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Question-1

Evaluate

Solution:
= 3 + 3 = 6.

Question-2

Evaluate

Solution:
= = π -

Question-3

Evaluate

Solution:
= π = π (1) = π

Question-4

Evaluate

Solution:
=

                = =

Question-5

Evaluate

Solution:
= =

                        = =

Question-6

Evaluate

Solution:
Evaluating the function at 0, we get it of the form .

Then, we try to rewrite the function cancelling the factors which are causing the limit to be of the form .

Hence,

=

=

=

=

=

=

Question-7

Evaluate

Solution:
Evaluating the function at 2, we get it of the form 0/0. Hence,

=

                     =

                     =

                     =

                     =

                     = .

Question-8

Evaluate

Solution:
Evaluating the function at 3, we get it of the form 0/0. Hence,

=

                     =

                     =

                     =

                     =

                     =

                     =

Question-9

Evaluate

Solution:
= = b

Question-10

Evaluate

Solution:
=]

                =

                = ÷

                = ÷

                =

                = 2.

Question-11

Evaluate , a + b + c 0

Solution:
= = 1

Question-12

Evaluate

Solution:
Evaluating the function at –2, we get it of the form

Hence, =

                         =

                         = = -

Question-13

Evaluate

Solution:
We have

= = =

Question-14

Evaluate = ab, 0

Solution:
=

              =

              = 1.

              =

              = .

Question-15

Evaluate

Solution:
Put y = π - x. So as x π , y 0, we have

= = = .

Question-16

Evaluate

Solution:
= =

Question-17

Evaluate

Solution:
=

                  =

                  =

                  =

                  = 2( 1 + 1) = 4.

Question-18

Evaluate

Solution:
=

                  =

                  =

                  =

                  = +

                  = +

                  = = .

Question-19

Evaluate

Solution:
= = = 0

Question-20

Evaluate a,b,a+b 0

Solution:
=

                   =

                   =

                   =

                   =

                   =

                   = = 1

Question-21

Evaluate

Solution:
=

 

                         =

                         =

                         =

                         =

Question-22

Evaluate

Solution:
Put y =x - So, as x , y 0, we have,

=

 =

 =

 =

 =

 =

 = 1. 2 = 2

 

Question-23

Find and , where f(x) =

Solution:
Limit as x 0

where have f(x) =

For x > 0

Right hand limit = =

                       = 3(0 + 1) = 3

For x < 0,

Left hand limit = =

                         = 2(0) + 3 = 3

As =

We have =3

Now, limit as x 1

For x > 1,

Right hand limit =

                       =

                       = 3(1 + 1) = 6

For x < 1,

Left hand limit = =

                     = 2(-1) + 3 = 1

As

does not exist.

Question-24

Find where f(x) =

Solution:
For x > 1

Right hand limit =
           
                       = (-x2 – 1)

                       = -(1)2 - 1 = -2

For x < 1

Left hand limit =

                     =

                     = (1)2-1 = 0

As f(x)

f(x) does not exist.

Question-25

Find where f(x) =

Solution:
Right hand limit =

                      =

For x < 0,

Left hand limit =

                     = = -1

As

does not exist.

Question-26

Find , where f(x) =

Solution:
We have x 0 i.e., x 0- and x 0+

Now x 0-

x 0+

i.e., does not exist.

Question-27

Find where f(x) = = - 5.

Solution:
=

putting x = 5 – h

             =

[h 0+ as x 5-]

 

            =

 

            = = 0

 

=

 

Putting x = 5 + h

 

             =

 

[h 0+ as x 5+]

 

=

 

             = = 0

 

Hence = 0

Question-28

Suppose f(x) = and if = f(I). What are the possible values and a and b?

Solution:
Given: = f(l)

Hence = = f(l)

= = 4

 

a + b = b – a = 4

 

we have two equations i.e.,

 

a + b = 4

and b – a = 4

 

Solving both the equations, we get

 

a = 0 , b = 4

Question-29

f(x) = (x – a1) (x – a2)…. (x – an) what is ? For some a a1, a2….an, compute .

