for classes xi & xii.

1. Limits and Continuity
We discuss a number of functions. Our aim is to isolate an
important property of a function called continuity.
1. Let f (x) = sin(x). This is defined for all x .
[Recall we use radians automatically in order to have the
derivative of sin x being cos x.]
2. Let f (x) = log(x). This is defined for x > 0, and so naturally has
a restricted domain. Note also that the domain is an open set.
3. Let f (x) = when x a, and suppose f (a) = 2a.
4. Let f (x) =
5. Let f (x) = 0 if x < 0, and let f (x) = 1 for x 0.
6. Let f (x) = sin when x 0 and let f (0) = 0.
7. In each case we are trying to study the behaviour of the
function near a particular point. In example 1, the function is
well behaved everywhere, there are no problems, and so there
is no need to pick out particular points for special care. In
example 2, the function is still well behaved wherever it is
defined, but we had to restrict the domain. In all of what

2. follows, we will assume the domain of all of our functions is
suitably restricted.
8. We won't spend time in this course discussing standard
functions. It is assumed that you know about functions such as
sin x, cos x, tan x, log x, exp x, tan-1x and sin-1x, as well as the
``obvious'' ones like polynomials and rational functions -- those
functions of the form p(x)/q(x), where p and q are polynomials.
In particular, it is assumed that you know these are
differentiable everywhere they are defined. We shall see later
that this is quite a strong piece of information. In particular, it
means they are examples of continuous functions. Note also
that even a function like f (x) = 1/x is continuous, because,
wherever it is defined (ie on - {0}), it is continuous.
9. In example 3, the function is not defined at a, but rewriting the
function
10. = x + a if x a,
11. we see that as x approaches a, where the function is not
defined, the value of the function approaches 2a. It thus seems
very reasonable to extend the definition of f by defining f (a) =
2a. In fact, what we have observed is that
12. = (x + a) = 2a.
Definition 4.3 Say that f (x) tends to l as x a iff given > 0,
there is some > 0 such that whenever 0 < | x - a| < ,
then | f (x) - l| < .
Example : Let f (x) = for x 2. Show how to
define f (2) in order to make f a continuous function at 2.
13. Solution. We have

3. = = (x2 + 2x + 4)
14. Thus f (x) (22 + 2.2 + 4) = 12 as x 2. So defining f (2) =
12 makes f continuous at 2, (and hence for all values of x).
15. [Can you work out why this has something to do with the
derivative of f (x) = x3 at the point x = 2?]
One sided limits
Definition Say that f (x) = l, or that f has a limit from
the left iff given > 0, there is some > 0 such that
whenever a - < x < a, then | f (x) - f (a)| < .
There is a similar definition of ``limit from the right'', writen
as f (x) = l
Example Define f (x) as follows:-
f (x) =
Calculate the left and right hand limits of f (x) at 2.
Solution. As x 2 -, f (x) = 3 - x 1 +, so the left hand limit is 1.
As x 2 +, f (x) = x/2 1 +, so the right hand limit is 1. Thus
the left and right hand limits agree (and disagree with f (2),
so f is not continuous at 2).
Note our convention: if f (x) 1 and always f (x) 1 as x 2 -,
we say that f (x) tends to 1 from above, and write f (x) 1 + etc.

4. Proposition If f (x) exists, then both one sided limts
exist and are equal. Conversely, if both one sided limits exits
and are equal, then f (x) exists.
Proposition (Continuity Test) The function f is continuous
at a iff both one sided limits exits and are equal to f (a).
Example : Let f (x) = Show that f is
continuous at 1. [In fact f is continuous everywhere].
Solution. We use the above criterion. Note that f (1) = 1. Also
f (x) = x2 = 1 while f (x) = x = 1 = f (1).
so f is continuous at 1.
Exercise Let f (x) = Show that f is
continuous at 0. [In fact f is continuous everywhere]
Example : Let f (x) = | x|. Then f is continuous in .
Solution. Note that if x < 0 then | x| = - x and so is continuous,
while if x > 0, then | x| = x and so also is continuous. It remains
to examine the function at 0. From these identifications, we see
that | x| = 0 +, while | x| = 0 +. Since 0 + = 0 - = 0
= | 0|, by the 4.12, | x| is continuous at 0
Results giving Coninuity

