01 Continuity and Differentibility Part-I

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Maths-C and D SPARK SUMMIT-12

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CONTINUITY AND DIFFERENTIBILITY

INTRODUCTION:

Geometricaly the definition of continuity of a function x = a implies that the graph of thefunction has no break at x = a. The graph cannot jump immediately to a point above orbelow the line y = (a).

y = f(x)

O ax

f(x)

Continuous function at x = a Discontinuous function at x = a

y = f(

x)

O ax

f(x)

CONDITIONS FOR A FUCNITONS TO BE CONTINUOUS

(i) Left hand limit of f(x) at x = a– exists.

(ii) Right hand limit of f(x) at x = a+ exists.

(iii) Left hand limit of f(x) = Right hand limit of f(x) = Value of function (at x = a).

Note: A function which is not continuous at x a is known as discontinuous at x = a.

y = f(x) y = f(x) y = f(x)

f(a)

y=f(x)

O O Oa a ax x x

f(x) f(x) f(x)

Left hand limit f(x) Right hand limit f(x)Limit f(x) does not exist x a f(x) is not continuous at x = a

Left hand limit f(x) Right hand limit f(x) Limit f(x) exist at x = a f(x) is

continuous x = a

f(a)

Left hand limit f(x) Right hand limit f(x)The funciton f(x) is continuous at x = a

f(a)

Let f be a real function and a be in the domain of f. We say f is continuous at a, if

x 0 x a x aLimf x f a Limf x Limf x f a

Hence f(x) is continuous if x aLim f x

exists and equals f(a).

The function f(x) is said to be continuous at x = a from left if x 0 x 0Limf x f a Limf x f a

DISCONTINUITY OF A FUNCTION:

A function f(x), which is not continuous at a point x = a, is said to be discontinuous atx = a. The function f(x) can be discontinuous at a point x = a in any one of the followingaways.

• f(a) is not defined

• Either x 0 x 0Limf x or Limf x

or both non existing ro infinite

• LHL and RHL both exist but unequal, i.e. x 0 x 0Limf x Limf x

• LHL and RHL both exist and equal but not equal to f(a), i.e., x 0 x 0Limf x Limf x f a

This particular kind is called REMOVABLE DISCONTINUITY

2 LOCUS LEARNERSS TREE,HYDERABAD, INDIA

SPARK SUMMIT-12 Maths-C and D

Continuity of a Function in an Interval:

• A function f(x) is said to be continuous in an open interval (a,b), if f(x) is continuousat every point of the interval.

• A function f(x) is said to be continuous in a closed interval [a,b] if f(x) is continuous in(a,b). In addition f(x) is continuous at x = a from right and f(x) is continuous at x = bfrom left.

SOME CONTINUOUS FUNCTION: Every constant function is continuous at all points.Proof : Let f(x) = c, where c is constant.

The domain of a constant function is R. Let a be an arbitrary numbe in D, then

f(a) = c, x a x alimf x limc c

From (1) and (2), we have x alimf x f a

Thus, f(x) is continuous at x = a, for all x R .

Hence, f(x) is continous. Since, a is an arbitrary number in R.Therefore, f(x) is continuous everywhere in R.

The identity function is continuous.

Proof: Lef f(x) = x, for every x RLet a be an arbitrary real number.

Then, f(a) = a .....(1) and x a x alim f x limx a

.....(2)

From (1) and (2), we get x alim f x f a a

Thus, f(x) is continuous at x = a, a R . Hence the identity function is continuous in R.

A polynomial function is everywhere continuous.

Proof: Let 2 n0 2 2 nf x a a x a x ...a x , n I, n 0, x R be a polynomial function.

We shall prove the theorem by induction on n.

Step 1: When n = 0, we get 0f x a

which is a constant function therefore is continuous.

When n = 0, we get 0 1f x a a x

which is the sum of constant function and a multiple of the identity function. Itbeing the sum of two continuous functions is continuous everywhere.

Step 2: Let every polynomial function of degree at most n be everywhere continuous.Consider a general polynomial function of degree (n + 1), be

2 n n 10 1 2 n n 1 n 1g x a a x a x ...a x a x , where a 0

n 1 n0 1 2 n n 1g x a x a a x ..... a x a x

It is the sum of a constant function a0 (which is everywhere continuous) and the productof identity function x (which is everywhere continuous) and the polynomial function

n1 2 n 1a a x ...a x of degree at most n (which is everywhere continuous by induction

assumption). Therefore, g(x) is everywhere continuous.Hence, by the principle of mathematical induction, a polynomial function is every wherecontinuous.

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If f and g be continuous functions in D, then,

(i) fg is continuous (ii) f – g is continuous

(iii) c f is continuous. (iv) fg is continuous.

(v) fg is continuous in D except those points where g x 0

(vi) 1is continuous on D x : f x 0

f

Proof: Let a be an arbitrary number in D. Since f and g are continuous on D. so they arecontinuous at a also.

x a x alim f x f a and limg x g a

(i) x a x alim f g x limg x f a g a f g a

.....(1)

And (f + g)(x) = (f + g)(a) .....(2)

From (1) and (2), we have (f + g) is continuous at x = a.

Since a is an aribtrary number in D. Hence, (f + g) is continuous in D.

(ii) x a x a x a x alim f g x lim f x g x limf x limg x f a g a f g a

.....(1)

f g x f g a at x a .....(2)

From (1) and (2) , we get x alim f g x f g a

Thus, (f – g) is continuous at x = a. Since a is an aribitrary number in D.

Hence, (f – g) is continuous in D.

(iii) x a x a x alim c f x lim c f x c limf x c f a

.....(1)

c f x c f x c f a at x a c f a .....(2)

From (1) and (2) , we have x alim c f x c f a

Thus, (c f) is continuous at x = a. Since a is an aribitrary number in D.

Therefore, (c f) is continuous in D.

(iv) x a x a x a x alim f g x lim f x g x lim f x limg x

f a g a f g a .....(1)

At x = a, f g x f g a .....(2)

From (1) and (2), we get x alim f g x f g a

Thus, f g is continuous at x = a. Since a is an aribitrary number in D.

Therefore, f g is continuous in D..



(v)

x a

x a x ax a

limf xf x f af flim x lim a

g g x limg x g a g

.....(1)

f fAt x a, x a

g g

.....(2)

From (1) and (2), we have x a

f flim x a

g g

Thus, 1f

is continuous at x = a. Since a is aribtrary point in D such that f a 0 .

Hence, 1f

is continuous on D x : f x 0 .

If f is continuous on its domain d, then |f| is also continous on D.Proof. Recall that |f| (known as absolute function) is defined as :|f|(x) = |f(x)|

Let a be an aribitrary real number in D. Then, f is continuous at a.

f x f a at x a .....(1) x alim f x f a

Now, x a x alim f x lim f x

[By definition of |f|] x alimf x f a f a

.....(2)

From (1) and (2), we have x alim f x f a

|f| is continuous at x = a.

Since a si a arbitrary point in D. Therefore, |f| is continuous in D.

Remark: The converse of the above theorem not be true.

The composition of two continuous functions is a continuous function.

