# NCERT Solutions For Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2

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Textbook | NCERT |

Board | CBSE |

Category | NCERT Solutions |

Class | Class 12 |

Subject | Maths |

Chapter | Chapter 5 |

Exercise | Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.2 |

Number of Questions Solved | 10 |

## NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2

**NCERT TEXTBOOK EXERCISES**

**Ex 5.2 Class 12 Maths Question 1.**

sin(x² + 5)

**Solution:**

Let y = sin(x2 + 5),

put x² + 5 = t

y = sint

t = x²+5

$\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx }$

$\frac { dy }{ dx } =cost.\frac { dt }{ dx } =cos({ x }^{ 2 }+5)\frac { d }{ dx } ({ x }^{ 2 }+5)$

= cos (x² + 5) × 2x

= 2x cos (x² + 5)

**Ex 5.2 Class 12 Maths Question 2.**

cos (sin x)

**Solution:**

let y = cos (sin x)

put sinx = t

∴ y = cost,

t = sinx

∴ $\frac { dy }{ dx } =-sin\quad t,\frac { dt }{ dx } =cos\quad x$

$\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } =(-sint)\times cosx$

Putting the value of t, $\frac { dy }{ dx } =-sin(sinx)\times cosx$

$\frac { dy }{ dx } =-[sin(sinx)]cosx$

**Ex 5.2 Class 12 Maths Question 3.**

sin(ax+b)

**Solution:**

let = sin(ax+b)

put ax+bx = t

∴ y = sint

t = ax+b

$\frac { dy }{ dt } =cost,\frac { dt }{ dx } =\frac { d }{ dx } (ax+b)=a$

$Now\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } =cost\times a=acos\quad t$

$\frac { dy }{ dx } =acos(ax+b)$

**Ex 5.2 Class 12 Maths Question 4.**

sec(tan(√x))

**Solution:**

let y = sec(tan(√x))

by chain rule

$\frac { dy }{ dx } =sec(tan\sqrt { x } )tan(tan\sqrt { x } )\frac { d }{ dx } (tan\sqrt { x } )$

$\frac { dy }{ dx } =sec(tan\sqrt { x } ).tan(tan\sqrt { x } ){ sec }^{ 2 }\sqrt { x } .\frac { 1 }{ 2\sqrt { x } }$

**Ex 5.2 Class 12 Maths Question 5.**

$\ \frac { sin(ax+b) }{ cos(cx+d) }$

**Solution:**

y = $\ \frac { sin(ax+b) }{ cos(cx+d) }$= v/u

u = sin(ax+b)

**Ex 5.2 Class 12 Maths Question 6.**

cos x³ . sin²(x^{5}) = y(say)

**Solution:**

Let u = cos x³ and v = sin² x^{5},

put x³ = t

**Ex 5.2 Class 12 Maths Question 7.**

$2\sqrt { cot({ x }^{ 2 }) } =y(say)$

**Solution:**

$2\sqrt { cot({ x }^{ 2 }) } =y(say)$

**Ex 5.2 Class 12 Maths Question 8.**

cos(√x) = y(say)

**Solution:**

cos(√x) = y(say)

$\frac { dy }{ dx } =\frac { d }{ dx } cos\left( \sqrt { x } \right) =-sin\sqrt { x } .\frac { d\sqrt { x } }{ dx }$

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

**Ex 5.2 Class 12 Maths Question 9.**

Prove that the function f given by f (x) = |x – 1|,x ∈ R is not differential at x = 1.

**Solution:**

The given function may be written as

$f(x)=\begin{cases} x-1,\quad if\quad x\ge 1 \ 1-x,\quad if\quad x<1 \end{cases}$

$R.H.D\quad at\quad x=1\quad =\underset { h\rightarrow 0 }{ lim } \frac { f(1+h)-f(1) }{ h }$

**Ex 5.2 Class 12 Maths Question 10.**

Prove that the greatest integer function defined by f (x)=[x], 0 < x < 3 is not differential at x = 1 and x = 2.

**Solution:**

(i) At x = 1

$R.H.D=\underset { h\rightarrow 0 }{ lim } \frac { f(1+h)-f(1) }{ h }$

## NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exercise 5.2 PDF

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### Other Chapter of Class 12 Maths Chapter 5 Continuity and Differentiability

**Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1****Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2****Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3****Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4****Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5****Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.6****Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7****Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8**