 


Solution:
Since f(x) is a polynomial of degree n, the could be found by putting x - a1 in f(x).

 

Thus = (a1 - a1) (a1 - a2)….. (a1 - an) = 0

 

Similarly,

 

                 = (a - a1) (a1 - a2)….. (a - an)

Question-30

If f(x) = For what value(s) of a does

Solution:
Let us examine the limit of f(x) at x = 0

 

=

 

         =

         = = 1

 

 

does not exist.

Question-31

If the function f(x) satisfies = π , evaluate

Solution:
According to the question:

 

 

= π

 

or = π

 

or f(1) – 2 = 0

or f(1) = 2

 

Now = f(1) = 2.

Question-32

If f(x) = , For what integers m and n does both and exist?

Solution:

For a limit of exist at a point, left hand limit most be equal to the right hard limit at that point.

 

Now, for to exist, we have

 

=

 

         = m. 0+n = n

 

=

 

         = n.0+m = m

 

Hence, n = m. Thus exist for all m = n.

 

For to exist, we have

 

=

 

         = n(1 + 0) + m = m + n

 

= f(l +n)

 

         = n(1 + 0) + m = m + n

 

As m + m = m + n,

 

exists for all integral values of m and n.

Question-33

Find the derivative of x2 – 2 at x = 10.

Solution:
We have f(x) = x2 – 2

 

f(10) =

         =

 

         =

 

         =

 

         = = 20.

Question-34

Find the derivative of 99x at x = 100.

Solution:
We have f(x) = 99x

 

f’(100) =

 

          =

 

          = = = 99.

Question-35

Find the derivative of x at x = 1.

Solution:
We have f(x) = x

 

f(x) =

f(1) =

      =

      = = 1

Question-36

 (i)  x3 – 27
(ii) (x – 1) (x – 2)
(iii)
(iv)

 

Solution:
(i) Let f(x) =x3 – 27

 

=

 

     =

 

     =

 

     = =

 

     = 3x2

(ii) Let f(x) = (x – 1) (x – 2)

                 = x2 – 2x – x + 2 = x2 – 3x + 2

 

=

 

     =

 

     =

 

     = = = 2x – 3

 

(iii) Let f(x) =

=

     =

     =

 

     =

 

     =

 

     =

 

     = = =

 

(iv) Let f(x) =

= =

 

     =

 

     =

 

     =

 

     =

 

     = =

Question-37

For the function f(x) = .

Solution:
Here, LHS = f'(1) =

= = 100 × 1 = 100 …..(i)

R.H.S = 100 f'(0) =

= 100

= 100 × 1 = 100 ….(i)

From (i) and (ii)

L.H.S. = R.H.S or f'(1) = 100 f'(0).

Question-38

Find the derivative of xn + axn – 1 + a2xn – 2 + …..+ an – 1x + an for some fixed number a.

Solution:
f(x) = xn + axn – 1 + a2 + xn – 2+……+ an – 1x + an

 

Hence,

 

f’(x) = nxn – 1 = a(n – 1)xn – 2 + a2(n – 2) xn – 3 + ….+an – 1 + 0

 

      = nxn – 1 + a(n – 1) xn – 2 + a2(n – 2)xn – 3 +…..+ an – 1

Question-39

(iii)

 


Solution:
(i) Let f(x) = (x – a) (x – b)

 

Using product rule, we have

 

= (x – a)

 

       = (x – a)

 

       = (x – a) [1 – 0] + (x – b) [1 – 0]

 

       = (x – a) + (x – b) = 2x - (a + b)

 


(ii) Let f(x) = (ax2 + b)2

 

=

 

Using product rule, we have


      = (ax2 + b)

      = (ax2 + b)

      = (ax2 + b)

 

      = (ax2 +b)[a(2x2 – 1)] + (ax2 + b) [a(2x2 – 1)]

 

      = (ax2 +b)[2ax]+ (ax2 + b) [(2ax]

 

      = 4ax (ax2 + b)

 

 

(iii) Let f(x) =

Using quotient rule, we have.