5. Just as for sequences, building continuity directly by
calculating limits soon becomes hard work. Instead we give a
result that enables us to build new continuous functions from
old ones, just as we did for sequences. Note that if f and g are
functions andk is a constant, then k.f, f + g, fg and
(often) f /g are also functions.
Proposition Let f and g be continuous at a, and let k be a
constant. Then k.f, f + g and fg are continuous at f. Also, if g(a)
0, then f /g is continuous at a.
Proof. We show that f + g is continuous at a. Since, by
definition, we have (f + g)(a) = f (a) + g(a), it is enough to show
that
(f (x) + g(x)) = f (a) + g(a).
Pick > 0; then there is some such that if | x - a| < ,
then | f (x) - f (a)| < /2. Similarly there is some such that
if | x- a| < , then | g(x) - g(a)| < /2. Let = min( , ),
and pick x with | x - a| < . Then
| f (x) + g(x) - (f (a) + g(a))| | f (x) - f (a)| + | g(x) - g(a)| < /2
+ /2 = .
This gives the result
Note: Just as when dealing with sequences, we need to know
that f /g is defined in some neighbourhood of a. This can be
shown using a very similar proof to the corresponding result
for sequences.
**Proposition Let f be continuous at a, and let g be continuous
at f (a). Then gof is continuous at a

6. Proof. Pick > 0. We must find > 0 such that if | x - a|
< , then g(f (x)) - g(f (a))| < . We find using the given
properties of f and g. Since g is continuous at f (a), there is
some > 0 such that if | y - f (a)| < , then | g(y) - g(f (a))|
< . Now use the fact that f is continuous at a, so there is
some > 0 such that if | x - a| < , then | f (x) - f (a)| < .
Combining these results gives the required inequality.
Example : The function f : x sin3x is continuous.
Solution. Write g(x) = sin(x) and h(x) = x3. Note that each
of g and h are continuous, and that f = goh. Thus f is
continuous.
Example: Let f (x) = tan . Show that f is
continuous at every point of its domain.
Solution. Let g(x) = . Since -1 < g(x) < 1, the function
is properly defined for all values of x (whilst tan x is undefined
when x = (2k + 1) /2 ), and the quotient is continuous, since
each term is, and since x2 + a2 0 for any x. Thus f is
continuous, since f = tanog.

7. **Example : Suppose that sin(1/x) = l; in other words,
assume, to get a contradiction, that the limit exists. Let xn= 1/( n);
then xn 0 as n , and so by assumption, sin(1/xn) = sin(n )=
0 l as n . Thus, just by looking at a single sequence, we see that
the limit (if it exists) can only be l. But instead, consider the
sequence xn = 2/(4n + 1) , so again xn 0 as n . In this
case, sin(1/xn) = sin((4n + 1) /2) = 1, and we must also have l = 1.
Thus l does not exist.
Note: Sequences often provide a quick way of demonstrating
that a function is not continuous, while, if f is well behaved on
each sequence which converges to a, then in fact f is continuous
at a. The proof is a little harder than the one we have just
given, and is left until next year.
**Infinite limits
There are many more definitions and results about limits. First
one that is close to the sequence definition:
Definition Say that f (x) = l iff given > 0, there
is some K such that whenever x > K, then | f (x) - l| < .
Example : Evaluate .
Solution. The idea here should be quite familiar from our
sequence work. We use the fact that 1/x 0 as x . Thus

8. = as x .
LIMITS
Let f and g be two real valued functions with the same domain
such that f(x)≤g(x)
For all x in domain of definition, for some a, if both
≤
Identities involving calculus
[ Angle B= , A→C as →0 ⇨
CA→0 and

9. BA→BC as →0]
These can be seen from looking at the diagrams.
Sine and angle ratio identity
Proof: Area of OCD < Area of sector OAC < Area of OAB
½ r2sinx < ½ r2x < ½ r2tanx [In OAP, AB = OA tanx]
, so
, so
, or

10. , so
, but
, so
(By sandwich theorem)
Cosine and angle ratio identity
Proof:

11. The limits of those three quantities are 1, 0, and 1/2, so the
resultant limit is zero.
Cosine and square of angle ratio identity
Proof:
As in the preceding proof,
The limits of those three quantities are 1, 1, and 1/2, so the
resultant limit is 1/2.
INFORMAL APPROACH TO LIMIT
Consider a function f(x) = .
It is 0/0 form (known as indeterminate form) at x =2, this function
is defined ∀ x except x = 2.