Proof: Let f and g be two real functions such that gof exists. Then,

Then, a Domain f f a Range f

f a Domain g Range f Domain g

Since f and g are continuous on their domains,

Therefore, a Domain f and f a Domain g

f is continuous at x = a and g is continuous at f(a)

x a x f alim f x f a and lim g y g f a

x a f x f alim f x f a and lim g f x g f a , where y f x

x alim g f x g f a

x a f x f a

x alim gof x gof a

gof is continous at x a .

Since a is an arbitrary piont in its domain, Hence, gof is continuous

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The logarithmic function is continuous in its domain.

Proof: Let f(x) = logcx, where c > 0, be the logarithmic function. Domain of f is 0,

Let a ab an arbitrary point in 0, . Then,

Right hand limit (at x = a). h 0x a

lim f x lim f a h

[Putting x = a + h]

c ch 0 h 0

hlimlog a h lim log a 1

a

c

c c ch 0 h 0 h 0

hlog 1h ha

lim log a log 1 lim log a limha aa

c

c c ch 0 h 0

hlog 1

halog a lim lim log a logc 0 log a f a

h aa

.....(1)

Similarly, x alim a f a

.....(2) c cf x log x f a log a .....(3)

From (1), (2) and (3), we get x a x alim f x lim x f a

So, f(x) is continuous at x = a. But a is a real number in 0,

Hence, log x is continuous in 0, .

The exponential function ax, a > 0 is everywhere continuous.

Proof: Let f(x) = ax, we have

x

x x

x 0 x 0 x 0

a 1lim a lim a 1 1 lim x 1

x

x

x 0 x 0 x 0

a 1lim lim x lim 1

x

x

ex 0

a 1lim log a

x

elog a 0 1 0 1 1

Let c be an arbitrary real number. Then,

Left hand limit (at x = c). 1 h 0x c

lim f x limf c h

[Putting x = c – h]

c h c h c c c ch 0h 0 h 0 h 0

1 1 1lim a lim a a a lim a a a f c

1a a

0a 1

Right hand limit (at x = c).

h 0x c

lim f x lim c h

[Putting x = c + h]

c h c h c h

h 0 h 0 h 0lim a lim a a a lim a

c 0 c ca a a 1 a f c 0a 1

Left hand limit (at x = c) = Right hand limit (at x = c) = f(c)

So, f(x) is continuous at x = c. Since c is an arbitrary real number.

Corollary: ex is everywhere continuous.



SOLVED EXAMPLES

1. Find the points at which the function 2

3x + 7f x =x 5x + 6

is continuous.

Sol: The function 2 2

3x 7 3x 7f xx 5x 6 x 3x 2x 6

=

3x 7 3x 7x x 3 2 x 3 x 2 x 3

The function is not defined if x2 – 5x + 6 = 0 at x = 2 and x = 3

The numerator of the function is continuous at every point. Also the function in the denominatoris continuous at every point.

f(x) is continuous at every point R, except for x = 2 and x = 3, where is not defined.

f(x) is continuous on the points R – { 2, 3}.

2. Show that the function f(x) = 2x – |x| is continuous at x = 0.

Sol : We have, 2x x, when x 0

f x2x x , when x 0

or x, when x 0f x

3x, when x 0

f(0) = 0 .....(1)Left hand limit (at x = 0)

= x 0 hx 0

lim f x lim 2x x

[Putting x = 0 – h as x 0, h 0 ]

= h 0lim 3 0 h 3 0 0 0

.....(2)

And right hand limit (at x = 0) = x 0 hx 0

lim f x lim x

= h 0lim 0 h 0 0 0

.....(3) [Putting x = 0 + h ]

From (1), (2) and (3), we have x 0 x 0lim f x lim f x f 0

Hence, function f is continuous at x = 0.

3. Is the function f defined by x, if x 1f x

5, if x 1

, continuous at x = 0? at x = 2?

Sol: We have, x, if x 1f x

5, if x 1

At x = 0. First note that the function is defined at the given point x = 0 and its value is 0.Then find the limit of the function at x = 0.

Left hand x 0 h h 0x 0

lim f x lim x lim 0 h 0

; Right hand x 0 h h 0x 0


Also, f(0) = 0. Thus, L.H.L. = R.H.L. = f(0).

At x = 1. First note that the function is defined at the given point x = 1 and its value is 1.

Then find the limit of the function at x = 1.




lim f x lim 5 lim 5 5

Thus, L.H.L. R.H.L. Hence, f is not continuous at x = 1.

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At x = 2. First note that the function is defined at the given point x = 2 and its value is 2.

Then find the limit of the function at x = 2.





Also, f(2) = 5. Here, L.H.L. = R.H.L. Hence, f is not continuous at x = 2.

4. If sin 3x

,when x 0f x x

1, when x 0

, find whether f is continuous at x = 0.

Sol: We have, sin 3x

,when x 0f x f 0 1x

1, when x 0

.....(1)

and x 0 x 0 x 0

sin 3x sin 3xlim f x lim lim 3

x 3x

3x 0

sin 3x3 lim 3 1 3

x .....(2) x 0

sin xlim 1

x

From (1) and (2), we have x 0lim f x f 0

. Therefore, f is not continuous at x = 0.

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

ax 1, if x 3f x

bx 3, if x 3

is continuous at x = 3.

Sol: We have, ax 1, if x 3f x

bx 3, if x 3

First note that the function is defined at the given point x = 3 and its value is 3a + 1.Then find the limits of the function at x = 3.


lim f x lim ax 1 lim a 3 h 1 3a 1

Right hand x 3 h h 0x 3

lim f x lim bx 3 lim b 3 h 3 3b 3

For continuity, Left hand limit = Right hand limit

2 23a 1 3b 3 3a 3b 2 a b a b

3 3

6. For what values of k, is the following function continuous at x = 0?

2

1 cos 4x, x 0

f x 8xk, x 0

[CBSE 2000 C]

Sol: Left hand 2x 0 x 0

1 cos 4xlim f x lim

8x

2222

2 2x 0 h h 0 h 0

2sin 2 0 h2sin 2x sin2hlim lim lim 1 1

2h8x 8 0 h



Right hand 2 22

2 2 2x 0 h h 0 h 0 h 0x 0

1 1 2sin 2h1 cos 4x 2sin 2h sin 2hlim f x lim lim lim lim 1

2h8x 8h 8h

Also, f(0) = k

For continuity at x = 0, x 0lim f x f 0 1 k k 1

.

7. A function f(x) is defined as 1x

1 , if x ¹ 0f x = 1+ e

0, if x = 0

. Is the function continuous at

x = 0 ?

Sol: Left hand limit = 1 1/0 hx 0 h h 0x 0x

1 1lim f x lim lim1 e1 e

[Putting x = 0 – h]

= h 0

1/h

1 1 1lim 11 1 1 01 1e e

.....(1)

Right hand limit = 1 1/0 hx 0 h h 0x 0x

1 1lim f x lim lim1 e1 e

= 1h 0h

1 1lim 01 e1 e

.....(2)

From (1) and (2), we have

Left hand limit of f(at x = 0) Right hand limit of f(at x = 0) = f(0) x 0lim f x does not exist

Hence the function f is not continuous at x = 0.