 

                 =

 

                 =

 

                 =

                 =

 

                 =

                 =

Question-40

Find the derivative of for some constant a.

Solution:
Let f(x) =

 

Using quotient rule, we have

 

f’(x) =

 

      =

 

      =

 

      =

 

      =

Question-41

(i) 2x - 
(ii) (5x3 + 3x – 1)(x – 1)
(iii) x-3(5 + 3x)
(v) x-4(3 – 4x-5)
(vi)
f

 

Solution:
(i) Let f(x) = 2x -

 

f’(x) = =

 

      = = 2

 

(ii) Let f(x) = (5x3 + 3x – 1)(x – 1)

 

f’(x) = (x – 1) + (5x3 + 3x – 1)

 

      = (x – 1)

      = (x – 1)

      = (x – 1) [5(3x3-1) – 3] + (5x3 + 3x – 1)

 

      = (x – 1) [15x2 – 3] + (5x3 + 3x – 1)

      = 15x3 – 3x – 15x2 + 3 + 5x2 + 3x – 1

      = 20x3 – 15x2 + 2

 

(iii) Let f(x) = x-3(5 + 3x)

 

f’(x) =

 

      = x-3

 

       = x-3

 

     = x-3

 

     = 3x-3 - (15 – 9x) x-4

 

     = =

 

     =

 

(iv) Let f(x) = x5(3 – 6x-9) = 3x5 – 6x-4

 

f’(x) =

 

      = 

 

      =

 

      = (3)(5) x4 – (6) (-4) x-5


      = 15x4 + 24x-5

 

(v) Let f(x) = x-4(3 – 4x-5)

 

f’(x) = x-4

 

       = x-4

 

       = x-4[0 – (4)(-5)x6] + (3 - 4x-5)(-4)x-5

      = x-4[20 x6] – x-5(12 - 16x5)(-4)x-5

      = 20 x-10 – 12x-5 + 16x-10 = 36 x-10 – 12 x-5

 

(vi) Let f(x) =

 

f’(x) = = -

 

      = -

 

      =

     = -

 

     = - =

Question-42

Find the derivative cos x from first principles.

Solution:
Let f(x) = cos x. Then

 

=

 

                       = =

 

                       = 

                       = -sin x 1 = sin x

Question-43

(vii) 2 tan x – 7 sec x

 


Solution:
(i) Let f(x) = sinx cos x

 

Using product rule, we have

 

f'(x) = sin x

 

      = sin x (- sin x) + cos x (cosx)

 

      = - sin2x + cos2x = cos2x – sin2x

 

(ii) Let f(x) = sec x

 

f’(x) = = =

      = =

 

      = = sec x tan x.

 

(iii) Let f(x) = 5 sec x + 4 cos x

 

f’(x) =

 

      =

 

      = 5

 

      = 5 sec x tan x – 4 sin x

 

(iv) Let f(x) = cosec x

 

f’(x) = = =

 

      = = = -cosec x cot x

 


v) Let f(x) = 3 cot x + 5 cosec x

 

f’(x) =

 

      =

 

      =

 

      = 3

 

      = 3

 

      = 3

 

      = 3

 

      = -3 cosec2x – 5 cosec x cot x.

 

(vi) Let f(x) = 5 sinx – 6 cos x + 7

 

f’(x) =

 

      =

 

      = 5 cos x – 6(- sin x) + 0 = 5 cos x + 6 sin x

 

(vii) Let f(x) = 2(tan x – 7 sec x)

 

f’(x) =

      =

 

      = 2

 

      = 2

 

      = 2 - 7 sec x tan x

 

      = 2 sec2x – 7 sec x tan x





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