12. If x ≠2, then f(x) = = x+2.
The following table exhibits the values of f(x) at points which are
close to 2 on its two sides
( left & right on the real line).
x 1. 1. 1. 1. 1. 1. 1.9 2 2.0 2. 2. 2. 2. 2. 2.
4 5 6 7 8 9 9 1 1 2 3 4 5 6
f( 3. 3. 3. 3. 3. 3. 3.9 0/ 4.0 4. 4. 4. 4. 4. 4.
x) 4 5 6 7 8 9 9 0 1 1 2 3 4 5 6
Graph of f(x)
4
2
-2 0 2
It is evident from the above table and the graph of f(x) that as x
increases and comes
closer to 2 from left hand side of 2, the values of f(x) increase
and come closer to 4.
x→ 2- , f(x)→4 or , =4

13. From right hand side of 2 , using notation x→ 2+ , f(x)→4 or ,
=4
Question If f(x) = x sin(1/x) , x≠ = 0, x=0 , then test
the continuity of f(x) at x=0.
Answer L.H.L. = 0 [∵ -1≤ sin(1/x)≤1 and 0. Finite
value =0 ] R.H.L. =0 ⇨ f(x) is cts.
PROPERTIES OF INFINITY
(i) c.∞ → ∞ , if c > 0
(ii) c.∞ = 0 , c = 0
(iii) c.∞ →-∞ , c < 0.
(iv) c∞ = ∞ if c > 1
(v) =0,0≤c≤1
(vi) = 1 , c = 1.
(vii) = -∞ , = ∞.
INTERDETERMINATE FORMS
EXAMPLES:
( )
( )
( )
( - )
1∞ ( )
00 ( )
∞0 (

14. o PROBLEM 3 : Determine if the following function is
continuous at x=0 .
SOLUTION : Function f is defined at x=0 since
i.) f(0) = 2 .
The left-hand limit
=2.
The right-hand limit
=2.
Thus, exists with

15. ii.) .
Since
iii.) ,
all three conditions are satisfied, and f is continuous at x=0 .
SOLUTION : Function h is not defined at x=-1 since it leads to division
by zero. Thus,
i.) h(-1)
does not exist, condition i.) is violated, and function h is NOT
continuous at x = -1 .
o
PROBLEM 5 : Check the following function for continuity at
x=3 and x=-3 .
SOLUTION 5 : First, check for continuity at x=3 . Function f is defined
at x=3 since
i.) .
The limit

16. (Circumvent this indeterminate form by factoring the numerator and
the denominator.)
(Recall that A2 - B2 = (A-B)(A+B) and A3 - B3 = (A-B)(A2+AB+B2 ) . )
(Divide out a factor of (x-3) . )
=
,
i.e.,
ii.) .
Since,

17. iii.) ,
all three conditions are satisfied, and f is continuous at x=3 . Now,
check for continuity at x=-3 . Function f is not defined at x = -3 because
of division by zero. Thus,
i.) f(-3)
does not exist, condition i.) is violated, and f is NOT continuous at x=-3
.
o PROBLEM 11 : For what values of x is the following function
continuous ?
SOLUTION 11 : Consider separately the three component functions
which determine f . Function is continuous for x > 1 since it
is the quotient of continuous functions and the denominator is never
zero. Function y = 5 -3x is continuous for since it is a
polynomial. Function is continuous for x < -2 since it is the
quotient of continuous functions and the denominator is never zero.
Now check for continuity of f where the three components are joined
together, i.e., check for continuity at x=1 and x=-2 . For x = 1 function f
is defined since
i.) f(1) = 5 - 3(1) = 2 .

