Example 1:

 Given are two sets A {1, 2, -2, 3} and B = {1, 2, 3, 5}. Is the function f(x) = 2x - 1 defined from A to B?

Solution :

Out of all the ordered pairs, the ordered pairs which are related by the function f(x) = 2x - 1 are {(1, 1), (2, 3), (3, 5) But for (-2) in A, we do not have any value in B. So, this function does not exist from
A->B.

Example 2:

A function f is defined as f: N -> N (where N is natural number set) and f(x) = x+2. Is this function ONTO?

Solution :

        Since, N = {1, 2, 3, 4, .........} and A = B = N

       For : A->B

        When x = 1                f(x) = 3

        When x = 2                f(x) = 4

So f(x) never assume values 1 and 2. So, B have two elements which do not have any pre-image in A. So, it is not an ONTO function.

Example 3 :

Find the range and domain of the function f(x) = (2x+1)/(x-1) and also find its inverse.

Solution :

This function is not defined for x = 1. So, domain of the function is
R -{1}.

Now, for finding the range

Let,(2x+3)/(x-1) = y

        => 2x + 3 = yx - y

        => yx - 2x = y + 3

        => (y - 2)x = y + 3

        => x =(y+3)/(t-2)

So, y cannot assume value 2

Range of f(x) is R - {2}.

Inverse is y =(x+3/x-2) .

Example 4:

 Find domain and range of the function f(x) = (x²+2x+3)/(x²-3x+2)

Solution :

        This function can be written as : f(x) =(x²+2x+3)/(x-1)(x-2) .

        So, domain of f(x) is R - {1, 2}

        For range, let (x²+2x+3)/(x²-3x+2) = y

                => (1 - y)x² + (2x + 3y) x + 3 - 2y = 0

for x to be real, Discriminant of this equation must be > 0

        D > 0

       => (2 + 3y)² - 4(1 - y)(3 -2y) > 0

       => 4 + 9y² + 12y - 4(3 + 2y² - 5y) > 0

       => y² + 32y - 8 > 0

       => (y + 16)² - 264 > 0

       => y < - 16 - √264  or  y  > -  16 + √264.

Example 5 :

Find the period of following functions

(a)        cos3 x + sin 5 x

(b)        |cos x| + |sin2 x|

(c)         x - |x|.

Solution :

(a)    f(x) = cos 3x + sin 5x

        period of cos 3x = 2∏/3  and period of sin 5x = 2∏/53

         L.C.M. of 2∏/3 and 2∏/5  is 2p

         So period of f(x) is 2p.

Note:     Let g(x) = cos 3x

                g((2∏/3)+x)  = cos3 ((2∏/3)+x)

                                   = cos (2∏ + 3x)

                                   = cos 3x

                                   = g(x)

(b)    f(x) = |cos x| + |sin2 x|

               Period of |cos x| = ∏

                Period of |sin 2 x| = ∏/2

                So, period of f(x) is ∏

(c)       f(x) = x [x]

                Let T be the period of this function

               => f(T + x) = f(x)

               => T + x - [T + x] = x - [x]

               => T = [T + x] - [x]                     .......... (1)

               => T = integer - integer

                        = integer

Let T = 1  (Therefore 1 is the smallest positive integer)

Equation (1) becomes

                1 = [1 + x] - [x]

which is true for all x ε R

Period of f(x) is 1.

Example 6:

Show that the inverse of a linear fraction function is always a linear fraction function (except where it is not defined).

Solution:

Let, f(x) = (a+bx)/(c+dx) be the said linear fraction function.

Let at some x it attains value y, so,

                 (a+bx)/(c+dx) = y

                => a + bx - cy - dxy = 0

                => a - cy + x (b - dy) = 0

                => x = (cy-a)/(b-dy).

Which is again a linear fraction function defined in R except

                 at x = -c/d and y = b/d

and inverse of the given function is, y = (cx-a)/(b-dx).

 Example 7:

If graph of function f(x) is as shown in the figure given below, then plot the graph of |f(x)|.

         f(x) + 1, f(x + 2) and f^-1 (x)

Solution:

(a)    |f(x)| will reflect the graph of f(x) below x axis to the (-) ve y axis side. So the graph will be as shown in the figure given below.

(b)    f(x) + 1 will just shift the graph by one unit position up. So the required graph is as shown in the figure given below.

(c)    f(x + 2) will shift the graph of f(x) by two units to left, the graph will be as shown in the figure given below.

(d)    f^-1(x) is obtained by reflection of graph f(x) on the line y = x as shown in the figure given below.

Example 8:

Show that the following functions are even

         (a)    f(x) = x²/(2x²-1) + x²/2 + 1

         (b)    f(x) = (a^x+a^-x)/2

         (c)    f(x) = x² - |x|

Solution:

        (a)    f(x) = x²/(2x²-1)  +  x²/2  +  1

        so, f(-x) = (-x)²/(2(-x)²-1)  +  (-x)²/2  +  1

                     = x²/(2x²-1)  +  x²/2  +  1 = f(x)

        so, f(x) in sum function.

        (b)    f(x) = (a^x+a^-x)/2

               =>  f(-x) = (a^-x+a^x)/2 = f(x)

        so, f(x) is even function

        (c)    f(x) = x² - |x|

                => f (-x) = (-x)² - |-x| = x² - |x| = f(x)

        so, f(x) is even function.

Example 9:

        Show that following functions are odd.

        (a)    f(x) = (e^x-1)/(e^x+1)

        (b)    f(x) = log((1-x)/(1+x))

        (c)    f(x) = √(1+x+x²) - √(1-x+x²)

Solution:

        (a)    f(x) = (e^x-1)/(e^x+1)

               => f(x) = (e^-x-1)/(e^-x+1) = (1-e^x)/(1+e^x)

               = -((e^x-1)/(e^x+1))  = -f(x)

               =>  f(-x) = -f(x)

               =>  so, f(x) is an odd function

        (b)    f(x) = log ((1-x)/(1+x))

                => f(-x) = log((1+x)/(1-x)) log((1-x)/(1+x))^-1

               => -log((1-x)/(1+x))

               => f(-x) = -f(x)

        so, f(x) is an odd function.

        (c)    f(x) = √(1+x+x²) - √(1-x+x²)

                f(-x) = √(x²-x+1) - √(1+x+x²)

                       = -[√(1+x+x²) - √(1-x+x²)]

                f(-x) = -f(x)

        so, f(x) is an odd function

Example 10:

        If f(x) = 1 + x; 0 < x < 2

                   = 3 - x; 2 < x < 3

        Determine

(a)       g(x) = f(f(x))

(b)       f(f(f(x)))

(c)        f([x])

(d)       [f(x)]

Where [ ] represents the greatest integer function.

Solution:

 (b)    Let 0 < x < 1

                        f(f(f(x)))

                        = f(2 + x) 2 < 2 + x < 3

But we observe that there is no single definition f(f(x)) for this interval.

Therefore we reduce the interval 0 < x < 1 to 0 < x < 1.

Let 0 < x < 1

        f(f(f(x)))

        = f(2 + x); 2 < x + 2 < 3

        = 3 - (2 + x)

        = 1 - x

Let 1 < x < 2

        = f(2 - x); 0 < 2 - x < 1

        = 1 + 2 - x

        = 3 - x

Let 2 < x < 3

        = f(f(f(x)))

        = f(4 - x); 1 < 4 - x < 2

        = 1 + (4 - x)

        = 5 - x


Let x = 0

        f(f(f(x))) = f(f(1)) = f(2) = 3

(c)            f([x])

        Let 0 < x < 1

                f[x] = f(0) = 1

        Let 1 < x < 2

                f[x] = f(1) = 2

        Let 2 < x < 3

                f[x] = f(2) = 3

        Let x = 3

                f([x]) = f(3) = 0

(d)            [f(x)]

        First draw the graph of y = f(x)

                Let    0 < x < 1

                        1 < f(x) < 2 => [f(x)] = 1

                Let    1 < x < 2

                        2 < f(x) < 3 => [f(x)] = 2

                Let    x = 2

                        f(x) = 3

                        [f(x)] = 3

                Let    2 < x < 3

                        0 < f(x) < 1 => [f(x)] = 0

Example 11:

        If x² + y² = 1

        prove that - √2 < x + y <√2 .



Solution:

        Since, x² + y² = 1 => x = cos θ, y = sin θ

        Consider,

                x + y = cos θ + sin θ

                = √2((1/√2)sinθ + (1/√2)cosθ )

                = √2sin((∏/4) + θ)

Recall :     sin((∏/4)+θ)  can take maximum value 1 and minimum value -1.

              =>|√2 sin((∏/4)+θ)| ≤ √2

              => - √2 < x + y < √2.          Hence proved.

Example 12:

        Check the invertibility of the function f(x) = (e^x - e^-x); and then find its inverse.

Solution:

        We have

               f(x) = e^x - e^-x; x ε R

                lim_x->α  f(x) = α

                lim_x->-α  f(x) = -α

                f'(x) = e^x + e^-x > 0 ∀ ε R

               Therefore f : R -> R

                f(x) = e^x - e^-x is a bijective function

                Therefore f(x) is invertible

        Now, f(x) = y = t - 1/t [where t = e^x]

                => t² - 1 = ty

                => t² - ty - 1 = 0

                => t = (y+√(y²+4))/2  [t cannot be negative]

        Now

                t = e^x

                => e^x = (y+√(y²+4))/2

               => x = log_e ((y+√(y²+4))/2)

              Therefore Inverse of y = e^x - e^-x is y = log_e ((x+√(x²+4))/2)

Example 13:

        If f(x) = ((1-x)/(1+x))  x ≠ 1 and x ε R.

        Then show that

        (i)     f(1/x) = -f(x), x ≠ 0

        (ii)    f(f(x)) + f(f(1/x)) > 2 for x > 0.

Solution:

        Since, x² + y² = 1 => x = cos θ, y = sin θ

        Consider,

                x + y = cos θ + sin θ

                = √2((1/√2)sinθ + (1/√2)cosθ )

                = √2sin((∏/4) + θ)

Recall :     sin((∏/4)+θ)  can take maximum value 1 and minimum value -1.

              =>|√2 sin((∏/4)+θ)| ≤ √2

              => - √2 < x + y < √2.          Hence proved.

Example 12:

        Check the invertibility of the function f(x) = (e^x - e^-x); and then find its inverse.

Solution:

        We have

               f(x) = e^x - e^-x; x ε R

                lim_x->α  f(x) = α

                lim_x->-α  f(x) = -α

                f'(x) = e^x + e^-x > 0 ∀ ε R

               Therefore f : R -> R

                f(x) = e^x - e^-x is a bijective function

                Therefore f(x) is invertible

        Now, f(x) = y = t - 1/t [where t = e^x]

                => t² - 1 = ty

                => t² - ty - 1 = 0

                => t = (y+√(y²+4))/2  [t cannot be negative]

        Now

                t = e^x

                => e^x = (y+√(y²+4))/2

               => x = log_e ((y+√(y²+4))/2)

              Therefore Inverse of y = e^x - e^-x is y = log_e ((x+√(x²+4))/2)

Example 13:

        If f(x) = ((1-x)/(1+x))  x ≠ 1 and x ε R.

        Then show that

        (i)     f(1/x) = -f(x), x ≠ 0

        (ii)    f(f(x)) + f(f(1/x)) > 2 for x > 0.

Solution:

                f(x) = (1-x)/(1+x), x ≠ 1 and x ε R

                => f(1/x) = (1-(1/x))/(1+(1/x)) = (x-1)/(x+1), x ≠ 0

                => - f(x)

        Now f(f(x)) = (1-(1-x)/(1+x))/(1+(1-x)/(1+x)) = (2x)/2 = x

        and   f(f(1/x)) = (1-(x-1))/(1+x))/(1+(x-1)/(1+x)) = 2/2x = 1/x

                => f(f(x)) + f(f(1/x)) = x + 1/x

                = (√x-(1/√x))² + 2

        R.H.S. = 2 + a positive number   >  2

        so f(f(x)) + f(f(1/x)) > 2

Example 14:

        Let A = R - {3},

B = R - {1}, let f: A -> B be defined by f(x) = (x-2)/(x-3). Is f bijective? Give reasons.

Solution:

(a)    Let us test the function for injectivity

        Let    x₁, x₂ ε A and f(x₁) = f(x₂)

                =>(  x₁-2)/(x₁-3) = (x₂-2)/(x₂-3)

                => x₁ = x₂

        Therefore f is one-one function (injective)                  .........(1)

(b)    Let us test the function for surjectivity

Let y be any arbitrary element of B and suppose there exists an x such that f(x) = y

        (x-2)/(x-3) = y => x = (3y-2)/(y-1)

since y ≠ 1, x is real

Also, x ≠ 3, for if x = 3, then 3 = (3y-2)/(y-1)

or 3y - 3 = 3y - 2 => - 3 = - 2, which is false

Thus x = (3y-2)/(y-1) ε A such that f(x) = y i.e. ∀ y ε B, we have x ε A.

and so f is surjective

This proves that f is bijective.

 Example 15:

        Show that if an odd function is invertible, then its inverse is also an odd function.

Solution:

        Let y = f(x) be an odd function

        Then

                f(-x) = -f(x) = -y

        Since it is invertible, so we can write

                x = g(y)

        Where g(x) = f^-1 (x)

        Consider,

                g(-y) = g(-f(x))

                = g(f(-x)) = -x = -g(y)

        So g(x) is also an odd function.

Example 16:

        Sketch the graph of each of the following functions

        (a)    f(x) = x⁴ - 2x² + 3

        (b)    f(x) = 2x/(1+x²) 

        (c)    f(x) = sin2x - 2sinx.

Solution:

(a)     y = f(x) = x⁴ - 2x² + 3

        (i)     Domain of f(x) is R

        (ii)    f(x) is even so graph will be symmetrical about y axis.

        (iii)    y = x⁴ - 2x² + 3 = (x² - 1)² + 2.

                So minimum value of y is at x² = 1(x = + 1).

        (iv)   When x = 0 the value of y = 3

                The graph of the function is as shown in fig.

(b)    y = f(x) = 2x/(1+x²).

        (i)     Domain = R

        (ii)    f(x) = -f(x), so function is odd the graph is not symmetric about any axis but symmetric about origin.

      So it is sufficient to consider only. x > 0

        (iii)    y = 0 when x = 0 there is no other point of intersection with co-  ordinate axes.

(iv)   As (x - 2)² > 0

       => x² + 1 > 2x

        So 2x/(x²+1) < 1 and equality holds at x = 1. Also from 0 to 1 the function increases and from 1 to a it decreases. So the graph is as shown in fig.

(c)            y = f(x) = sin² x - 2sinx

        (i)     Domain of y is R

        (ii)    0 < (sin x - 1)² < 4

                => 0 < sin² x - 2sinx + 1 < 4

                => -1 < sin² x - 2sinx < 3

        (iii)    f(x) has period 2∏ so it is

        Sufficient to draw the graph for domain [0, 2∏]

        (iv)   y = 0 for x = 0, n∏

Note :       More about increasing/decreasing we shall study in Module 5.

Example 17:

        Solve (x)² = [x]² + 2x

        Where [x] represents greatest integer less than or equal to x.

        (x) represents integer just greater than or equal to x.

Solution:

 Method 1:

 Case I :

        Let x = n ε I

        => Given equation becomes:

                n² = n² + 2n

                => n = 0

Case II:

        Let x ε I

        i.e. n < x < n + 1

        Given equation becomes:

                (n - 1)² = n² + 2x

                 => x = n + 1/2, n ε I

Therefore  x = 0 or x = n + 1/2; n ε I

 Method 2:

Case I :

        x I

        x = [x] + {x}; where {x} represent fraction part of x.

                x = (x) - (1 - {x})

        (x + 1 - {x})² = (x - {x})² + 2x (Using given equation)

                => (x + 1 - {x})² + 1 + 2 (x - {x})² = (x - {x})² + 2x

                => 1 - 2 {x} = 0

                => {x} = 1/2

                x = n + 1/2, n ε I

        Also, x = 0, by observation.

Example 18:

        Find the set x if the function f:[2, α] -> x where f(x) = 5 - 4x + x² is bijective.

Solution:

                y = x² - 4x + 5

                = (x - 2)² + 1

        When x = 2, y = 1

        As   x ε [2, α) then y  ε  [1, α]

        Therefore Set X ≡ [1, α)