Analysis Exam Solutions

Introduction :

       Mathematical analysis otherwise called as analysis. Analysis mainly concern with the notion of a limit. Analysis exam solutions includes the theory of differentiation, integration, analytic functions and infinite series. These analysis exam theories are often studied in the context of real numbers, complex numbers, and real and complex functions. Analysis exam solutions are useful for students to prepare for analysis exam.

 

Analysis exam solutions:

 

Example 1:

Find `lim_(x->0)` 3x + | x | / 7x − 5 | x | , if it exists.

Solution:
Rf(0) = `lim_(x->0)`   3x + | x | / 7x − 5 | x |

          = `(3x + x) / (7x - 5x)` (since x > 0, | x | = x)

          =`lim_(x->0+) `     `(4x) /(2x) `

          =  2 .

L f(0) = `lim_(x->0-)`    3x + | x | / 7x − 5 |x |

          =` lim_(x->0-)`    3x − x / 7x − 5(− x)     (since x < 0, | x | = − x)

         =   `lim_(x->0-)`   `(2x) /(12x)`

         =    `lim_(x->0-)`   (1/6) = `1/6`

Since Rf(0) ≠ Lf(0), the limit does not exist.


Example 2:

⌠(tanx + cotx) 2dx = ⌠(tan2x + 2tanx cotx + cot2x) dx

                            = ⌠[(sec2x − 1) + 2 + (cosec2x − 1)] dx

                            = ⌠(sec2x + cosec2x)dx

                            = tanx + (− cotx) + c

                            = tanx − cotx + c

 

Analysis exam solutions:

 

Example 3:

Evaluate analysis exam`lim_(x->3)`` (x^2 + 8x + 11) / (x^2 - 9). `

Solution:
Let f(x) = `(x^2 + 8x + 11) / (x^2 - 9)` .

This is of the form f(x) = g(x) / h(x) ,

where g(x) = x2 + 8x + 11 and h(x) = x2 − 9.

            g(3) = 44 ≠ 0 and the h(3) = 0.

            f(3) is equal to `(g(3))/(h(3))` = `44/ 0` .

Hence

`lim_(x->3)`  `(x^2 + 8x + 11) / (x^2 - 9)` does not exist.

 

Example 4:

Differentiate (x2 + 7x + 2) (ex − logx) with respect to x.

Solution: Let y = (x2 + 7x + 2) (ex − logx)

                    y′ =` d/dx ` [x2 + 7x + 2) (ex − logx)]

                       = (x2 + 7x + 2) `d/dx` (ex − logx) + (ex − logx)`d/dx` (x2 + 7x + 2)
 
                       = (x2 + 7x + 2)` d/dx` (ex) − `d/dx` (logx) + (ex − logx)` d/dx ` (x2) + `d/dx` (7x) + `d/dx` (2)

                       = (x2 + 7x + 2) (ex`1/ x` ) + (ex − logx) (2x + 7 + 0)

                       = (x2 + 7x + 2) (ex`1/ x` )+ (ex − logx) (2x + 7) .