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Arguments are also called as parameters

● Decomposing complex problems into simpler pieces. 

● Reducing duplication of code.

The constant variable can be declared using const keyword. The const keyword makes variable , value stable. The constant variable should be initialized while declaring. The const modifier enables to assign an initial value to a variable that cannot be changed later inside the body of the function.

Syntax:

(const )

Python’s built in function help ( ) is very useful. When it is provided with a program-name or a module-name or a function-name as an argument, it displays the documentation of the argument as help. It also displays the docstrings within its passed-argument’s definition. For example : 

help (math) will display the documentation related to module math.

def printme (str): 

“This prints a passed string into this function” 

print str 

return

Python includes following functions that perform trignometric calculations.

A default argument is an argument that assumes a default value, if a value is not provided in the function call.

#!/usr/bin/python

# Function definition is here

def printinfo(name,age=35):

“This prints a passed info into this function”

print “Name: “,

name; print “Age “,

age return;

# Now you can call printinfo function

 printinfo(age=50,name=”miki”);

 printinfo(name=”miki”);

When the above code is executed, it produces following result:

Name: miki 

Age 50 

Name:miki 

Age 35

The statement return [expression] exits a function.

The pre-defined functions that are always available for use are known as python’s built-in functions. For example : 

len (), type (), int (), raw-input () etc.

Given as f = {(1, −1), (4, −2), (9, −3), (16, 4)} and g = {(−1, −2), (−2, −4), (−3, −6), (4, 8)}

f: {1, 4, 9, 16} → {-1, -2, -3, 4} and g: {-1, -2, -3, 4} → {-2, -4, -6, 8}

Co-domain of f = domain of g

Therefore, gof exists and gof: {1, 4, 9, 16} → {-2, -4, -6, 8}

(gof)(1) = g(f(1)) = g(−1) = −2

(gof)(4) = g(f(4)) = g(−2) = −4

(gof)(9) = g(f(9)) = g(−3) = −6

(gof)(16) = g(f(16)) = g(4) = 8

Therefore, gof = {(1, −2), (4, −4), (9, −6), (16, 8)}

Since, the co-domain of g is not same as the domain of f.

Therefore, fog does not exist.

Given function f: R → R be defined by f(x) = x³ − 3

Let us prove that f⁻¹ exists

Injectivity of f:

Let x and y be two elements in domain (R),

Such that, x³ − 3 = y³ − 3            

⇒ x³ = y³        

⇒ x = y

Therefore, f is one-one.

Surjectivity of f:

Let y be in the co-domain (R)

Such that f(x) = y

⇒ x³ – 3 = y

⇒ x³ = y + 3 

⇒ x = ³√(y + 3) in R

⇒ f is onto.

Therefore, f is a bijection and, hence, it is invertible.

Find f ^-1

Let f^-1(x) = y ...(1)

⇒ x = f(y)

⇒ x = y³ − 3

⇒ x + 3 = y³

⇒ y = ³√(x + 3) = f^-1(x) [from (1)]

Therefore, f^-1(x) = ³√(x + 3)

Now, f^-1(24) = ³√(24 + 3)

= ³√27

= ³√3³

= 3

And f^-1(5) = ³√(5 + 3)

= ³√8

= ³√2³

= 2

(i) Given as A = {0, −1, −3, 2}; B = {−9, −3, 0, 6} and f(x) = 3x.

Therefore, f = {(0, 0), (-1, -3), (-3, -9), (2, 6)}

Here, the different elements of the domain have different images in the co-domain.

Thus, this is one-one.

Range of f = Range of f = B

Therefore, f is a bijection and,

Clearly, f ^-1 exists.  

Hence, f ^-1= {(0, 0), (-3, -1), (-9, -3), (6, 2)}

(ii) Given as A = {1, 3, 5, 7, 9}; B = {0, 1, 9, 25, 49, 81} and f(x) = x²

Therefore, f = {(1, 1), (3, 9), (5, 25), (7, 49), (9, 81)}

Here, the different elements of the domain have different images in the co-domain.

Thus, f is one-one.

This is not onto because the element 0 in the co-domain (B) has no pre-image in the domain (A)

⇒ f is not a bijection.

Therefore, f ^-1does not exist.

Option : (D) 

[x]²-5[x]+6 = 0 

([x]-2)([x]-3) = 0 

if [x] = 2 

2≤x<3 

And, 

if [x] = 3 

3≤x<4 

Therefore,

x ∈ [2,4]

π² ≈ 9.8596

[π²] = 9 and

[-π²] = -10

Now, 

f(x) = sin[π²] x + sin[-π²]x 

= sin 9x - sin 10x 

Now, 

Checking values of f(x) at given points..

f(\(\frac{\pi}{2}\)) = sin 9(\(\frac{\pi}{2}\)) - sin 10(\(\frac{\pi}{2}\))

= 1-0 

= 1 

Option A is correct.. 

f(π) = sin 9π - sin 10π 

= 0 - 0 

= 0

f(\(\frac{\pi}{4}\)) = sin 9(\(\frac{\pi}{4}\)) - sin 10(\(\frac{\pi}{4}\))

= \(\frac{1}{\sqrt 2}\) - 1

Option B & C are incorrect..

Option : (A)

Given,

f(x) = |x|/x 

When -4≤x<0,

f(x) = \(-\frac{x}{x}\)

= -1 

When 0<x≤4,

f(x) = \(\frac{x}{x}\)

= 1 

R(f) = {-1, 1}

g(x) = x² + x – 2

(gof) (x) = 4x² – 10x + 4 

= (2x – 3)² + (2x – 3) – 2 

= g(2x – 3) = g(f(x)) 

∴ f(x) = 2x – 3 

(gof) (x) = 4x² – 10x + 4 

= (-2x + 2)² + (-2x + 2) – 2 

= g(-2x + 2) 

= g(f(x)) 

∴ f(x) = -2x + 2

f(x) = 1 + 2x + 4x 

Since, 2x > 0, 4x > 0 

∴ f(x) > 1 

∴ Range of f = (1, ∞)

f(x) = e^x , g(x) = log x 

(fog) (x) = f(g(x)) 

= f(log x) 

= e^{log x} 

= x 

(i) f(x) = ln (x – 5)

f is defined, when x – 5 > 0

∴ x > 5

∴ Domain of f = (5, ∞)

(ii) f(x) = log₁₀ (x² – 5x + 6)

x² – 5x + 6 = (x – 2) (x – 3)

f is defined, when (x – 2) (x – 3) > 0

∴ x < 2 or x > 3

Solution of (x – a) (x – b) > 0 is x < a or x > b

where a < b

∴ Domain of f = (-∞, 2) ∪ (3, ∞)