Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.3

Chapter 5 of the Class 12 NCERT Mathematics textbook, "Continuity and Differentiability," introduces the concepts of continuity and differentiability of functions, which are fundamental in calculus. Exercise 5.3 focuses on applying these concepts to solve problems related to the continuity and differentiability of functions.

NCERT Solutions for Class 12 - Mathematics Part I - Chapter 5 Continuity and Differentiability - Exercise 5.3

This section provides detailed solutions for Exercise 5.3 from Chapter 5 of the Class 12 NCERT Mathematics textbook. The exercise involves problems that require students to determine whether given functions are continuous and differentiable at specified points or over given intervals. Solutions are presented step-by-step to help students understand the application of continuity and differentiability concepts.

Find \frac{dy}{dx}  of the following:

Question 1. 2x + 3y = sin x

Solution:

On differentiating both sides w.r.t. x, we get

2 + 3 \frac{dy}{dx}  = cos x

3 \frac{dy}{dx}  = cos x - 2

\frac{dy}{dx}  = (cosx - 2)/3 

Question 2. 2x + 3y = sin y

Solution:

On differentiating both sides w.r.t. x, we get

2 + 3 \frac{dy}{dx}  = cos y\frac{dy}{dx}

(cosy - 3) \frac{dy}{dx}  = 2

\frac{dy}{dx} = \frac{2}{(cos⁡y - 3)}

Question 3. ax + by² = cos y

Solution:

On differentiating both sides w.r.t. x, we get

a + b * 2y(\frac{dy}{dx} ) = -sin y * \frac{dy}{dx}

(2by + siny) \frac{dy}{dx}  = -a

\frac{dy}{dx} = \frac{-a}{2by+siny}     

Question 4. xy + y² = tan x + y

Solution:

On differentiating both sides w.r.t. x, we get

(x *\frac{dy}{dx}  + y) + 2y\frac{dy}{dx}  = sec²x + \frac{dy}{dx}

(x + 2y - 1)\frac{dy}{dx}  = sec²x - y

\frac{dy}{dx} = \frac{(sec^2x-y)}{(x+2y-1)}

Question 5. x² + xy + y² = 100

Solution:

On differentiating both sides w.r.t. x, we get

2x + (x\frac{dy}{dx}  + y) + 2y\frac{dy}{dx}  = 0

(x + 2y) * \frac{dy}{dx}  = -(2x + y)

\frac{dy}{dx} = \frac{-(2x + y)}{(x + 2y)}  

Question 6. x³ + x²y + xy² + y³ = 81

Solution:

Differentiate both sides w.r.t. x

3x²+(x²\frac{dy}{dx}  + y * 2x) + (x * 2y * \frac{dy}{dx}  + y²) + 3y² * \frac{dy}{dx}  = 0

(x² + 2xy + 3y²)\frac{dy}{dx}  = -(3x² + 2xy + y²)

\frac{dy}{dx} = \frac{3x^2 + 2xy + y^2}{x^2 + 2xy + 3y^2}

Question 7. Sin²y + cos xy = π

Solution:

Differentiate both sides w.r.t. x

2 sin y * \frac{dy}{dx}  (sin⁡y) - sin(xy) * \frac{dy}{dx}  xy = 0

2sin y * cosy \frac{dy}{dx}  - sin(xy)(x * \frac{dy}{dx}  + y) = 0

(2sin cos y - sin (xy) - x)) \frac{dy}{dx}  = y(xy)

\frac{dy}{dx} = \frac{y sin⁡(xy)}{2 sin⁡ y cos⁡ y - x sin⁡ (xy)}

\frac{dy}{dx} = \frac{ysin(xy)}{sin2y - x sinxy}

Question 8. sin² x + cos² y = 1

Solution:

 2 sin x * \frac{dy}{dx}  (sin ⁡x) + 2 cos y * \frac{dy}{dx}  (cos ⁡y) = 0

2 sin x * cos x + 2 cos y*(-sin y) * \frac{dy}{dx}  = 0

2 sin x * cos x - 2 cos x - 2 cos y sin y * \frac{dy}{dx}  = 0

Sin(2x) - sin(2y) - \frac{dy}{dx}  = 0

\frac{dy}{dx} = \frac{sin(2x)}{sin(2y)}

Question 9. y = sin^-1(\frac{2x}{(1 + x²)}

Solution:

Put x = tanθ          

θ = tan^-1x

y = sin^{-1}\frac{2tanθ}{1+tan^2θ}

y = sin^-1(sin 2θ)          

y = 2θ

y = 2tan^-1x         -(1)

On differentiating eq(1), we get

\frac{dy}{dx} = \frac{2}{(1 + x^2)}

Question 10. Y=tan^{-1}\frac{(3x-x^3)}{(1-3x^2 )}  , -1/√3 < x < 1/√3

Solution:

Put x = tanθ         

θ = tan^-1x

y = tan^{-1}(\frac{3tanθ - tan^3θ}{1-3tan^2θ})

y = tan^-1(tan 3θ)         

y = 3θ

y = 3tan^-1x          -(1)

On differentiating eq(1), we get

\frac{dy}{dx} = \frac{3}{(1 + x^2)}

Question 11. y = cos^{-1}(\frac{1-x^2}{1+x^2})  , 0 < x < 1

Solution:

Put x = tanθ          

θ  = tan^-1 x

y = cos^{-1}(\frac{1-tan^2θ}{1+tan^2θ})

y = cos^-1(cos 2θ)   

y = 2θ 

y = 2tan^-1x         -(1)

On differentiating eq(1), we get

\frac{dy}{dx} = \frac{2}{(1 + x^2)} 

Question 12. y = sin^{-1}(\frac{1-x^2}{1+x^2})  , 0 < x < 1

Solution:

Put x = tanθ       

θ = tan^-1x

y = sin^{-1}(\frac{1 - tan^2θ}{1 + tan^2θ})

y = sin^-1(cos 2θ)

y = sin^-1(sin (π/2 - 2θ))         

y = π/2 - 2θ

y = π/2 - 2 tan^-1x  

\frac{dy}{dx} = \frac{(-2)}{(1 + x^2)} 

Question 13. y = cos^{-1}(\frac{2x}{1+x^2})  , -1 < x < 1

Solution:

Put x = tanθ     

θ = tan^-1x

y = cos^{-1}(\frac{2tanθ}{1 + tan^2θ} )        

y = cos^-1(sin 2θ)

y = cos^-1(cos (π/2 - 2θ))         

y = π/2 - 2θ

y = π/2 - 2tan^-1x  

\frac{dy}{dx} = \frac{(-2)}{(1 + x^2)} 

Question 14. y = sin^{-1}(2x\sqrt{1-x^2})  , -1/√2 < x < 1/√2

Solution:

Put x = sinθ     

θ =  sin^-1 x

y = sin^-1(2sinθ√(1 - sin²θ))

y = sin^-1(sin 2θ) = 2θ

y = 2sin^-1x

\frac{dy}{dx} = \frac{2}{\sqrt{(1 - x^2)}}

Question 15. y = sec^{-1}(\frac{1}{2x^2-1})  , 0 < x < 1/√2

Solution:

Put x = tanθ

y = sec^{-1}(\frac{1}{2 cos^2 - 1})  

y = sec^-1(1/cos2θ))

y = sec^-1(sec2θ) = 2θ

y = 2cos^-1x

\frac{dy}{dx}   = \frac{-2}{\sqrt{1 - x^2} }  

Summary

Exercise 5.3 focuses on problems related to the continuity and differentiability of functions. Students are required to determine whether functions are continuous and differentiable at specific points or over certain intervals. Key concepts include:

• Continuity: A function is continuous at a point if it is defined at that point, the limit exists, and the limit equals the function's value.
• Differentiability: A function is differentiable at a point if its derivative exists at that point.
• Application: The exercise applies these concepts to various functions, analyzing their behavior to determine continuity and differentiability.