Contents

- 1 NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability
- 1.1 Chapter 5 Continuity and Differentiability Exercise 5.1
- 1.2 Chapter 5 Continuity and Differentiability Exercise 5.2
- 1.3 Chapter 5 Continuity and Differentiability Exercise 5.3
- 1.4 Chapter 5 Continuity and Differentiability Exercise 5.4
- 1.5 Chapter 5 Continuity and Differentiability Exercise 5.5
- 1.6 Chapter 5 Continuity and Differentiability Exercise 5.6
- 1.7 Chapter 5 Continuity and Differentiability Exercise 5.7
- 1.8 Chapter 5 Continuity and Differentiability Exercise 5.8

These Solutions are part of NCERT Solutions for Class 12 Maths . Here we have given NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 12 |

Subject |
Maths |

Chapter |
Chapter 5 |

Chapter Name |
Continuity and Differentiability |

Exercise |
Ex 5.1, Ex 5.2, Ex 5.3, Ex 5.4,Ex 5.5,Ex 5.6, Ex 5.7,Ex 5.8 |

Number of Questions Solved |
121 |

Category |
NCERT Solutions |

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

### Chapter 5 Continuity and Differentiability Exercise 5.1

**Question 1.**

Prove that the function f (x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.

**Solution:**

(i) At x = 0. lim_{x–>0} f (x) = lim_{x–>0} (5x – 3) = – 3 and

f(0) = – 3

∴f is continuous at x = 0

(ii) At x = – 3, lim_{x–>3} f(x)= lim_{x–>-3} (5x – 3) = – 18

and f( – 3) = – 18

∴ f is continuous at x = – 3

(iii) At x = 5, lim_{x–>5} f(x) = lim_{x–>5} (5x – 3) = 22 and

f(5) = 22

∴ f is continuous at x = 5

**Question 2.**

Examine the continuity of the function f(x) = 2x² – 1 at x = 3.

**Solution:**

lim_{x–>3} f(x) = lim_{x–>3} (2x² – 1) = 17 and f(3)= 17

∴ f is continuous at x = 3

**Question 3.**

Examine the following functions for continuity.

(a) f(x) = x – 5

(b) f(x) = , x≠5

(c) f(x) = ,x≠5

(d) f(x) = |x – 5|

**Solution:**

(a) f(x) = (x-5) => (x-5) is a polynomial

∴it is continuous at each x ∈ R.

**Question 4.**

Prove that the function f (x) = x^{n} is continuous at x = n, where n is a positive integer.

**Solution:**

f (x) = x^{n} is a polynomial which is continuous for all x ∈ R.

Hence f is continuous at x = n, n ∈ N.

**Question 5.**

Is the function f defined by continuous at x = 0? At x = 1? At x = 2?

**Solution:**

(i) At x = 0

lim_{x–>0-} f(x) = lim_{x–>0-} x = 0 and

lim_{x–>0+} f(x) = lim_{x–>0+} x = 0 => f(0) = 0

∴ f is continuous at x = 0

(ii) At x = 1

lim_{x–>1-} f(x) = lim_{x–>1-} (x) = 1 and

lim_{x–>1+} f(x) = lim_{x–>1+}(x) = 5

∴ lim_{x–>1-} f(x) ≠ lim_{x–>1+} f(x)

∴ f is discontinuous at x = 1

(iii) At x = 2

lim_{x–>2} f(x) = 5, f(2) = 5

∴ f is continuous at x = 2

**Find all points of discontinuity off, where f is defined by**

**Question 6.**

**Solution:**

at x≠2

**Question 7.**

**Solution:**

**Question 8.**

Test the continuity of the function f (x) at x = 0

**Solution:**

We have;

(LHL at x=0)

**Question 9.**

**Solution:**

**Question 10.**

**Solution:**

**Question 11.**

**Solution:**

At x = 2, L.H.L. lim_{x–>2-} (x³ – 3) = 8 – 3 = 5

R.H.L. = lim_{x–>2+} (x² + 1) = 4 + 1 = 5

**Question 12.**

**Solution:**

**Question 13.**

Is the function defined by a continuous function?

**Solution:**

At x = 1,L.H.L.= lim_{x–>1-} f(x) = lim_{x–>1-} (x + 5) = 6,

R.HL. = lim_{x–>1+} f(x) = lim_{x–>1+} (x – 5) = – 4

f(1) = 1 + 5 = 6,

f(1) = L.H.L. ≠ R.H.L.

