference. Whenever you are stuck solving a problem in this chapter, you can refer to the solution for finding the right approach. Hence, the learning process will not face any hurdle when it comes to doubts. The solution will provide the right path to follow and complete the exercise.

Clearing Doubts

The students will face doubts while solving the problems in Exercise 5.6 Class 12 Maths. These doubts can be quickly clarified by using the Class 12 Maths NCERT Solutions Chapter 5 Exercise 5.6. These solutions are prepared by the teachers to help the students so that they can complete the syllabus and get time to practice it. You will not have to wait to get the problems solved when you have an easy explanation of all the problems at your fingertips.

Assessing Your Problem - Solving Skills

Another reason for using the NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.6 is to get a medium to assess your problem-solving skills. As mentioned earlier, differentiability and continuity is an important chapter to cover for the Class 12 syllabus and a future professional course. Hence, you need to assess your skills here. Refer to the solutions to find out whether you are on the right track or not. Diagnose the weaknesses beforehand and work on them. Make them your strengths and sharpen your skills.