18. The right-hand limit
=
(Circumvent this indeterminate form one of two ways. Either factor
the numerator as the difference of squares, or multiply by the
conjugate of the denominator over itself.)
=2.
The left-hand limit
=
= 5 - 3(1)
=2.
Thus,
ii.) .
Since

19. iii.) ,
all three conditions are satisfied, and function f is continuous at x=1 .
Now check for continuity at x=-2 . Function f is defined at x=-2 since
i.) f(-2) = 5 - 3(-2) = 11 .
The right-hand limit
=
= 5 - 3( -2)
= 11 .
The left-hand limit
=
= -1 .
Since the left- and right-hand limits are different,
ii.) does NOT exist,
condition ii.) is violated, and function f is NOT continuous at x=-2 .
Summarizing, function f is continuous for all values of x EXCEPT x=-2 .

20. o PROBLEM 13 : Determine all values of the constants A and
B so that the following function is continuous for all values
of x .
SOLUTION 13 : First, consider separately the three components which
determine function f . Function y = Ax - B is continuous for for
2
any values of A and B since it is a polynomial. Function y = 2x + 3Ax +
B is continuous for for any values of A and B since it is a
polynomial. Function y = 4 is continuous for x > 1 since it is a
polynomial. Now determine A and B so that function f is continuous at
x= -1 and x= 1 . First, consider continuity at x= -1 . Function f must be
defined at x= -1 , so
i.) f(-1)= A(-1) - B = - A - B .
The left-hand limit
=
= A (-1) - B
=-A-B.
The right-hand limit
=
= 2(-1)2 + 3A(-1) + B

21. = 2 - 3A + B .
For the limit to exist, the right- and left-hand limits must exist and be
equal. Thus,
ii.) ,
so that
2A - 2B = 2 ,
or
(Equation 1)
A-B=1.
Now consider continuity at x=1 . Function f must be defined at x=1 , so
i.) f(1)= 2(1)2 + 3A(1) + B = 2 + 3A + B .
The left-hand limit
=
= 2(1)2 + 3A(1) + B
= 2 + 3A + B .
The right-hand limit
=
=4.

22. For the limit to exist, the right- and left-hand limits must exist and be
equal. Thus,
ii.) ,
or
(Equation 2)
3A + B = 2 .
Now solve Equations 1 and 2 simultaneously. Thus,
A - B = 1 and 3A + B = 2
are equivalent to
A = B + 1 and 3A + B = 2 .
Use the first equation to substitute into the second, getting
3 (B + 1 ) + B = 2 ,
3B+3+B=2,
and
4 B = -1 .
Thus,
and

23. .
For this choice of A and B it can easily be shown that
iii.)
and
iii.) ,
so that all three conditions are satisfied at both x=1 and x=-1 , and
function f is continuous at both x=1 and x=-1 . Therefore, function f is
continuous for all values of x if and .
o PROBLEM 14 : Show that the following function is continuous for
all values of x .
SOLUTION 14 : First describe f using functional composition. Let g(x) =
-1/x2 and h(x) = ex . Function h is well-known to be continuous for all
values of x . Function g is the quotient of functions continuous for all
values of x , and is therefore continuous for all values of x except x=0 ,
that x which makes the denominator zero. Thus, for all values of x
except x=0 ,

24. f(x) = h ( g(x) ) = e g(x) = e -1/x2
is a continuous function (the functional composition of continuous
functions). Now check for continuity of f at x=0 . Function f is defined
at x=0 since
i.) f(0) = 0 .
The limit
(The numerator approaches -1 and the denominator is a positive
number approaching zero.)
,
so that
=0,
i.e.,

25. ii.) .
Since
iii.) ,
all three conditions are satisfied, and f is continuous at x=0 . Thus, f is
continuous for all values of x .
o PROBLEM 15 : Let
Show that f is continuous for all values of x .
SOLUTION 15 : First show that f is continuous for all values of x .
Describe f using functional composition. Let , , and
2
k(x) = x . Function h is well-known to be continuous for all values of x
. Function k is a polynomial and is therefore continuous for all values
of x . Function g is the quotient of functions continuous for all values
of x , and is therefore continuous for all values of x except x=0 , that x
which makes the denominator zero. Thus, for all values of x except
x=0 ,
is a continuous function (the product and functional composition of
continuous functions). Now check for continuity of f at x=0 . Function f
is defined at x=0 since
i.) f(0) = 0 .