=> f is not continuous at x = 1

At x = c < 1, lim_{x–>c} (x + 5) = c + 5 = f(c)

At x = c > 1, lim_{x–>c} (x – 5) = c – 5 = f(c)

∴ f is continuous at all points x ∈ R except x = 1.

**Discuss the continuity of the function f, where f is defined by**

**Question 14.**

**Solution:**

In the interval 0 ≤ x ≤ 1,f(x) = 3; f is continuous in this interval.

At x = 1,L.H.L. = lim f(x) = 3,

R.H.L. = lim_{x–>1+} f(x) = 4 => f is discontinuous at

x = 1

At x = 3, L.H.L. = lim_{x–>3-} f(x)=4,

R.H.L. = lim_{x–>3+} f(x) = 5 => f is discontinuous at

x = 3

=> f is not continuous at x = 1 and x = 3.

**Question 15.**

**Solution:**

At x = 0, L.H.L. = lim_{x–>0-} 2x = 0 ,

R.H.L. = lim_{x–>0+} (0)= 0 , f(0) = 0

=> f is continuous at x = 0

At x = 1, L.H.L. = lim_{x–>1-} (0) = 0,

R.H.L. = lim_{x–>1+} 4x = 4

f(1) = 0, f(1) = L.H.L.≠R.H.L.

∴ f is not continuous at x = 1

when x < 0 f (x) = 2x, being a polynomial, it is

continuous at all points x < 0. when x > 1. f (x) = 4x being a polynomial, it is

continuous at all points x > 1.

when 0 ≤ x ≤ 1, f (x) = 0 is a continuous function

the point of discontinuity is x = 1.

**Question 16.**

**Solution:**

At x = – 1,L.H.L. = lim_{x–>1-} f(x) = – 2, f(-1) = – 2,

R.H.L. = lim_{x–>1+} f(x) = – 2

=> f is continuous at x = – 1

At x= 1, L.H.L. = lim_{x–>1-} f(x) = 2,f(1) = 2

∴ f is continuous at x = 1,

R.H.L. = lim_{x–>1+} f(x) = 2

Hence, f is continuous function.

**Question 17.**

Find the relationship between a and b so that the function f defined by

is continuous at x = 3

**Solution:**

At x = 3, L.H.L. = lim_{x–>3-} (ax+1) = 3a+1 ,

f(3) = 3a + 1, R.H.L. = lim_{x–>3+} (bx+3) = 3b+3

f is continuous ifL.H.L. = R.H.L. = f(3)

3a + 1 = 3b + 3 or 3(a – b) = 2

a – b = or a = b + , for any arbitrary value of b.

Therefore the value of a corresponding to the value of b.

**Question 18.**

For what value of λ is the function defined by

continuous at x = 0? What about continuity at x = 1?

**Solution:**

At x = 0, L.H.L. = lim_{x–>0-} λ (x² – 2x) = 0 ,

R.H.L. = lim_{x–>0+} (4x+ 1) = 1, f(0)=0

f (0) = L.H.L. ≠ R.H.L.

=> f is not continuous at x = 0,

whatever value of λ ∈ R may be

At x = 1, lim_{x–>1} f(x) = lim_{x–>1} (4x + l) = f(1)

=> f is not continuous at x = 0 for any value of λ but f is continuous at x = 1 for all values of λ.

**Question 19.**

Show that the function defined by g (x) = x – [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.

**Solution:**

Let c be an integer, [c – h] = c – 1, [c + h] = c, [c] = c, g(x) = x – [x].

At x = c, lim_{x–>c-} (x – [x]) = lim_{h–>0} [(c – h) – (c – 1)]

= lim_{h–>0} (c – h – (c – 1)) = 1[∵ [c – h] = c – 1]

R.H.L. = lim_{x–>c+} (x – [x])= lim_{h–>0} (c + h – [c + h])

= lim_{h–>0} [c + h – c] = 0

f(c) = c – [c] = 0,

Thus L.H.L. ≠ R.H.L. = f (c) => f is not continuous at integral points.

**Question 20.**

Is the function defined by f (x) = x² – sin x + 5 continuous at x = π?