8. A function f(x) is defined by 2

22

2

x 1, for x 1f x x 1

0 , for x 1

. Discuss the continutity of

f(x) at x = 1 .

Sol: We have, 2

22

2

x 1, for x 1f x x 1

0 , for x 1

22

22

1, for 0 x 1

x 11 1

0, for 1 x 2x 1

2x 1x 1

1Lim f x Lim

x 1

So, x 1lim f x

does not exist. Hence, f(x) is not continuous at x = 1

9. If

3x

2

4 1, x 0

f x sin x / 4 log 1 x /3

k , x 0

is continuous at x = 0, find k .

Sol: Since f(x) is continuous at x = 0. Therefore, x 0Lim f x f 0

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3x

2x 0

4 1Lim k

x xsin log 1

4 3

3x

x 02

2

4 112

xLim k

xsin log 1 x /34x / 4 x /3

312 log 4 k 3k 12 log 4

10. Discuss the continuity of the following functions(i) f(x) = sin x + cos x (ii) f(x) = sin x – cos x (iii) f(x) = sinx × cos x

Sol: We have proved tha sin x and cos x are the continuous functions.(i) f(x) = sin x + cos x being the sum of two continuous functions is continuous wherever

it is defined.(ii) f(x) = sin x – cos x being the difference of two continuous functions is continuous it

is defined.(iii) f(x) = sin x × cos x being the product of two continuous functions is continuous

wherever it is defined.

11. If a function f is defined as x 2

, x 2f x x 20, x 2

. Show that f is everywhere

continuous except x = 2.Sol: We have,

x 21, x 2

x 2x 2x 2,x 2f x f x 1, x 2x 2 x 2

0, x 2 0; x 2

x 2 ,x 2x 2

x 2, x 2

Case I: When x < 2, we have f(x) = –1, which, being a constant function, is continuous ateach point x < 2.

Case II: Also, When x < 2, we have f(x) = 1, which, being a constant function, iscontinuous at each point x > 2.

Case III: Let us consider the point x = 2 we have, left hand limit (at x = 2).

x 2 x 2lim f x lim 1 1

.....(1)

and right hand (at x = 2) x 2 x 2lim f x lim 1 1

.....(2) and f(2) = 0 .....(3)

From (1), (2) and (3), we have x 2 x 2

Left limit f x Right limit of f x f 2

Thus, f(x) is not continuous at x = 2.Hence, f(x) is everywhere continuous, except at x = 2.



CLASSROOM WORKSHEET - 1

1. Examine that sin |x| is a continuous function. [NCERT]

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

3. 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. [NCERT]

4. For what value of k is the following function continuous at x = 2?

2x 1, x 2

f x k, x 23x 1, x 2

5. For what value of k is the function 2kx , x 2

f x3, x 2

continuous at x = 2. [NCERT]

6. Discuss the continuity of the function f(x) = sin2x + x2 – 2x at, the point x = 0.

7. If 1xsin , x 0

f x = x0. x = 0

. Find whether f is continuous at x = 0.

8. Show that sin x

, x 0f x x

x 1, x 0

is a continuous function. [NCERT]

9. Discuss the continuity of the function f(x) = sin |x|.

10. Discuss the continuity of the function f(x) = |x| + |x – 1| in interval [–1, 2].

11. Show that sec x is continuous function.

12. Show that x 4

, x 4f x x 40, x 4

is continuous at each point except 4.

13. Find the value of , so that 3x 8, x 5f x

2 , x 5

is continuous.

14. Find the value of a, b so that 2, if x 3

f x ax b, if 3 x 5

9, if x 5

is continuous.

15. If

2

2

2

x, 0 x 1

a

f x a, 1 x 2

2b 4b, 2 x

x

is continuous 0 x , find a, b..

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HOME WORKSHEET - 1

1. Show that the function f(x) = 2x – |x| is continuous at x = 0. [CBSE 2002]

2. If x

, x 0f x ,x0, x 0

find whether f(x) is continuous at x = 0. [NCERT]

3. If the function 3ax b, if x 1

f x 11, if x 15ax 2b, if x 1

is continuous at x = 1, find the values of a and b.

[CBSE 2000]

4. Find the value of k so that the function f is continuous at the indicated point

k cos x

, if x2x 2f x at x

23, if x

2

. [NCERT]

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

[NCERT]

6. Find the value of k so that the 2kx , x 1

f x at x 14, x 1

functions f are continuous at the

indicated points. [CBSE 2007]

7. A function f(x) is defined as 1+ x when x < 2f x =

5 - x when x 2

Is the function continuous at x = 2 ?

8. For the function 2

x, x 0f x a, x 0

x , x 0

find the value of ‘a’ that will make it continuous at x = 0.

9. If f(x)=2

2

sin axf(x)x

for x 0,f(0) 1 is continuous at x=0 then a=

10. A function ‘f’ is defined as sin(a 1)x sin x

x

when 0x and 1 22 1/2

3/2

x bx xbx

when

x > 0, the value of ‘a’ when f is continuous at x=0 b 0 is

11. The value of f(0) so that f(x) = 2/x1 5x is continuous at x = 0 is

12. If kx

2

e 1 sinkxf(x) , x 0 and f(0) 9

4x

, then find k is



13. Discuss the continuity of the function f(x) at x = 0 if 2x 1, if x 0f x

2x 1, if x 0

.

[CBSE 2002]

14. Show that the function sin x

, when x 0f x x

2, when x 0

is continuous.

15. Find the value of so that x 5, x 2f x

x 1, x 2

is continuous.

16. Find the value of a, b so that 1, if x 0

f x ax b, if 3 x 57, if 5 x

is continuous.

17. If 2a ax b, 0 x 2

f x 3x 2, 2 x 42ax 5b, 4 x 8

is continuous on [0, 8] find the values of a and b..

OBJECTIVE TYPE QUESTIONS

1. The function 1/x

1/x

x.e 1f(x) sin for x 0,1 e x

f(0) = 0 at x = is

1) continuous 2) dis-continuous 3) not determinant 4) none

2. Let f(x) = 1cos cos x then f(x) is

1) continuous at x 2) discontinuous at x= 3) discontinuous at x= 4) none of these

3. Let a function f be difined by x | x |

f(x)x

for x 0 and f(0)=2, then f is

1) continuous no where 2) continuous every where3) continuous for all x except x=1 4) continuous for all x except x=0

4. If 2

2

x , for x rationalf x

x , for x irrational

, then

1) f is continuous at x = 0 and x = 1/22) f is discontinuous at x = 0 and x = 1/23) f is continuous at x=0 ans discontinuous at x = 1/24) f is discontinuous at x=0 and continuous at x = 1/2

5. The function 2

5x 4 for 0 x 1f(x) 4x 3x for 1 x 2

3x 4 for x 2

1) continuous at x = 1 2) dis-continuous at x = 13) continuous at x = 2 4) dis-continuous at x = 2

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6. The function 3x 27f(x) for x 3, f(3) 27x 3

at x = 3 is

1) continuous 2) dis-continuous 3) not determined 4) none

7. Column-I Column-II

a) f(x) |x| |x 1| p) f(x) is discontinuous at x = 1

b) f(x) [x] |1 x| q) f(x) is continuous at x = 1

c) 2

|x 3|, x 1f(x) x 3x 13,x 1

4 2 4

r) f(x) is continuous every where

d)

2

2

x ,for x rationalf(x)

x , for x irrational

s) f(x) is continuous at x = 0

8. Column -I Column -II

a) xe is continuous on p)(0,)

b) logx is continuous on q) R-{ 0 }

c) cosx is continuous on r) R (2n 1) ;n z2

d) secx is continuous on s) R

9. The 2

2x 3f(x)2x 5x 2

is discontinuous at x = a, x = b then 2(a + b) = ______

SYNOPSIS

DIFFERENTIABILITY AT A POINT:

Let f(x) be a real valued function defined on an open interval (a, b) where c a,b .