26. The limit does not exist since the values of oscillate
between -1 and +1 as x approaches zero. However, for
so that
.
Since
,
it follows from the Squeeze Principle that
ii.) .
Since
iii.) ,
all three conditions are satisfied, and f is continuous at x=0 . Thus, f is
continuous for all values of x .
o PROBLEM 1 : Compute .
SOLUTION1 : Note that DOES NOT EXIST since values of

27. oscillate between -1 and +1 as x approaches 0 from the left.
However, this does NOT necessarily mean that does not
exist ! ? #. Indeed, x3 < 0 and
for x < 0. Multiply each component by x3, reversing the inequalities
and getting
or
.
Since
,
it follows from the Squeeze Principle that
.
o PROBLEM 2 : Compute .

28. SOLUTION2 : First note that
,
so that
and
.
Since we are computing the limit as x goes to infinity, it is reasonable
to assume that x+100 > 0. Thus, dividing by x+100 and multiplying by
x2, we get
and
.
Then
=

29. =
=
= .
Similarly,
= .
Thus, it follows from the Squeeze Principle that
= (does not exist).
DIFFERENTIATION OF INVERSE TRIGONOMETRIC FUNCTIONS
None of the six basic trigonometry functions is a one-to-one function.
However, in the following list, each trigonometry function is listed
with an appropriately restricted domain, which makes it one-to-one.
1. for
2. for
3. for
4. for , except

30. 5. for , except x = 0
6. for
Because each of the above-listed functions is one-to-one, each has an
inverse function. The corresponding inverse functions are
1. for
2. for
3. for
4. arc for , except
5. arc for , except y = 0
6. arc for
In the following discussion and solutions the derivative of a function
h(x) will be denoted by or h'(x) . The derivatives of the above-
mentioned inverse trigonometric functions follow from trigonometry
identities, implicit differentiation, and the chain rule. They are as
follows.
1.
2.
3.
4. arc
5. arc

31. 6. arc
Important points: (1)when exists but either f© does not
exist or ≠f(c), we say that f has a reomovable discts.
;otherwise f has non- reomovable discts.
(2) composition of real valued fns. of f and g (fog) is defined at c. if g
is cts. At c and if f is cts. At g© then fog is cts. At c, fog(x)=f(g(x)) is
defined whenever the range of g is subset of domain of f. [Ex. sin(x²)]
Some questions on differentiability:
1. Show that f(x)=x2 is diff. at x=1, find f’(1)
2. Prove that f(x) = [x], 0<x<3 is not diff. at x=1 & 2.
3. Discuss the continuity & diff. of f(x)=
4. Show that f(x)= is not diff. at x=2.
5.For which value of a &b the function f(x) = is diff. at x=0
Solutions:
1. R f’(1) = = =2
L f’(1)= = =2 ⇨ L.H.D = R.H.D=2
2. R f’(1) = = =1-1/h=0{∵ [1+h]=1}
L f’(1)= = = 0-1/-h=∞ ⇨R.H.D
≠L.H.D ⇨ f is not diff. at x=1.
At x=2, similarly f is not diff. at x=2, ∵ R.H.D= 0≠ L.H.D=∞ {[2+h]=2 &
[2-h] = 1}
3. L.H.lt =0 = R.H.lt ⇨ f is cts. At x=0

32. L.H.D = =0= R.H.D ⇨ diff.
4. R.H.D. = = -1
L.H.D = = 1 ⇨ not diff.
5. Given L.H.D= R.H.D ⇨ =
=
a=2 & b=0 [ ∵ f is cts. ⇨b=0]
ASSIGNMENT(continuity & differentiability) (XII)
**Question 1 Determine a and b so that the function f given by
f(x) = , x<п/2
=a, x=п/2
= , x>п/2
Is continuous at x=п/2.
Answer [a = 1/3 , b = 8/3]
**Question 2 Find k such that following functions are continuous at
indicated point
(i) f(x) = at x=0
(ii) f(x) = (2x+2 - 16)/(4x – 16) , x≠2
= k, x = 0 at x=2.