**Solution:**

Let f(x) = x² – sinx + 5,

**Question 21.**

Discuss the continuity of the following functions:

(a) f (x) = sin x + cos x

(b) f (x) = sin x – cos x

(c) f (x) = sin x · cos x

**Solution:**

(a) f(x) = sinx + cosx

**Question 22.**

Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

**Solution:**

(a) Let f(x) = cosx

**Question 23.**

Find all points of discontinuity of f, where

**Solution:**

At x = 0

**Question 24.**

Determine if f defined by is a continuous function?

**Solution:**

At x = 0

**Question 25.**

Examine the continuity of f, where f is defined by

**Solution:**

**Find the values of k so that the function is continuous at the indicated point in Questions 26 to 29.**

**Question 26.**

**Solution:**

At x =

L.H.L =

**Question 27.**

**Solution:**

**Question 28.**

**Solution:**

**Question 29.**

**Solution:**

**Question 30.**

Find the values of a and b such that the function defined by

to is a continuous function.

**Solution:**

**Question 31.**

Show that the function defined by f(x)=cos (x²) is a continuous function.

**Solution:**

Now, f (x) = cosx², let g (x)=cosx and h (x) x²

∴ goh(x) = g (h (x)) = cos x²

Now g and h both are continuous ∀ x ∈ R.

f (x) = goh (x) = cos x² is also continuous at all x ∈ R.

**Question 32.**

Show that the function defined by f (x) = |cos x| is a continuous function.

**Solution:**

Let g(x) =|x|and h (x) = cos x, f(x) = goh(x) = g (h (x)) = g (cosx) = |cos x |

Now g (x) = |x| and h (x) = cos x both are continuous for all values of x ∈ R.

∴ (goh) (x) is also continuous.

Hence, f (x) = goh (x) = |cos x| is continuous for all values of x ∈ R.

**Question 33.**

Examine that sin |x| is a continuous function.

**Solution:**

Let g (x) = sin x, h (x) = |x|, goh (x) = g (h(x))

= g(|x|) = sin|x| = f(x)

Now g (x) = sin x and h (x) = |x| both are continuous for all x ∈ R.

∴f (x) = goh (x) = sin |x| is continuous at all x ∈ R.

**Question 34.**

Find all the points of discontinuity of f defined by f(x) = |x|-|x+1|.

**Solution:**

f(x) = |x|-|x+1|, when x< – 1,

f(x) = -x-[-(x+1)] = – x + x + 1 = 1

when -1 ≤ x < 0, f(x) = – x – (x + 1) = – 2x – 1,

when x ≥ 0, f(x) = x – (x + 1) = – 1

### Chapter 5 Continuity and Differentiability Exercise 5.2

**Differentiate the functions with respect to x in Questions 1 to 8.**

**Question 1.**

sin(x² + 5)

**Solution:**

Let y = sin(x2 + 5),

put x² + 5 = t

y = sint

t = x²+5

= cos (x² + 5) × 2x

= 2x cos (x² + 5)

**Question 2.**

cos (sin x)

**Solution:**

let y = cos (sin x)

put sinx = t

∴ y = cost,

t = sinx

∴

Putting the value of t,

**Question 3.**

sin(ax+b)

**Solution:**

let = sin(ax+b)

put ax+bx = t

∴ y = sint

t = ax+b

**Question 4.**

sec(tan(√x))

**Solution:**

let y = sec(tan(√x))

by chain rule

**Question 5.**

**Solution:**

y = =

u = sin(ax+b)

**Question 6.**

cos x³ . sin²(x^{5}) = y(say)

**Solution:**

Let u = cos x³ and v = sin² x^{5},

put x³ = t

**Question 7.**

**Solution:**

**Question 8.**

cos(√x) = y(say)

**Solution:**

cos(√x) = y(say)

**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

**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

### Chapter 5 Continuity and Differentiability Exercise 5.3

**Find in the following**

**Question 1.**

2x + 3y = sinx

**Solution:**

2x + 3y = sinx

Differentiating w.r.t x,

=>

**Question 2.**

2x + 3y = siny

**Solution:**

2x + 3y = siny

Differentiating w.r.t x,

=>

**Question 3.**

ax + by² = cosy

**Solution:**

ax + by² = cosy

Differentiate w.r.t. x,

=>

**Question 4.**

xy + y² = tan x + y

**Solution:**

xy + y² = tanx + y

Differentiating w.r.t. x,

**Question 5.**

x² + xy + y² = 100

**Solution:**

x² + xy + xy = 100

Differentiating w.r.t. x,

**Question 6.**

x³ + x²y + xy² + y³ = 81

**Solution:**

Given that

x³ + x²y + xy² + y³ = 81

Differentiating both sides we get

**Question 7.**

sin² y + cos xy = π

**Solution:**

Given that

sin² y + cos xy = π

Differentiating both sides we get

**Question 8.**

sin²x + cos²y = 1

**Solution:**

Given that

sin²x + cos²y = 1

Differentiating both sides, we get

**Question 9.**

**Solution:**

put x = tanθ

**Question 10.**

**Solution:**

put x = tanθ

**Question 11.**

**Solution:**

put x = tanθ

**Question 12.**

**Solution:**

put x = tanθ

we get

**Question 13.**

**Solution:**

put x = tanθ

we get

**Question 14.**

**Solution:**

put x = tanθ

we get

**Question 15.**

**Solution:**

put x = tanθ

we get

### Chapter 5 Continuity and Differentiability Exercise 5.4

**Differentiate the following w.r.t.x:**

**Question 1.**

**Solution:**

**Question 2.**

**Solution:**

x=sint

**Question 3.**

**Solution:**

**Question 4.**

**Solution:**

**Question 5.**

**Solution:**

**Question 6.**

…

**Solution:**

…

**Question 7.**

**Solution:**

y =

**Question 8.**

log(log x),x>1

**Solution:**

y = log(log x),

put y = log t, t = log x,

differentiating

**Question 9.**

**Solution:**

let

**Question 10.**

cos(log x+e^{x}),x>0

**Solution:**

y = cos(log x+e^{x}),x>0

put y = cos t,t = log x+e^{x}

### Chapter 5 Continuity and Differentiability Exercise 5.5

**Differentiate the functions given in Questions 1 to 11 w.r.to x**

**Question 1.**

cos x. cos 2x. cos 3x

**Solution:**

Let y = cos x. cos 2x . cos 3x,

Taking log on both sides,

log y = log (cos x. cos 2x. cos 3x)

log y = log cos x + log cos 2x + log cos 3x,

Differentiating w.r.t. x, we get

**Question 2.**

**Solution:**

taking log on both sides

log y = log

**Question 3.**

(log x)^{cosx}

**Solution:**

let y = (log x)^{cosx}

Taking log on both sides,

log y = log (log x)^{cosx}

log y = cos x log (log x),

Differentiating w.r.t. x,

**Question 4.**

x – 2^{sinx}

**Solution:**

let y = x – 2^{sinx},

y = u – v

**Question 5.**

(x+3)^{2}.(x + 4)^{3}.(x + 5)^{4}

**Solution:**

let y = (x + 3)^{2}.(x + 4)^{3}.(x + 5)^{4}

Taking log on both side,

logy = log [(x + 3)^{2} • (x + 4)^{3} • (x + 5)^{4}]

= log (x + 3)^{2} + log (x + 4)^{3} + log (x + 5)^{4}

log y = 2 log (x + 3) + 3 log (x + 4) + 4 log (x + 5)

Differentiating w.r.t. x, we get

**Question 6.**

**Solution:**

let

let

**Question 7.**

(log x)^{x} + x^{logx}

**Solution:**

let y = (log x)^{x} + x^{logx} = u+v

where u = (log x)^{x}

∴ log u = x log(log x)

**Question 8.**

(sin x)^{x}+sin^{-1} √x

**Solution:**

Let y = (sin x)^{x }+ sin^{-1 }√x

let u = (sin x)x, v = sin^{-1} √x

**Question 9.**

x^{sinx} + (sin x)^{cosx}

**Solution:**

let y = x^{sinx} + (sin x)^{cosx} = u+v

where u = x^{sinx}

log u = sin x log x

**Question 10.**

**Solution:**

y = u + v

**Question 11.**

**Solution:**

Let u = (x cosx)^{x}

logu = x log(x cosx)

**Find of the functions given in Questions 12 to 15.**

**Question 12.**

x^{y} + y^{x} = 1

**Solution:**

x^{y} + y^{x} = 1

let u = x^{y} and v = y^{x}

∴ u + v = 1,

Now u = x

**Question 13.**

y^{x }= x^{y}

**Solution:**

y = x

x logy = y logx

**Question 14.**

(cos x)^{y} = (cos y)^{x}

**Solution:**

We have

(cos x)^{y} = (cos y)^{x}

=> y log (cosx) = x log (cosy)

**Question 15.**

xy = e^{(x-y)}

**Solution:**

log(xy) = log e^{(x-y)}

=> log(xy) = x – y

=> logx + logy = x – y

**Question 16.**

Find the derivative of the function given by f (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) and hence find f'(1).

**Solution:**

Let f(x) = y = (1 + x)(1 + x^{2})(1 + x^{4})(1 + x^{8})

Taking log both sides, we get

logy = log [(1 + x)(1 + x^{2})(1 + x^{4})(1 + x^{8})]

logy = log(1 + x) + log (1 + x^{2}) + log(1 + x^{4}) + log(1 + x^{8})

**Question 17.**

Differentiate (x^{2} – 5x + 8) (x^{3} + 7x + 9) in three ways mentioned below:

(i) by using product rule

(ii) by expanding the product to obtain a single polynomial.

(iii) by logarithmic differentiation.

Do they all give the same answer?

**Solution:**

(i) By using product rule

f’ = (x^{2} – 5x + 8) (3x^{2} + 7) + (x^{3} + 7x + 9) (2x – 5)

f = 5x^{4} – 20x^{3} + 45x^{2} – 52x + 11.

(ii) By expanding the product to obtain a single polynomial, we get

**Question 18.**

If u, v and w are functions of w then show that

in two ways-first by repeated application of product rule, second by logarithmic differentiation.

**Solution:**

Let y = u.v.w

=> y = u. (vw)

### Chapter 5 Continuity and Differentiability Exercise 5.6

**If x and y are connected parametrically by the equations given in Questions 1 to 10, without eliminating the parameter. Find .**

**Question 1.**

x = 2at², y = at^{4}

**Solution:**

**Question 2.**

x = a cosθ,y = b cosθ

**Solution:**

**Question 3.**

x = sin t, y = cos 2t

**Solution:**

**Question 4.**

**Solution:**

**Question 5.**

x = cos θ – cos 2θ, y = sin θ – sin 2θ

**Solution:**

**Question 6.**

x = a(θ – sinθ), y = a(1 + cosθ)

**Solution:**

**Question 7.**

**Solution:**

**Question 8.**

**Solution:**

**Question 9.**

x = a sec θ,y = b tan θ

**Solution:**

x = a sec θ,y = b tan θ

**Question 10.**

x = a(cosθ+θsinθ), y = a(sinθ-θcosθ)

**Solution:**

x = a(cosθ+θsinθ), y = a(sinθ-θcosθ)

**Question 11.**

If show that

**Solution:**

Given that

### Chapter 5 Continuity and Differentiability Exercise 5.7

**Find the second order derivatives of the functions given in Questions 1 to 10.**

**Question 1.**

x² + 3x + 2 = y(say)

**Solution:**

**Question 2.**

x^{20} = y(say)

**Solution:**

**Question 3.**

x.cos x = y(say)

**Solution:**

**Question 4.**

log x = y (say)

**Solution:**

**Question 5.**

x^{3} log x = y (say)

**Solution:**

x^{3} log x = y

**Question 6.**

e^{x} sin5x = y

**Solution:**

e^{x} sin5x = y

**Question 7.**

e^{6x} cos3x = y

**Solution:**

e^{6x} cos3x = y

**Question 8.**

tan^{-1} x = y

**Solution:**

**Question 9.**

log(logx) = y

**Solution:**

log(logx) = y

**Question 10.**

sin(log x) = y

**Solution:**

sin(log x) = y

**Question 11.**

If y = 5 cosx – 3 sin x, prove that

**Solution:**

Hence proved

**Question 12.**

If y = cos^{-1} x, Find in terms of y alone.

**Solution:**

**Question 13.**

If y = 3 cos (log x) + 4 sin (log x), show that

**Solution:**

Given that

y = 3 cos (log x) + 4 sin (log x)

**Question 14.**

**Solution:**

Given that

**Question 15.**

If y = 500e^{7x} + 600e^{-7x}, show that .

**Solution:**

we have

y = 500e^{7x} + 600e^{-7x}

**Question 16.**

If e^{y}(x+1) = 1,show that

**Solution:**

**Question 17.**

If y=(tan^{-1} x)² show that (x²+1)²y_{2}+2x(x²+1)y_{1}=2

**Solution:**

we have

y=(tan^{-1} x)²

### Chapter 5 Continuity and Differentiability Exercise 5.8

**Question 1.**

Verify Rolle’s theorem for the function

f(x) = x² + 2x – 8,x∈ [-4,2]

**Solution:**

Now f(x) = x² + 2x – 8 is a polynomial

∴ it is continuous and derivable in its domain x∈R.

Hence it is continuous in the interval [-4,2] and derivable in the interval (- 4,2)

f(-4) = (-4)² + 2(-4) – 8 = 16 – 8 – 8 = 0,

f(2) = 2² + 4 – 8 = 8 – 8 = 0

Conditions of Rolle’s theorem are satisfied.

f'(x) = 2x + 2

∴ f’ (c) = 2c + 2 = 0

or c = – 1, c = – 1 ∈ [-4,2]

Thus f’ (c) = 0 at c = – 1.

**Question 2.**

Examine if Rolle’s theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle’s theorem from these example?

(i) f(x) = [x] for x ∈ [5,9]

(ii) f (x) = [x] for x ∈ [-2,2]

(iii) f (x) = x² – 1 for x ∈ [1,2]

**Solution:**

(i) In the interval [5, 9], f (x) = [x] is neither continuous nor derivable at x = 6,7,8 Hence Rolle’s theorem is not applicable

(ii) f (x) = [x] is not continuous and derivable at -1, 0, 1. Hence Rolle’s theorem is not applicable.

(iii) f(x) = (x² – 1),f(1) = 1 – 1 = 0,

f(2) = 22 – 1 = 3

f(a)≠f(b)

Though it is continous and derivable in the interval [1,2].

Rolle’s theorem is not applicable.

In case of converse if f (c)=0, c ∈ [a, b] then conditions of rolle’s theorem are not true.

(i) f (x) = [x] is the greatest integer less than or equal to x.

∴f(x) = 0, But fis neither continuous nor differentiable in the interval [5,9].

(ii) Here also, theough f (x) = 0, but f is neither continuous nor differentiable in the interval [-2,2].

(iii) f (x)=x² – 1, f'(x)=2x. Here f'(x) is not zero in the [1,2], So f (2) ≠ f’ (2).

**Question 3.**

If f: [-5,5] –>R is a differentiable function and if f (x) does not vanish anywhere then prove that f (- 5) ≠ f (5).

**Solution:**

For Rolle’s theorem

If (i) f is continuous in [a, b]

(ii) f is derivable in [a, b]

(iii) f (a) = f (b)

then f’ (c)=0, c e (a, b)

∴ f is continuous and derivable

but f (c) ≠ 0 =>f(a) ≠ f(b) i.e., f(-5)≠f(5)

**Question 4.**

Verify Mean Value Theorem, if

f (x) = x² – 4x – 3 in the interval [a, b], where a = 1 and b = 4.

**Solution:**

f (x) = x² – 4x – 3. It being a polynomial it is continuous in the interval [1,4] and derivable in (1,4), So all the condition of mean value theorem hold.

then f’ (x) = 2x – 4,

f’ (c) = 2c – 4

f(4)= 16 – 16 – 3 = – 3,

f(1)= 1 – 4 – 3 = – 6

Then there exist a value c such that

**Question 5.**

Verify Mean Value Theorem, if f (x)=x^{3} – 5x^{2} – 3x in the interval [a, b], where a = 1 and b = 3. Find all c ∈ (1,3) for which f’ (c) = 0.

**Solution:**

f (x)=x^{3} – 5x^{2} – 3x,

It is a polynomial. Therefore it is continuous in the interval [1,3] and derivable in the interval (1,3)

Also, f'(x)=3x²-10x-3

**Question 6.**

Examine the applicability of Mean Value theroem for all three functions given in the above Question 2.

**Solution:**

(i) F (x)= [x] for x ∈ [5,9], f (x) = [x] in the interval [5, 9] is neither continuous, nor differentiable.

(ii) f (x) = [x], for x ∈ [-2,2],

Again f (x) = [x] in the interval [-2,2] is neither continous, nor differentiable.

(iii) f(x) = x²-1 for x ∈ [1,2], It is a polynomial. Therefore it is continuous in the interval [1,2] and differentiable in the interval (1,2)

f (x) = 2x, f(1) = 1 – 1 = 0 ,

f(2) = 4 – 1 = 3, f'(c) = 2c

We hope the NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability help you. If you have any query regarding NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability, drop a comment below and we will get back to you at the earliest.

## Leave a Reply