Then f(x) is said to be differentiable or derivable at x = c,

iff, x c

f x f clim

x c

exists finitely..

This limit is called the derivative or differential coefficient of the function f(x) at

x = c, and is denoted by f c or D f(c) or x c

d f xdx .

Thus, f(x) is differentiable at x = c. ; x c

f x f clim

x c

exists finitely



x c x c

f x f c f x f clim lim

x c x c

;

h 0 h 0

f c h f c f c h f clim lim

h h

Hence,

h 0h c

f x f c f c h f clim lim

x c h

is called the left hand derivative of f(x) at

x = c and is denoted by f c or LF c .

While,

h 0h c

f x f c f c h f clim lim

x c h

is called the right hand derivative of f(x) at x = c

and is denoted by f c or Rf c .

Thus, f(x) is differentiable at x = c. L f c = R f c .

If f c R f c , we say that f(x) is not differentiable at x = c.

DIFFERENTIABILITY IN A SET:

• A function f(x) defined on an open interval (a,b) is said to be differentiable or derivable inopen interval (a,b) if it is differentiable at each point of (a,b).

• A function f(x) defined on [a,b] is said to be differentiable or derivable at the end points aand b if it is differentiable from the right at a and from the left at b.

In other words

x a

f x f alim

x a

and

x b

f x f blim

x b

both exists.

“If f is derivable in the open interval (a,b) and also at the end points a and b, then f is saidto be derivable in the closed interval [a,b]”.

A function f is said to be a differentiable function if it is differentiable at every point of itsdomain.”

DIFFERENTIABILITY OF A FUNCTION:

The function f(x) is said to be differentiable at x = a, if

h 0 h 0

f a h f a f a h f alim lim or R f ' a L f ' a

h h

The common value of R f ' a and L f ' a is denoted by f ' a and is known as derivative of

f(x) at x = a.

If howover R f ' a L f ' a we say that f(x) is not differtiable at x = a.

Note:

• On both sides of the equation h is taken as positive and very small.

• Every differentiable function is continuous. But, every continuous function need not bedifferentiable.

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When we say d

sinx cos xdx

.

Here, we derive the formula from the first principles

h 0 h 0

sin x h sinx sin x h sinxdsinx lim lim cos x

dx h h

Whenever we define a derivative we put a condition provided the limit exists if the limit doesnot exist

E.g.h 0 h 0

d x 0 h 0 0 h 0 d xlim lim is not differentiable at x 0.

dx h h dx

DERIVATIVE OF f(x) AT THE POINT x = a

Derivative of funtion f(x) with respect to x at the point x = a is defined as

h 0

f a h f alim

h

Provided the limit exists and denoted by f ' a .

A function is said to be differentiable at x = a if it has a derivative there. A function is said tobe differentiable on an interval if it is differentiable at every point of the interval.

EXISTENCE OF THE DERIVATIVE:

The limit must be the same whether h approaches 0 from the right or from the left.Then the derivative exists. If these limits are different, the derivative does not exist and thefunction will not be derivative.

The function is not derivable i.e. f ' does exist if either of the left handed or righthanded limits:

(i) does no exist, or (ii) both exist but they are not equal.

GEOMETRICAL MEANING OF DIFFERENTIABILITY AT A POINT x = a:

Let us draw a curve y = f(x). The coordiantes of a point P on the curve are [a, f(a)]. Letpoint Q [a + h, f(a + h)] on the right side of P, and R[a – h, f(a – h)] on the left side of P.

Draw perpendiculars PS and RT.

Slope of chord f a h f aQS

PQPS h

; Slope of chord

f a h f aPTPR

RT h

f(x)

Q

S

XQ

Rh h

T

P

[(a–h, f(a–h)]

[(a+h, f(a+h)]

[a, f(a)]

When Q tends to P, then the chord PQ becomes a tangent at P. As h 0 . Points Qand R both tend to P from right hand side and from left hand side respectively. Slope ofthe tangent is given by



Q P R Plim Slope of chord PQ lim Slope of chord PR

h 0 h 0

f a h f a f a h f alim lim

h h

Now f(x) is differentiable at x = a. if

h 0 h 0

f a h f a f a h f alim lim

h h

Slope of the tangent at P, which is limiting position of the chords drawn on the leftside of P is the same as the slope of the tangent at P.

Thus f(x) is differentiable at the point P, iff there exists a unique tangent at P.

Every differentiable function is continuous. But, every continuous function neednot be differentiable.

Proof: Let the function f be differentiable at x = a. Then by definition:

h 0

f a h f alim f ' a

h

.....(1)

Now, f a h f af a h f a h

h

h 0 h 0 h 0

f a h f alim f a h f a lim lim h

h

h 0lim f a h f a f ' a 0 0

[From (1)] h 0lim f a h f a

Therefore f is continuous at a. Since a is chosen arbitrary so f is continuous everywhere.Hence, every differentiable function is continuous.If order to show that a continuous function need not be differentiable, it is sufficient togive an example of a function which is continuous but no differentiable.Consider f(x) = |x| at x = 0.

For continuity: Left hand limit at x = 0 h 0 h 0 h 0 h 0lim f 0 h lim f h lim h lim h 0

and Right hand limit at x = 0. h 0 h 0 h 0 h 0lim f 0 h lim f h lim h lim h 0

Here, Left limit = Right limit = f(0) = 0. f is continuous at x = 0.

B A

xx` O

135º

f(x)

f(x)=|x|

f(x)=

|x|

Again for differentiable:

h 0 h 0 h 0 h 0

0 h 0f 0 h f 0 hR f ' 0 lim lim lim lim 1 1

h h h

.....(1)

and h 0 h 0 h 0

f 0 h f 0 hL f ' 0 lim lim lim 1 1

h h

.....(2)

From (1) and (2), we get R f ' 0 L f ' 0 .

Therefore it is not differentiable at x = 0. Hence a continuous function need not bedifferentiable.

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SOLVED EXAMPLES

1. Show that f(x) = x2 is differentiable at x = 1 and find f ' 1 .

Sol: We have, 2f x x 2 2 2

h 0 h 0 h 0

f 1 h f 1 1 h 1 1 h 2h 1R f ' 1 lim lim lim

h h h

2

h 0 h 0 h 0

h h 2h 2hlim lim lim h 2 2

h h

.....(1) [Putting h = 0]

Now, 2 2 2

h 0 h 0 h 0

f 1 h f 1 1 h 1 1 h 2h 1L f ' 1 lim lim lim

h h h

h 0 h 0

h h 2lim lim h 2 2

h

.....(2) [Putting h = 0]

From (1) and (2), we have L f ' 1 R f ' 1 2

This shows that f(x) is differentiable at x = 1 and f ' 1 2 .

2. If f is defined by 2f x x 2x 7, find f ' 3 .

Sol: We have, 2f x x 2x 7

2 2

h 0 h 0

3 h 2 3 h 7 3 2 3 7f 3 h f 3R f ' 3 lim lim

h h

2

h 0 h 0 h 0

h h 89 h 6h 6 2h 7 9 6 7lim lim lim h 8 8

h h

.....(1)

and 2 2

h 0 h 0

3 h 2 3 h 7 3 2 3 7f 3 h f 3L f ' 3 lim lim

h h

2 2

h 0 h 0 h 0 h 0

h h 89 h 6h 6 2h 7 9 6 7 h 8hlim lim lim lim 8 h 8

h h h

.....(2)

From (1) and (2), we have L f ' 3 R f ' 3 8 .

3. Discuss the differentiability of f(X) at x = 0, where 1/ x

1 / x

e 1x ; x 0

f x e 10 ; x 0

.

Sol:

1/ x

1 / x

h 0 h 0 h 0

e 1h 0f 0 h f 0 f h f 0 e 1R f ' 0 lim lim lim f x 0 at x 0h h h

1 /h 1/h

1/h 1/hh 0 h 0

e 1 1 elim lim

e 1 1 e

(Multiplying by 1/he )



h 0

1 e 1 0lim 1

1 01 e

.....(1)

1 1When h 0, and e 0

h 0

and 1 /h

1 /h

h 0 h 0 h 0

h e 1 0f 0 h f 0 f h f 0 e 1L f ' 0 lim lim lim

h h h

1/h

1 /hh 0

e 1 e 1 0 1lim 1

0 1e 1 e 1

.....(2)

1e 0 and

0

From (1) and (2), we get R f ' 0 L f ' 0

Hence, f(x) is not differentiable at x = 0.

4. Prove that x is continuous at x = 0 but not differntiable at x = 0.

Sol: Let f(x) = x , Df = R.

x 0 x 0Lt f x Lt x 0 f 0

f is continuous at x = 0.

Now h 0

f 0 h f 0f 0 Lt

h

h 0 h 0

h 0 h 0Lt Lth h

h 0

hLt 1h

and h 0

f 0 h f 0f 0 Lt

h

h 0 h 0

h 0 h 0Lt Lth h

h 0

hLt 1h

Hence f 0 f 0 f is not differentiable at x = 0.

5. Discuss the differentiability of the following functions at x = 0 ?

(i) cos x x (ii) cos x x (iii) sin x x (iv) sin x x

Sol: cos x cos x or cos x . Thus is any case cos x cos x for all x R . Since h x x is

not differentiable at x = 0, so cos x x cos x x is not differentiable at x = 0.

sinx x if x 0f x sin x x

sin x x if x 0

Now f 0 2, f 0 2 so f is not differentiable at x = 0. Finally

sinx x if x 0g x sin x x

sin x x if x 0

In this case g 0 0 and g 0 0 . Thus sin x x is differentiable at x = 0.

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

2. Prove that the function f given by f x x 1 , x R is not differentiable at x = 1.

[NCERT]

3. For what values of a and b the function 2x , if x 1

f x2ax b if x 1

is differentiable at x = 1.

4. Discuss the differentiability of the function 2x , x 0

f xsinx, x 0

at the point x = 0.

5. Show that the function f(x) defined as 1

x cos , x 0f x x

0, x 0

is continuous at a point

x = 0 but not derivable at that point.

6. Examine for continuity and differntiability the function 1xsin , x 0

f x x0 , x 0 at x 0.

HOME WORKSHEET - 2

1. Discuss the continuity and differentiablity of 2 1

x sin , if x 0f x x

0, if x 0

. [NCERT]

2. Discuss trhe differentiability of f(x) at the given point 2x , if x 1

f x2x 1, if x 1

at x = 1.

3. Discuss trhe differentiability of f(x) at the given point 2

12x 13, x 3f x at x 3

2x 5, x 3

.

4. For what choice of a and b is the function.

2x , x c

f xax b , x c

; differentiable at x = c.

5. Show that xf x1 x

is differentiable for all x R .


SPARK SUMMIT-12 Maths-C and DOBJECTIVE TYPE QUESTIONS

1. The set of all points where the function f(x) = x x is differentiable is

(1) , (2) , 0 0, (3) 0, (4) 0,

2. If 1

x 2 for x -2

f x tan x 2 2 for x = -2

then

(1) f is continuous at x = –2 (2) f is not derivable at x = –2

(3) f is not continuous at x = –2 (4) f is derivable at x = –2

3. If x for 0 x 2f x

2 for x 2

, then f(x) is not differentiable at x = ___________

(1) 1 (2) 2 (3) 3 (4) 4

4. If 3 x, for x 0f x

3-x, for x<0

, then f(x) is not differentiable at x = ____________

(1) 0 (2) 1 (3) 2 (4) 3

5. If f is an even function and f1(x) exists then f1(0) = ____________

(1) 0 (2) 1 (3) –1 (4) f(0)

6. The set of points where the function f x x x is differentiable is

(1) , (2) R – {0} (3) 0, (4) None of these


a) x x p) continuous in (–1, 1)

b) x q) differentiable in (–1, 1)

c) x x r) differentiable in (0, 1)

d) x 1 s) not differentiable in (–1, 1)


a) 3f x x is p) continuous in (–1, 1)

b) f x x is q) differentiable in (–1, 1)

c) 1f x sin x is r) differentiable in (0, 1)

d) 1f x cos x is s) not differentiable in (–1, 1)

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SYNOPSIS

INTRODUCTION: Now we will learn differentiation of inverse trigonometric functions,exponential function and logarithmic functions.

INCREMENT: Let y = f(x) be a function of x. A small change in the value of x is called theincrement in x. And the change in the value of y corresponding to a small change in thevalue of x is called the increment in y.

Let y = x2, when x = 3, we have y = 9.

If x changes from 3 to 3.1 then increment in x = 3.1 – 3 = 0.1

The corresponding value of y will be (3.1)2 = 9.61 and the increment in y in this case= 9.61 – 9 = 0.61.

A small change in the value of x, as it increases or decreases from one value x = x1 toanother value x = x2 is denoted by x (read as “delta x”) and is called increment in x.

Hence 2 1x x x .

If should be remembered that x is not the product of and x but is simply a symbol for

a small change in x. Similarly, a small change in y or increment in y is y .

DIFFERENTIAL COEFFICIENT:

The limit of incremental ratio i.e. ylim and x

x

approaches zero is called the differential

coefficient of y with respect to x and is denoted by dydx

.

x 0 x 0

f x x f xdy y dlim f x lim

dx x dx x

STANDARD FORMULAE ALREADY DONE:

S.no Function Derivative

1 xn n xn–1

2 sin x cos x

3 cos x – sin x

4 tan x sec2x

5 cot x –cosec2x

6 sec x sec x tan x

7 cosec x –cosec x cot x

8 dyuv

dx

dv duu v

dx dx

9 d udx v

2

du dvv u

dx dxv


SPARK SUMMIT-12 Maths-C and DDIFFERENTIAL COEFFICIENT OF A FUNCTION OF A FUNCTION (CHAIN RULE)

Let y be a function of u and u a function of x.

Let x, u and y be the corresponding increments in x, u and y respectively..

x 0 x 0 x 0

y y u y y ulim lim lim as x 0 hence u 0

x u x x u x

dy dy dudx du dx

If y = f(t), = g(u), u = h(x). Then dy dt dudt du dx

E.g. Let y = sin x2 y sin u, Putting u = x2 dy

cosudu

2 duu x 2x

dx

Now, 2 2dy dy du dycosu 2x cos x 2x 2x cos x

dx du dx dx .

DIFFERENTIATION OF IMPLICIT FUNCTIONS

Implicit function. An equation of the form f(x, y) = 0, in which y is not expressible directlyin terms of, is known as an implicit function of x and y.

(or)

If a function in x and y is given, in such a way that x and y cannot be separated inleft hand side right hand side easily, then the function is called implicit function.

Let us consider,

(i) y – x + 2 = 0 (ii) y – sin xy = (iii) ax2 + 2hxy + by2 = 0

Here, (i) in first example y can be expressed in terms of x. It is explicit function.

In example (ii) and (iii) it is not easy way to solve for y i.e., y and x cannot beseparated on left hand side and right hand side easily. They are implicit functions.

We should keep in mind that derivative of x w.r.t ‘x’ is 1 and derivative of y w.r.t. ‘x’ is dy

.dx

E.g. 3 2d dy dy 3y , sin y cos y

dx dx dy 2 2d d dy

y 2y. But y 2ydy dx dx

SOLVED EXAMPLES

1. Find the derivative of abslute value functions.

Sol: Let 1

2 2 2y x x x by definition

Differentiating w.r.t. x, we get

1

2 22

2 2

dy 1 d 1 x xx x 2x

dx 2 dx x2 x x

d x

x , x 0dx x

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2. Differentiate : (i) 7(3x 8) (ii) 2(3x 2x 9)

Sol: (i) Let 7y (3x 8) , 7y u Putting u 3x 8 du

3 ....(1)dx

6dy7u ......(2)

du

We know that, 6dy dy du

. 7u .(3)dx du dx

[Using (1) and (2)]

6 621u 21(3x 8) [ u 3x 8]

(ii) Let 2y 3x 2x 9 ; y udy 1dx 2 u

Putting 2 du

u 3x 2x 9 6x 2dx

Thereforedy dy du 1

. .(6x 2)dx du dx 2 u

2

1.(6x 2)

2 3x 2x 9

2[ u 3x 2x 9]

2

(3x 1)

3x 2x 9

3. Differentiate w.r.t.’x’ : (i) 5sinx (ii) 5sin x (iii) 2sin(x 3x 6)

Sol: (i) Let 5y sinx ,dy

y sinu cosudu

Putting 5 4du

u x 5xdx

We know that 4 4 5 5dy dy du. cosu.(5x ) 5x cos x [ u x ]

dx du dx

(ii) Let 5 5y sin x (sinx) , 5y u 4dy

5udu

Putting u sinx du

cos xdx

We know that, 4 4dy dy du. 5u .(cos x) 5(sinx) (cos x) [ u sinx]

dx du dx

45sin x.cos x

(iii) Let 2y sin(x 3x 6) , y sinudy

cosudu

Putting 2u x 3x 6 du

2x 3dx

We know that,

2 2dy dy du. cosu.(2x 3) [cos(x 3x 6)].(2x 3) [ u x 3x 6]

dx du dx

2(2x 3)cos(x 3x 6)



4. Find dydx

if 2x + 3y = sin y..

Sol: We have, 2x + 3y = sin y

Differentiating both sides with respect to x, we get

dy dy dy dy dy dy 22 3 cosy 3 cosy 2 cosy 3 2

dx dx dx dx dx dx cosy 3


1. Different iate the funct ions with respect to x: ( i) sin(x2 + 5) (ii) sin (ax + b). [NCERT]

2. Find the differential coefficient of (i) cos(sin x) (ii) sin[cos(tan x) [NCERT]

3. Differentiate the following functions with respect to x: 2

3 2

5xy sin 2x 3

1 x

[CBSE 2000 C]

4. If y sin x sinx sinx ..... , prove that dy cosxdx 2y 1

.

5. If sin y = x sin (a + y), prove that 2sin a ydy

dx sina

.

6. Find dydx

in the following: [NCERT]

(i) 2sin y cosxy (ii) 2 2sin x cos y 1

7. If 2 2ax 2hxy by 0 , find dydx

.

8. 2

dy 1If x 1 y y 1 x 0, show that .

dx 1 x

[NCERT]

9. 2

2 22

dy 1 yIf 1 x 1 y a x y , then prove that .

dx 1 x

HOME WORKSHEET - 3

1. Differentiate the functions with respect to x: (i) cos x (ii)sec(tan x) .

Differentiate the following functions with respect to x:

2. (9 + 7x)6. 3. (3 – 4x)5. 4. (3x2 – x + 1)4.

5. cos(sin x2). 6. tan(2x + 3). 7.2

3 4x2 x

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8. If cosy = x cos(a + y), prove that 2cos a ydy

dx sin a

. [NCERT]

9. If 22 2x y xy , find dydx

. [CBSE 2009]

10. If siny = x cos(a + y), prove that 2cos a ydy

dx cosa

.

11. If y = x sin y, prove that dy y

x .dx 1 x cos y

SOLUTIONS

CLASSROOM WORK SHEET-1

3. Let x n, n I . Then, x is integer..

Left hand x nx n

x n

lim x x lim x n 1 n n 1 1

Right hand x nx n

x n

lim x x lim x n n n 0

Since left hand limit Right hand limit f(x) is discontinuous at all integers n.

When x is real number.

Now, let x = p, where n < p < n + 1, n being an integer. Then

Left hand x px p

x p

lim x x lim x n p n

Right hand x px p

x p

lim x x lim x n p n

f(p) = p – [p] = p – n. Hence, f(x) is continuous at all non-integral points p.

4. We have, 2x 1, x 2

f x k, x 23x 1, x 2

Left hand x 2 x 2 h h 0 h 0lim f x lim 2x 1 lim2 2 h lim 5 2h 5

.....(1)

Right hand x 2 x 2 h h 0 h 0lim f x lim 3x 1 lim3 2 h lim 5 3h 5

.....(2)

f(2) = k .....(3)

For the function f(x) to be continuous:



Left hand x 2lim f x

= Right hand x 2lim f x f 2

From(1), (2) and (3), we have, 5 = 5 = k

Hence, for the function f(x) to be continuous at x = 2, the value of k is 5.

5. Function f is continuous, when 3

k4

.

6. We have, f(x) = sin2x + x2 – 2x f(0) = sin20 + 0 – 2(0) = 0 + 0 – 0 = 0

Left hand limit at x = 0 =

2 2

x 0 hx 0lim f x lim sin x x 2x

= 22

h 0lim sin 0 h 0 h 2 0 h

[Putting x = 0 – h]

= 2 2

h 0lim sin h h 2h 0 0 0 0

Right hand limit at x = 0

= 2 2

x 0 hx 0lim f x lim sin x x 2x

= 22

h 0lim sin 0 h 0 h 2 0 h

[Putting

x = 0 + h] = 2 2

h 0lim sin h h 2h

= 0 + 0 – 0 = 0

From (1), (2) and (3), we get

Left hand limit (at x = 0) = Right hand limit (at x = 0) = f(0).

Hence the function f(x) = sin2x + x2 – 2x is continuous at x = 0.

7. We have, 1xsin , x 0

f x = x0, x = 0

f(0) = 0 .....(1)

Left hand limit (at x = 0)

x 0 h h 0x 0

1 1lim f x lim xsin lim 0 h sinx 0 h

[Putting x = (0 – h)]

= h 0

1lim hsinh

[ sin(–x) = –sinx] = 0 ..... (2)

[ 0 × any number = 0]

Add right hand limit (at x = 0)

x 0 hx 0

1lim f x lim xsinx

= h 0

1limhsin 0h

..... (3) [Putting x = 0 + h]

From (1), (2) and (3), we find that x 0 x 0lim f x lim f x f 0

Hence, f is continuous at x = 0.

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8. We have sin x

, x 0f x x

x 1, x 0

Case I: When x > 0, f(x) = x + 1, which is a polynomial function and is continuous.

Case II: When x < 0, sin xf x

x , we also know that sin x and x both are continuous

functions for all x. sin xf x is continuous for every x 0

x .

Case II: Let us now consider at the point x = 0.

Left hand limit (at x = 0). x 0 h h 0x 0

sin 0 hsin xlim f x lim lim

x 0 h

[Put x = 0 - h]

h 0 h 0

sin h sin hlim lim 1

h h

x 0

sin hlim 1

h

.....(1)

And Right hand limit (At x = 0).

x 0 h h 0x 0

lim f x lim x 1 lim 0 h 1

[Put x = 0 + h] = 1 .....(2)

Also, f(0) = 0 + 1 = 1 .....(3) f x x 1,x 0

From (1), (2) and (3), we have

Left hand limit (at x = 0) = Right hand limit (At x = 0) = f(0)Hence,f(x) is continuous at x = 0.And therefore f(x) is continuous at every point.

9. Everywhere continuous 10. Everywhere continuous

13.72

14.7 17

a , b2 2

15. a 1, b 1 and a 1, b 1 2

HOME WORK SHEET-1

1. We have 2x x, when x 0 x, when x 0

f x or f x2x x , when x 0 3x, when x 0

f 0 0 .....(1)

Left hand limit (at x = 0) x 0 hx 0

lim f x lim 2x x

[Putting x = 0 – h as x 0, h 0 ]

h 0lim 0 h 0 0 0

.....(2)

And right hand limit (at x = 0) x 0 hx 0

lim f x lim x

h 0lim 0 h 0 0 0

.....(3) [Putting x = 0 + h]

From (1), (2) and (3), we have x 0 x 0lim f x lim f x f 0

.

2. f is not continuous at x = 0. 3. a = 3 and b = 2



4. We have,

k cos x

, if x2x 2f x

3, if x2

Left hand

h 0x hx 22

k cos hk cos x 2

lim f x lim lim 12x 2 h

2

H 0 h 0

k sin h k sin h klim lim

2h 2 2h 2

h 0

sin hsince, lim 1

h

Right hand

h 0 h 0x hx 22

k cos hk cos x k sin h2

lim f x lim lim lim2x 2h2 h

2

h 0

k sin h klim

2 h 2

h 0

sin hlim 1

h

Also, f 32

Here for continuous, L.H.L. = R.H.L.k k k

f 3 3 k 62 2 2 2

6. k = 4

7. Left hand x 2 h x 2 hx 2

lim f x lim 1 x lim 1 2 h 3

..... (1)

Right hand x 2 h h 0x 2

lim f x lim 5 x lim 5 2 h 3

.....(2)

f(x) = 5 – x f(2) = 5 – 2 = 3

From (1), (2) and (3), we have

Left hand x 2

f x

= Right hand x 2lim f x f 2

The given function f(x) is continuous at x = 2

8. f(0) = a

Left hand x 0 h h 0 x 0x 0

lim f x lim x lim 0 h limh 0

Right hand 22 2

x 0 h h 0 h 0x 0lim f x lim x lim 0 h limh 0

For the function to be continuous : Left hand x 0x 0

lim f x Right hand lim f x f 0

From (1), (2) and (3), we have 0 = 0 = a. a 0

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9.2

22

Lt sin axa 1 a 1

x 0 x

10. LHL = RHL = f(0)

1/ 2

1/ 2

1 1sin(0 1) sin0 0 . .

x bxlt ltx x cx xx b x x

11 12

a c 3 1,2 2a c

15. 2 16. a = 3, b = –8 17. a = 3, b = –2


1. Key: 2

L.H.L. =

1/

1/

. 1sin0 1

x

x

Lt x ex e x

0 01 0

a

finite quantity which is not unique

( )0

Ltf x

x

does not exist and hence f(x) is discontinuous at x = 0

2. Key -1 ; 1

2 x, 2 xx, x 0

cos cos x x, 0 x2 x, x 2

f(x) is continuous at x

3. Key -4 ;

2 if x 0f(x) 0 if x 0

2 if x 0

; thus ( ) 2 0 ( )0 0

lt lt

f x f xx x

Hence, f is continuous everywhere except at x=0.

4. If 0x then 2 0x and 2 0x

ltf(x) 0 f(0)

x 0

; ( )f x is continuous at x=0

If 2 2x 1/2 then x 1/ 4 and x 1/ 4 ; lt

f(x)x 1 /2

does not exists

f is discontinuous at x=12

5. Key -1,3 ; (1 ) 5(1) 4 1f ; 1(1 ) 4.1 3 1 1f f(1) = 1

2(2 ) 4(2 ) 3(2) 16 6 10f ; (2 ) 3(2) 4 10f

f is continuous at x = 1 and x = 2



6. Key - 1 ; 3

2Lt Lt x 27f(x) 3.3 27 f(3)

x 3x 3 x 3

f(x) is continuous as x = 3

7. Key - (q,r,s), p,(q,r,s),(p,s)

(a) f(x) is continuous everywhere since |x| and |x-1| are continuous function.

(b) [x] is dis-continuous at x = 1 = 1 1 1 0 f(x) is dis-continuous at x = 1 ; similarly dis-continuous at 0

(c) Clearly, f(x) is continuous at x < 1; f(x) is dis-continuous at 1x

We check at x = 12Lt Lt x 3x 13 1 3 13

f(x) 24 2 4 4 2 4x 1 x 1

Lt Lt Ltf(x) | x 3| (x 3) (1 3) 2

x 1 x 1 x 1

f (1) = 2 f(x) is continuous at x = 1 also

(d) 2 2Lt

f(x) x (or) x 0x 0

2 2Ltf(x) x (or) x 0

x 0 ; f(x) is continuous at x = 0

Except 0, x may take rational (or) irrational except 0 f(x) as dis-continuous.

8. a) xe continuous on R b) logx is continuous on (0, )

c) cosx is continuous on R d) secx is continuous on R 2n 1 :n z2

9. Integer answer type :

2

2x 3f x

2x 5x 2

f(x) is discontinuous at x=a,b

1x ,2

2 ;

1 5a b 2

2 2 ; 2(a + b ) = 5

CLASSROOM WORK SHEET-2

3. a = 1, b = –1 4. Not differentiable

6. x 0 x 0

1Lt f x Lt xsin 0x

x 0

1Lt x 0 and sin 1 for all x R, x 0x

Also f(0) = 0 x 0Lt f x f 0

f is continuous at x = 0.

Now h 0

f 0 h f 0f 0 Lt

h

h 0 h 0

1hsinf h 0 hLt Lth h

h 0

1Lt sinh

, which does not exist f is not differentiable x = 0.

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HOME WORK SHEET-2

1. f(x) is continuous at x = 0. 2. Differentiable. 3. Differentiable.

4. Since f is differentiable at the point x = c, it must be continuous at this point

x c x cLt f x Lt f x f c

2 2

x c x cLt x Lt ax b c

c2 = ac + b = c2 c2 = ac + b _________ (1)

Also, f is differentiable at x = c f c f c

x 0 x 0

f c h f c f c h f cLt Lt

h h

2 2 2

x 0 x 0

c h c a c h b cLt Lt

h h

2 2 2 2

x 0 x 0

c h 2ch c ac ah b cLt Lth h

( ac + b = c2 by (1))

2

x 0 x 0

h h 2c ac ah b cLt Lth h

x 0 x 0Lt h 2c Lt a

h 0 2c = a and then from (1)

c2 = 2c.c + b c2 – 2 c2 = b b = – c2.

Hence a = 2 c and b = – c2.

5. By using def. of x , we get

x , x 01 x

f x 0 , x 0x , x 0

1 x

Clearly, f is differentiable for all x 0 R .

Let us examine it for differentiability at x = 0.

x 0 x 0

x 0f x f 0 1 xf 0 Lt Ltx 0 x

x 0

1Lt 11 x

and

x 0 x 0 x 0

x 0f x f 0 11 xf 0 Lt Lt Lt 1x 0 x 1 x

f 0 f 0

f is differentiable at x = 0. Hence f is differentiable for all x R .




1. Key - 1 ; 2

2

x ; x<0f x x x f x

x ; x 0

1 1 12x; x<0f x f 0 f 0 0

2x; x 0

f(x) is differentiable every where.

2. Key - 2,3 ; 1x 2 x 2

x 2Lt f x Lt 1

tan x 2

1x 2 x 2

x 2Lt f x Lt 1

tan x 2

f(x) is not continuous at x = –2 ; f(x) is not derivable at x = –2.

3. Key - 2 ; x for 0 x 2f x

2 for x 2

1 1

x 2 x 2

x 2 2 2f 2 Lt 1 ; f 2 Lt 0

x 2 x 2

f(x) is not differentiable at x = 2

4. Key - 1 ; 3 x, x 0f x

3-x, x<0

1 1

x 0 x 0

3 x 3 3 x 3f 0 Lt 1 ;f 0 Lt 1

x 0 x 0

f(x) is not differentiable at x = 0.

5. Key : 1

6. 2

2

x if x 0f x x x

x if x 0

Since x2 and –x2 are differentialbe functions, f(x) is differentiable, except possibly at x = 0.

Now h 0 h 0

f 0 h f 0 f hf 0 lim lim

h h

f 0 0

2

h 0 h 0

hlim lim h 0h

and h 0 h 0

f 0 h f 0 f h f 0f 0 lim lim

h h

2

h 0 h 0

hlim lim h 0h

Hence f is differentiable everywhere.

7. Key - (p,q,r), (p,r,s), (r,s), (p,q,r) ;

(a) x x is continuous and differentiable in (–1, 1)

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(b) x is continuous and not differentiable at x = 0

(c) x x is not continuous at x = 0.

(d) x 1 is continuous and differentiable in (–1, 1)

CLASSROOMWORKSHEET - 3

1. (i) Let 2y sin(x 5) dy

y sinu cosudu

Put 2 duu x 5 2x

dx .

We know that2dy dy du

. cosu.(2x) 2x.cos(x 5)dx du dx

(ii) Let y sin(ax b) dy

y sinu cosudu

Put u ax b =du d

(ax b) adx dx

We know thatdy dy du

. cosu.a acos(ax b)dx du dx

2. (i) Let y = cos(sin x)dy d

sin(sinx). (sin x) sin(sinx).cos xdx dx

[using chain rule](ii) Let y = sin [cos (tan x) ]

dy d dcos[cos(tanx)]. [cos(tanx)] cos[cos(tanx)][ sin(tanx)]. (tanx)

dx dx dx

2cos[cos(tanx)][sin(tanx)].sec x [Using chain rule]

3.

2

42 3

15 5x2 sin 4x 6

3 1 x

6. (i) We have, sin2y + cos xy =


dy dy2 sin y cos y sin xy x y 0

dx dx

dy dy dysin2y x sinxy y sin xy 0 sin2y x sin xy y sin xy

dx dx dx

dy y sin xydx sin2y x sinxy


SPARK SUMMIT-12 Maths-C and D(ii) We have, sin2x + cos2y = 1


dy2sinx cos x 2cos y siny 0

dx

dysin2x 2siny cos y 0

dx

dy dy sin2xsin2x sin2y 0

dx dx sin2y

7.ax hyhx by

HOME WORKSHEETS-3

1. (i) Let y cos x = y cosu = dy

sinudu

Putting u x 2u x

1 11

2 2du 1 1 1x x

dx 2 2 2 x

We know that dy dy du 1

. sinu.dx du dx 2 x

1sin x [ u x]

2 x

(ii) Let y sec(tan x) ; Put y secudy

secu tan udu

u tanv 2dusec v

dv = v x

dv 1dx 2 x

We know that,dy dy du dv

. .dx du dv dx

2 2

sec u tan u.sec v2 x

2 2sec(tanv) tan(tanv)sec v [ u tan v]

2 x

=2sec(tan x) tan(tan x)sec x

[ v x]2 x

2. 542 9 7x 3. 420 3 4x 4. 324 6x 1 3x x 1

5. 2 22x cos x sin sinx 6. 22 sec 2x 3 7.

322 3 4x

2 x

8. We have, cos y = x cos(a + y) cos y

xcos a y

Differentiating both sides with respect to y, we get

2 2

d dcos a y cos y cos y cos a y cos a y siny cos y sin a ydx dxdy dy

dy dycos a y cos a y

2cos a ydydx sin a

01 Continuity and Differentibility Part-I

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