33. Answer [ (i) k=1,(ii) k=1/2]
**Question 3 The function f is defined as
If f(x) is continuous on [0,8], find the values of a and b.
Answer [a=3,b=-2]
** Question 4 If f(x) = is continuous in
the [-1,1], find p.
Answer [p=-1]
**Question 5 Find the value of a and b such that the f(x) defined as
f(x) = is continuous for all
values of x in [0,п].
ANSWER [a=п/6 , b=-п/12]
** Question 6 Prove that = -4
[ Hint: Nr. Can be written as tanx(tanx-1)(tanx+1) =- [tanx(cosx-
sinx)(tanx+1)]/cosx
Cosx-sinx = cos )]
**Question 7 Prove that (i) = [ Hint:
put x= sinѲ]

34. (ii) ) = -3/2. [Hint: = & use
formula of ]
Question 8 f(x) = , =1 & =1, then p.t.
f(-2)=f(2)=1. [ Hint: =0]
Question 9 [Dr. = 2|sinx/2| & =1
|sinx/2| =+ve & -ve as x→0+ & x→0- , ⇨ limit does not exist]
Question 10 Show that the function
f(x) is continuous at x=0.
[Hint: use =1 , =1]
Question11 Show that f(x) = |x-3|,x∊R is cts. But not diff. at x=3.
[Hint:show L.H.lt=R.H.lt by |x-3| = x-3, if x ≥3 and –x+3, if x<3, L.hd=-1≠1(R.h.d)
QUESTION 12 Discuss the continuity of the fn. f(x) = |x+1|+|x+2|, at x
= -1 & -2 [Hint:f(x) = yes cts. At x=-1,-2
Question 13 Find the values of p and q so that f(x)
= is diff. at x = 1. [ answer is p=3 , q=5]
Question 14 For what choice of a, b, c if any , does the function
F(x) = becomes diff at x=1,2 & show that a=b=c=0.

35. Question15For what values a,b f(x)= is diff.at x=0
[Hint: L.H.d= 2 =1& R.H.d=a, since f‘(x)=0exists, a=2,b∊R]
ASSIGNMENT WITH HINTS (XI)
Question.1 Evaluate
**
[Hint
]
Question.2
[Hint =7/4]
Question.3 [use sin3x=3sinx-4sin³x, put
( ) =h ,answer is -4]
Question.4
[Hint
= 2]
Question.5 [hint: put tanx=sinx/cosx, answer is 1/2]

36. Question.6
[Hint: use sinx-siny=2cos(x+y)/2.sin(x-y)/2 , answer is 0]
Question.7 [ answer is ½]
Question.8 [ answer is 1/8]
Question.9 Let f(x) =
Find a,b so that and exist.
[Hint: exists
(ax+b)= = x2/2 2a+b=2……(1)
Similarly -2a+b = -1……(2)
a=3/4,b=1/2]
Question.10 and where [x] denotes the
integral part of x. Are they equal?
[HINT: = 3/3=1,x>3
= 3/2 ,x<3]
Question.11 and , are they equal?
-5 as -x -6

37. **Question.12 Test the continuity of the function at x=0, f(x) = ,
x ≠0
0, x=0 [Hint e-∞ =0,e∞ =∞]
Derivatives (XI)
BY FIRST PRINCIPLE
o Use the limit definition to compute the derivative, f'(x), for
.
SOLUTION :
(Get a common denominator for the expression in the numerator.
Recall that division by is the same as multiplication by .)
(Algebraically and arithmetically simplify the expression in the
numerator. It is important to note that the denominator of this

38. expression should be left in factored form so that the term can be
easily eliminated later.)
(The term now divides out and the limit can be calculated.)
o Use the limit definition to compute the derivative, f'(x), for
SOLUTION 6 :
(Recall a well-known trigonometry identity :

39. .)
(Recall the following two well-known trigonometry limits :
and .)
SOLUTION 3 : Differentiate . Apply the quotient rule.
Then

40. (Recall the well-known trigonometry identity .)
.
SOLUTION 4 : Differentiate . Apply the product rule.
Then

41. SOLUTION 5 : Differentiate . This is NOT a product of
functions. It's a composition of functions. Apply the chain rule. Then
.
SOLUTION 6 : Differentiate . Apply the product rule first,
followed by the chain rule. Then
(Ncert)limits & derivatives (xi)
Question 4:
Evaluate the Given limit:
Question 10:

42. Evaluate the Given limit:
Question 12:
Evaluate the Given limit:
Question 14:
Evaluate the Given limit:
Question 15:
Evaluate the Given limit:
Question 16:
Evaluate the given limit:
Question 17:
Evaluate the Given limit:
Question 18:
Evaluate the Given limit:

43. Question 20:
Evaluate the Given limit:
Question 21:
Evaluate the Given limit:
Question 22:
Question 23:
Find f(x) and f(x), where f(x) =
Question 24:
Find f(x), where f(x) =
Question 25:
Evaluate f(x), where f(x) =
Question 26:

44. Find f(x), where f(x) =
Question 28:
Suppose f(x) = and if f(x) = f(1) what are
possible values of a and b?
Question 30:
If f(x) = .
For what value (s) of a does f(x) exists?
Question 31:
If the function f(x) satisfies , evaluate
.
Question 32:

45. If . For what integers m and n does
and exist?
ANSWERS: 4. 19/2, 10. 2, 12. -1/4, 14. A/B, 15. 1/п, 16.
1/п, 17. 4, 18. (a+1)/b, 20. 1, 21. 0, 22. 2, 23. 3,6, 24.
Limit does not exist at x=1, 25. Limit does not exist at
x=0, 26. Limit does not exist at x=0, 28. A=0,b=4, 30.
f(x) exists for all a≠0, 31. 2, 32. we need
m=n; and exists for any integral value of m & n.
DERIVATIVES
Question 1:
Find the derivative of the following functions from first
principle:
(i) –x (ii) (–x)–1 (iii) sin (x + 1)
(iv)
Solution of (iv) : use formula of first principle(ab-initio)

46. We have f’(x) = =
= =
sin(x-п/8) [ , cosA-cosB= -2sin(A+B)/2.sin(A-B)/2]
Question 11:
Find the derivative of the following functions:
(i) sin x cos x (ii) sec x (iii) 5 sec x + 4 cos x
(iv) cosec x (v) 3cot x + 5cosec x
(vi) 5sin x – 6cos x + 7 (vii) 2tan x – 7sec x
Solution of (i) (2 sin x cos x)/2= sin2x/2 , derivative is ½
(2cos2x) = cos2x or can use Leibnitz product rule.
Question 16:
Find the derivative of the following functions :
Solution: by quotient rule or take derivative after simplification
It can be written as = =tan(
Derivative is - sec2( .
Question 17:
Find the derivative of the following functions:

47. Solution: multiple & divide by cosx, we get = -tan(п/4+x)
Derivative is - sec2(п/4+x) [simplest method]
Question 18:
Find the derivative of the following functions
Solution: can be written as = tan2(x/2)
Derivative is 2 tan(x/2) . sec2(x/2) . ½ = tan(x/2) . sec2(x/2).
Question 23:
Find the derivative of the following functions (x2 + 1) cos x
Question 24:
Find the derivative of the following functions: (ax2 + sin x)
(p + q cos x)
Question 25:
Find the derivative of the following function
Question 26:
Find the derivative of the following function :
Question 27:

48. Find the derivative of the following function:
Question 28:
Find the derivative of the following functions :
Question 29:
Find the derivative of the following functions : (x + sec x)
(x – tan x)
Question 30:
Find the derivative of the following functions: