Subject: Optional Mathematics

If f: A → B be a function from set A to set B and g: B → C be the function set B to set C, then the new function from set A to set C is called composite function of F and G. It is denoted by gof. The algebraic expressions in which the power of the variables is whole numbers are called polynomials. Rules of polynomials: 1 . Multiply each term of one polynomial by each term of the order. 2 . Arrange the terms in ascending or descending order.

Let A and B be two non-empty sets, then, every subset of cartesian product A\(\times \)B is a relation from A to B. Thus, a relation in which every element of set A is related (or associated) with a unique element of set B is said to be a function from A to B. Such as function is denoted by f : A→ B, which is read as "f is a function from A to B".

Symbolically, it follows that f: A → B if;

(i) f ⊆A×B

(ii) for each given x ∈ A, there exist a unique y∈B such that (x, y) ∈ f.

If an element y of B is associated with an element x of A, then y is an image of x under f and x is called the pre-image of y under f. We can also write y=f(x) and read f(x) as "f of x" or "f at x". A function is usually denoted by f, g, F, G, etc. A function is a relation in which, no two different ordered pairs have the same first component.

Let f and g be the function from A to B and from B to C, respectively, where A= {a1,b2,c3}, B = {c3,d4,e5} and C = {e5, f6, g7,h8,i9} and these are defined as follows:

Here, if f : A →B, a1 maps into c3 ; b2 maps into d4 ; and c3 maps into e5.

If g: B→ C, c3 maps into f6 ; d4 maps into g7 ; and e5 maps into h8.

thus we may write c3= f (a1), d4 = f(b2), e5= f(c3) and f6= g(c3), g7 = g(d4), h8 = f(e5)

Then, we have f6 = g(c3) = g [f(a1)],

g7 = g(d4) = g[f(b2)]

and h8=g(e5) = g[f(e3)]

this defines a set of ordered pairs of a function from A into C. This function is called the (product) composite function of g and f. it is denoted by gof or simply by gof. the nation gof indicates that f is applied first and then g. Thus, the arrow diagram of the new function is alongside.

Let f: A→ B be a function from A to B defined by the following arrow.

thus, the function f = { (a,1), (b,2) (c,3) } is one -one into where the domain of f = {a, b, c} and the range of f= { 1,2,3 }.

If we interchange the domain set of 'f' into range set and range set of into domain set, the set of ordered pairs will become {(1, a), (2, b), (3, c)}. Let this function e g. Then the function.

g= {(1, a), (2, b) (3, c) } is again a one-one onto, where the domain of g= {1,2,3} and the range of g={a,b,c}

the arrow diagram of this function g is as follows.

in such case, the function g is called the inverse function of and vice-versa. the inverse of the function f(x) is written as f^{-1}read as 'f inverse ' and we have y=f(x) if and only if x = \(f^{-1}\) (y) for every x∈D(f) and every y∈R(f). let u consider a many-one function defined by the following arrow diagram.

Thus, the function h= {(1, 2), (2,2), (3, 3) }

Let, we interchange the domain set and range set of h. we will get a relation R which is given below.

R= {(2, 1), (2, 2), (3, 3) }. Then the arrow diagram of this relation is given in the adjoining figure.

In relation R, two ordered pairs (2,1) and (2, 2) have the same first element 2. So, R is not a function.

It is concluded that a function f: A→ B will have its inverse function \(f^{-1}\) B→ A if and only if f is a one -one onto function.

Definition: If f: A→ B is a one to one onto function from A to B, then there exist a function g: B→ A such that the range of f is the domain of g is called the inverse of 'f' denoted by \(f{-1}\) (f inverse) such that y= f(x) if and only x=\(f{-1}\) (y) for every x∈D (f) and every y∈R(f).

A function that can be defined as the root of a polynomial equation is known as an algebraic function. we shall discuss some example of functions defined on the set R real number onto itself and these functions f(x) are defined by means of an equation. for instance the algebraic functions. \begin{align}f: R→R\end{align}

$$f(x)= a_ox^n +a_1x^n-1+……………+ a_n-1 x+a_n$$

For all x€R where the right side is a polynomial of degree n.

We shall discuss some particular causes of this equation.

**Constant Function:** For n=0, we have f(x)=a_{o}. it is usually denoted by f(x)=c.

In other words, a function is said to be a constant function if all its function values are the same. The graph of constant function is a straight line parallel to the x-axis at a given distance

**Linear functions**: For n=1, we have $$f(x) = a_ox + a_1$$. It is usually denoted by y = ax+b

In another word, a function is said to be a linear function if the polynomial is degree one. The graph of it is a straight line with slope m=a and y-intercept = for instance, y=x+2 is an equation of a straight line.

**Identity function**: if a=1 and b=0 in the linear function f(x) = ax +b, then we have

$$F(x) = x$$ for all $$x€R$$

This function is called identity function and its graph is shown below. It bisects the angle between the axes of coordinates.

**Quadratic function**: for n = 2, we have $$f(x)= a_1x^2 + a_1x + a_2.$$ It is written as $$f(x)=ax^2 + bx + c$$

This is a quadratic function which is polynomial of degree 2 and its graph is a parabola.

For instance, $$f(x) = x^{2} + x -2, f(x) = 4x^2, f(x) = x^2 + 2$$, etc are quadratic functions.

The graph of these functions is shown below.

**Cubic function** : for n= 3, we have\((x) f(x) = a_ox^3 +a_1x^2 + a_2x + a_3\) It is written as \(f(x)=ax^3 + bx^2 +cx + d\)

This is a cubic function which is polynomial of degree 3. For instance $$f(x) = x^3$$ is a cubic function whose graph is shown alongside

A function defined as the function which associates each angle with the definite real number.So, the domain of the trigonometric function is the set of angles and its co-domain is the set of real numbers. traditionally trigonometric functions are defined for angles of a triangle.But these trigonometric functions can be defined for angles of any magnitude.

The trigonometric ratios for angles such as 30^{0}, 45^{0}, 60^{0}, etc. can be calculated with the help of elementary plane geometry. The following table shows the values of trigonometrical ratios from 0^{0} to 360^{0}.

θ |
0^{0} |
30^{0} |
45^{0} |
60^{0} |
90^{0} |
120^{0} |
135^{0} |
150^{0} |
180^{0} |

Sinθ |
0 | \(\frac{1}{2}\) | \(\frac{1}{\sqrt{2}}\) | \(\frac{\sqrt{3}}{2}\) | 1 | \(\frac{\sqrt{3}}{2}\) | \(\frac{1}{\sqrt{2}}\) | \(\frac{1}{2}\) | 0 |

Cosθ |
1 | \(\frac{\sqrt{3}}{2}\) | \(\frac{1}{\sqrt 2}\) | \(\frac{1}{2}\) | 0 | \(-\frac{1}{2}\) | \(-\frac{1}{\sqrt{2}}\) | \(-\frac{\sqrt{3}}{2}\) | -1 |

Tanθ |
0 | \(\frac{1}{\sqrt{3}}\) | 1 | \(\sqrt 3\) | ∞ | \(-\sqrt 3\) | -1 | \(-\frac{1}{\sqrt{3}}\) | 0 |

Polynomial is an algebraic expression consist of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents. 2x, 5x^{2} + 3, 2a^{3} - b^{2} + 3x - 2, etc are the examples of polynomials.

Simple Operations on Polynomials

In order to multiply two polynomials, following steps are used.

(i) Multiply each term of one polynomial by each term of the other and

(ii) Then add (or subtract) the like terms thus obtained.

(iii) Simplify and arrange the like terms in ascending order or descending order.

In multiplying two polynomials, the law of indices \(x^m×x^n=x^{m+n}\) is applied.

it is noted that the degree of the product of two polynomials is equal to the sum of the degrees of the polynomial factors.

Worked out with examples

**Example 1**.

If \(f(x)=3x^3v^4\: and\:g(x)=5x^2y^3\), find \(f(x).g(x)\)

Solution:

\begin{align*}f(x)\times g(x) &=3x^3y^4×5^2y^3=(3×5)\:(x^4×y^3) \\ &=15.x^{3+2}.y^{4+3}=15x^4y^7\end{align*}

**Example 2.**

If \(f(x)=x^2-3x+2\) and \(g(x)=x^3-x^2+2x+4\), find \(f(x).g(x)\).

Solution:

\begin{align*} f(x)\times g(x)&=(x^2-3x+2)(x^3-x^2+2x+4) \\ &=x^2(x^3-x^2+2x+4)-3x(x^3-x^2+2x+4)+2(x^3-x^2+2x+4) \\ &=x^5-x^4+2x^3+4x^2-3x^4+3x^3-6x^2-6x^2-12x+2x^3-2x^2+4x+8 \\ &=x^5+(-1-3)x^4+(2+3+1)x^3+(4-6-2)x^2 (-12+4)x+8 \\ &=x^5-4x^4+7x^3-4x^2-8x+8\end{align*}

\begin{align*}\frac{\frac{x^3-x^2+2x+4\\\times\:x^2-3x+2}{2x^3-2x^2+4x+8\\-3x^4+3x^3-6x^2-12x\\x^5-x^4+2x^3-4x^2\\}}{x^5-4x^4+7x^3-4x^2-8x+8}\\ \end{align*}

We know from arithmetic how to divide ad integer by another smaller integer. If 30 is divided by 7, the quotient is 4 and the remainder is 2. i.e., \begin{align*}7)30(4 \\ \frac{-28}{2}\end{align*}

Here, we observe that \(30=7×4+2.\)

this relation ca be stated as follows.

Dividend= Divisor × Quotient + Reminder

Similarly, we can divide polynomials.

let \(f(x)\) and g(x) be two polynomials such that g(x) is a polynomial of smaller degree than that of \(f(x)\) and g(x)≠0. Then, there exist unique polynomials Q(x) and R(x) such that

\(f(x)=g(x).Q(x)+R(x)\)

where \(F(x)\)=dividend, g(x)=divisor, Q(x) is quotient and R(x) is remainder.

If R(x) = 0, then the divisor g(x) is a factor of the dividend f(x). The other factor of f(x) is the quotient Q(x).

The relation f(x) = g(x). Q(x) + R(x).

The following steps are used to divide a polynomial by the other:

(i) Arrange the divided f(x) and divisor g(x) in standard form i.e. generally descending powers of variable x.

(ii) Divided the first term divided f(x) by the first term of divisor g(x) to get the first term of quotient Q(x).

(iii) Multiply each term of divisor g(x) by the first term of quotient Q(x) obtained in step (ii) and subtract the product so obtained from the dividend f(x).

(iv) Take the remainder obtained in step (iii) as new dividend and continue the above process until the degree of the remainder is less than of the divisor.

**Statement: If a polynomial f(x) is divided by x - a, then the remainder is f(a).**

**Proof:** If we divided f(x) by x - a, then we get Q(x) as quotient and R as remainder.

Then, f(x) = (x - a). Q(x) + R

Put x = a. Then,

or, f(a) = (a - a) Q (a) + R

or, R = f(a)

Hence, remainder = f(a) = the value of polynomial f(x) = a

**Statement: if f (x) is a polynomial and a is real number, then (x - a) is a factor of f(x) if f(a) = 0**

**Proof:**If we divide f(x) by x-a, then we get Q(x) as quotient and R as remainder.

Then, f(x) = (x - a). Q(x) + R ...........(i)

Put x = a. Then

f(a) = (a - a). Q(a) + R

or, f(a) = R

When f(a) = 0. Then R = 0.

Putting the value of R in (i) we get,

f(x) = (x - a). Q(x)

So, (x - a) is a factor of f(x).

Hence, (x - a) is a factor of f(x) if f(a) = 0.

Synthetic division is the process which helps us to find the quotient and remainder when a polynomial f(x) is divided by x - a.

**Application of synthetic division**

Let Q(x) and R be the quotient and remainder when a polynomial f(x) is divided by ax -b.

Then, f(x) = (ax - b). Q(x) + R = a(x - \(\frac{b}{a}\)). Q(x) + R .

Where a.Q(x) = g(x)

or, Q(x) = \(\frac{1}{a}\) g(x).

Here, g(x) and R are the quotient and reminder when f(x) is divide by (x - \(\frac{b}{a}\)).

This result leads us to conclude that process of synthetic division discussed earlier is also useful to find out the quotient and remainder when f(x) is divided by (ax - b)

Factor theorem and the synthetic division are very useful to find the factors of a polynomial.

Let us see the following example:

**Factorize: x ^{2} + x - 2**

Constant term of this polynomial is 4 and the possible factors of 4 are: ±1, ±2.

Since the degree of f(x) is 2, so there will be at most two factors.

When x = 1, f(1) = 1+1-2 =0

(x-1) is a factor.

When x = -1, f(1) = -1-1-2 =-4

(x+1) is not a factor.

when x = 2, f(2) = 4+2-2 = 4

(x+2) is not a factor

When x = -2, f(1) = 4-2-2 = 0

(x+2) is a factor.

∴ (x-1) and (x+2) is a factor.

∴ x^{2} + x - 2 = (x-1)(x+2).

But instead of finding all the factors by using factor theorem, the synthetic division can be used after getting one with the help of factor theorem.

Let f(x) = a_{n}x^{n} + a_{n-1}x^{n-1} + ..... + a_{o} be the polynomial in x. Then f(x) = 0 is called a polynomial equation in x.

ax + b = 0 is a linear equation.

zx^{2}+ bx + c = 0 is a quadratic equation.

ax\( ^3\) + bx^{2} + cx + d = 0 is a cubic equation

ax\(^4\) + bx\(^3\) + cx^{2} + dx + e = 0 is a big quadratic equation.

( Here, a,b,c,d,e are the real numbers).

If α is a real number such that f(α) = 0, thenα is called a root of the polynomial equation f(x) = 0.

Functions:

Let A and B be two non-empty sets, then, every subset of cartesian product A\(\times \)B is a relation from A to B. Thus, a relation in which every element of set A is related ( or associated ) with a unique element of set B is said to be a function from A to B. Such as function is denoted by f : A→ B, which is read as "f is a function from A to B".

Polynomials:

Polynomial is an algebraic expression consist of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents. An example of a polynomial of a single indeterminate x is x2 − 4x + 7.

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

If f= {(1,3), (0,0), (-1,-3) and g= {(2,3), (-3,-1), (3,5)} show the function g_{o}f by an arrow diagram and find it in ordered pair form.

f= {(1,3), (0,0), (-1,-3)}, g= {(0,2), (-3,-1), (3,5)}, the above relation shown in the mapping diagram as follow:

g_{o}f= g(3) = 5

g_{o}f = g(0) =2

g_{o}f= g(-3) = -1

g_{o}f = g {(0,2), (-1,-1), (1,5)} _{Ans}

If f = {(-2,4), (-1,8), (0,0), (1,4), (2,6)} and g = {(0,-2), (4,0), (6,1), (8,-2)}, show the function g_{o}f by an arrow diagram and find it ordered pair form.

f = {(-2,4), (-1,8), (0,0), (1,4), (2,6)},

g = {(o,-2), (4,0), (6,1), (8,-2)}

The mapping diagram is shown below:

g_{o}f(-2) = g(4) = 0

g_{o}f(-1) = g(8) = -2

g_{o}f(0) = g(0) = 0

g_{o}f(1) = g(4)=0

g_{o}f(2) = g(6) = 1

g_{o}f= {(-2,0), (-1,-2), (0,-2) , (1,0), (2,1)} _{Ans}

If f = {(p,1), (q, 2), (r,3), (s,4)} and g = {(1,5), (2,8), (3,-3), (4,6)}, show the function f_{o}g by an arrow diagram and find it in ordered pair form.

f = {(p,1), (q,2), (r,3), (s,4)}, g = {(1,5), (2,8), (3,-3), (4,6)}

The mapping diagram is shown below:

f_{o}g = {(5,p), (8,q), (-3,r), (6,s)} _{Ans}

If f = {(x,3), (y,4), (z,7)}, g = {(3,a), (4,b), (7,c)}, show the function fog by an arrow diagram and find it in ordered pair form.

f = {(x,3), (y,4), (z,7)}, g = {(3,a), (4,b), (7,c)}

The mapping diagram is shown below:

f_{o}g = {(a,x), (b,y), (c,z)} _{Ans}

If f(x) = x^{3 }and g(x) = x^{2 }find the fg(x).

f(x) = x^{3} and g(x) = x^{2}

fg(x)

= f(x^{2}) [\(\because\) g(x) = x^{2}]

= (x^{2})^{3 }[\(\because\) f(x) = x^{3}]

= x^{6}_{Ans}

If f(x) = \(\frac 1x\), x ∈ R, x ≠ 0, find the value of ff(2).

f(x) =\(\frac 1x\)

ff(x) = f \(\frac 1x\) [\(\because\) f(x) = \(\frac 1x\)]

= \(\cfrac{1}{

\cfrac{1}{x}}\) =1× \(\frac x1\) = x

∴ ff(2) = 2 _{Ans}

If f(x) = 3x^{2} + x - 28 and g(x) = 6-x are the two polinomials. Find the value of f(x) ⋅ g(x).

Here, f(x) = 3x^{2} + x - 28 and g(x) = 6-x

f(x)⋅ g(x)

= (3x^{2} + x - 28) (6-x)

= 18x^{2} + 6x - 168 -3x^{3}-x^{2} + 28x

= -3x^{3} + 17x^{2}+ 34x - 168 _{Ans}

Write down the range and the inverse function of the following functions:

f = {(2, \(\frac 12\)), (3, \(\frac13\)), (4,\(\frac 14\))}

Here,

f = {(2,\( \frac 12\)), (3, \(\frac13\)), (4,\(\frac 14\))}

Range = { \(\frac 12\), \(\frac 13\), \(\frac 14\)}

Inverse function (f^{-1}) = {(\(\frac 12\), 2), (\(\frac 13\), 3), (\(\frac 14\), 4)}_{Ans}

Find the inverse function of the following function:

- {(3,2), (-1,-7), (4,5), (6,8)}
- {x, 2x+1 : x ∈ R}

Here,

- {(3,2), (-1,-7), (4,5), (6,8)}

Let f = {(3,2), (-1,-7), (4,5), (6,8)}

Changing domain into range and range into domain.

f^{-1} = {(2,3), (-7,-1), (5,4), (8,6)}

- {x, 2x+1 : x∈ R}

Let y = f(x) = 2x+1

Interchange the place of x and y

x = 2y + 1

x-1 = 2y

2y = x-1

y = \(\frac {x-1}{2}\)

∴ f^{-1} = \(\frac{x-1}{2}\) _{Ans}

If f(4x-15) = 8x-27 then find ff(2).

f(4x-15) = 8x-27

Or,

f(4x-15) = 2(4x-15) + 3

∴ f(x) = 2x+3

ff(x) = f(2x+3) = 2 (2x+3) +3 = 4x +6+3 = 4x+9

ff(2) = 4 × 2 + 9 = 17 _{Ans}

If g(2x-5) = 8x-19 find gg(2).

g(2x-5) = 8x-19

g(2x-5) = 4(2x-5) +1

g(x) = 4x+1

gg(x) =g(4x+1) = 4(4x+1) +1 = 16x+4+1 = 16x+5

gg(2) = 16×2+5 = 32+5 =37 _{Ans}

If f(x+3) = 3x+5 then find f^{-1}.

f(x+3) = 3x+5

f(x+3) = 3(x+3) -4

f(x) = 3x-4

Let y = f(x) = 3x-4

Y = 3x-4

Interchanging the place of x and y in the above relation

x = 3y-4

Or, 3y = x+4

y = \(\frac {x+4}{3}\)

f^{-1} (x) = \(\frac {x+4}{3}\) _{Ans}

If g(x) = 2x and fog(x) = 6x-2 then find f(x)

fog(x) = 6x-2

Or, f[g(x)] = 6x-2

Or, f(2x) = 6x-2

Or, f(2⋅\( \frac x2\) = 6 \(\frac x2\)-2

f(x) = 3x-2 _{Ans}

If f(x) = 8x-3 and fg(x) = 20x+1 find g(x).

Let g(x) = ax+b

fog(x) = f[g(x)] = f(ax+b) = 8(ax+b) -3

fog(x) = 8ax +8b - 3 ------ (1)

fog(x) = 20x+1 ------------(2)

From eq^{n} (1) and (2)

8ax +8b -3 = 20x +1

Equating on both sides,

8a = 20

a = \(\frac {20}{8}\)

∴ a = \(\frac 52\)

8b -3 = 1

8b = 4

b =\(\frac 48\) = \(\frac 12\)

∴ b = \(\frac 12\)

Putting the value of a and b in g(x) = ax + b

g(x) = \(\frac 52\) x + \(\frac 12\) = \(\frac {5x+1}{2}\)_{Ans}

Find the inverse function of {(x, 2x+1) : x ∈R}.

Function f = {(x, 2x+1) : x∈R}

Let, y = 2x + 1

Interchanging the place of x and y

x = 2y + 1

or, x-1 = 2y

or, y = \(\frac {x-1}{2}\)

∴ f^{-1}= {(x, \(\frac {x-1}2\)) :x∈R } _{Ans}

If f(x) = 2x + 3 and g(x) = 5x^{2} find fg (x) and gf (x).

f(x) = 2x + 3, g (x) = 5x^{2}

fg (x)

= f (5x^{2}) [\(\because\) g(x) = 5x^{2}]

= 2×5x^{2}+ 3

= 10x^{2} +3 _{Ans}

Also,

gf(x)

= g (2x + 3) [\(\because\) f(x) = 2x + 3 ]

= 5 (2x + 3)^{2}_{Ans}

If f(x) = x^{2} + 5 and g(x) = x - 3 find fog (2).

fog (2)

= f(2x - 3) [\(\because\) g(x) = x - 3 ]

= f(-1)

= (-1)^{2} + 5 [\(\because\) f(x) = x^{2} + 5 ]

= 1 + 5

= 6 _{Ans}

Using synthetic division method. Find the remainder when x^{3} + 3x^{2} - 2x + 6 is divided by x - 3.

x^{3} + 3x^{2} - 2x + 6÷ (x - 3)

3 | 1 | 3 | -2 | 6 |

3 | 18 | 48 | ||

1 | 6 | 16 | 54 |

The remainder (R) = 54 _{Ans}

State and proved the remainder theorem.

If the polynomial f(x) is divided by x - a, the remainder is f(a)

Proof: Let, Q(x) be the quotient and R be the remainder when f(x) is divided by (x - a) then, f(x) = (x -a)⋅Q(x) + R.

When x = a, f(a) = (a -a)⋅Q(x) + R ∴ f(a) = R _{proved}

State and proved the factor theorem.

If the polynomial f(x) is divided by (x - a) and R = f(a) = 0, then (x - a) is a factor of f(x).

Proof: Let Q(x)

Let Q(x) be the quotient when f(x) is divided by (x - a), then f(x) = (x-a) Q(x) + R = (x-a) Q(x) + 0

f(x) = (x-a) Q(x)

∴(x-a) is a factor of f(x). _{proved}

Using synthetic division method find the quotient and remainder of the expression x^{4} - x^{3} - 3x^{2} -2x + 5 where x + 1 = 0.

x^{4} - x^{3} - 3x^{2} -2x + 5÷ (x + 1)

-1 | 1 | -1 | -3 | -2 | 5 |

-1 | 2 | 1 | 1 | ||

1 | -2 | -1 | -1 | 6 |

∴ Quotient (Q) = x^{3} - 2x^{2} - x -1

Remainder (R) = 6 _{Ans}

Using synthetic division method, find the quotient and remainder when x^{3} - 2x + 3 is divided by x - 1.

x^{3} - 2x + 3 is divided by x - 1.

1 | 1 | 0 | -2 | 3 |

1 | 1 | -1 | ||

1 | 1 | -1 | 2 |

Quotient (Q) = x^{2} - x -1

Remainder (R) = 2 _{Ans}

Divide the polynomial: f(x) = x^{3} +3x^{2} -4x +2 and g(x) = x + 1.

Here,

f(x)÷ g(x) = \(\frac {x^3 +3x^2 -4x +2}{x +1}\)

x + 1 | x^{3} + 3x^{2} - 4x +2 | x^{2} + 2x - 6 |

x - - | ||

2x^{2} - 4x + 2 | ||

2x - - | ||

-6x + 2 | ||

-6x - 6 + + | ||

8 |

∴ Quotient Q(x) = x^{2} + 2x - 6

Remainder R(x) = 8 _{Ans}

If x - k is a factor of the polynomial x^{3} - kx^{2} - 2x + k + 4, find the value of k.

Here,

x - k is factor of the polynomialx^{3} - kx^{2} - 2x + k + 4, then x = k must satisfies the given polynomial, so that the given polynomial should be zero.

x - k is a factor of f(k) then f(k) = 0

f(k) = k^{3} - k⋅ k^{2} - 2k + k + 4

or, 0 = k^{3} - k^{3} -k +4

or, 0 = -k +4

∴ k = 4 _{Ans}

If 2x + 1 is a factor of 2x^{3} + ax^{2} + x + 2, find the value of x.

Here,

f(x) =2x^{3} + ax^{2} + x + 2

2x + 1 is a factor of f(x) so f(x) = 0

2x + 1 = o

or, 2x = -1

∴ x = - \frac 12

2x^{3} + ax^{2} + x + 2 = 0

or, 2× (- \(\frac 12)^3\) + a⋅ (- \(\frac 12)^2\) - \(\frac 12\) + 2 = 0

or, 2× - \(\frac 18\)+ a⋅ \(\frac 14\) - \(\frac 12\) + 2 = 0

or, - \(\frac 14\) + \(\frac a4\) - \(\frac 12\) + \(\frac 21\) = 0

or, \(\frac {-1 +a -2 +8}{4}\) = 0

or, a + 5 = 0

∴ a = -5 _{Ans}

Find the quotient and remainder when f(x) = 4x^{3} -6x^{2} -5 +3x is divided by x - 2 using systentic division method.

Arrange the given polynomial in decreasing order

f(x) = 4x^{3} -6x^{2}+3x -5

The constant term with the sign changed is +2

∴ x - 2 = x - (+2)

Writes the coefficient of f(x) in decreasing order:

2 | 4 | -6 | +3 | -5 |

↓ | 8 | 4 | 14 | |

4 | 2 | 7 | 9 | |

Q | R |

∴ Quotient Q(x) = 4x^{2} +2x +7

Remainder R(x) = 9 _{Ans}

Find the quotient and remainder when f(t) = 7t^{4} -4t^{3} +6t +3 is divided by t -3 using synthetic method of division.

Arranging the given polynomial in descending order

f(t) = 7t^{4} -4t^{3} +6t +3

The constant term with sign changed is + 3,

t - 3 = t - (+3)

Write the coefficient in descending order;

+3 | 7 | -4 | 0 | 6 | 3 |

↓ | 21 | 51 | 153 | 477 | |

7 | 17 | 51 | 159 | 480 | |

Q | R |

∴ Quotient Q(t) = 7t^{3} + 17x^{2} +51x +159

Remainder R(t) = 480 _{Ans}

(x + 2) is a factor of x^{3} + kx^{2} - 4x + 12, find the value of x.

(x + 2) is a factor of x^{3} + kx^{2} - 4x + 12 or x + 2 = 0 or x = -2

Putting the value x = -2 in the given expression and f(-2) = 0

x^{3} + kx^{2} - 4x + 12 =0

or, (-2)^{3} + k(-2)^{2} - 4× (-2) + 12 = 0

or, -8 + 4k + 8 + 12 = 0

or, 4k = -12

or, k = - \(\frac {12}{4}\) = -3

∴ k = -3 _{Ans}

If x - 3 is a factor of x^{3} + 4x^{2} + kx - 30, find the value of k.

x - 3 is a factor of the given expression.

So, x=3 must satisfy the given expression, then f(3) = 0

f(x) = x^{3} + 4x^{2} + kx - 30

or, f(3) = 3^{3} + 4×3^{2} + k⋅3 - 30

or, 0 = 27 + 36 +3k -30

or, 3k + 33 = 0

or, 3k = -33

or k = -\(\frac {33}{3}\) = -11

∴ k = -11 _{Ans}

If (x - 2) is a factor of x^{3} - kx^{2} - x - 2, find the value of k.

f(x) = x^{3} - kx^{2} - x - 2

If (x - 2) is a factor of f (x) then

f(2) = 0

f(2) =2^{3} - k⋅2^{2} - 2- 2

or, 0 = 8 - 4k - 4

or, 0 = 4 - 4k

or, 4k = 4

or, k = \(\frac 44\)

∴ k = 1 _{Ans}

What do you mean by a factor theorem? If x + 1 is a factor of 2x^{3} - kx^{2} - 8x + 5, find the value of k.

If f(a0 = 0, the remainder when f(x) is divided by (x - a) is zero then (x - a) is factor of f(x).

x + 1 = 0

or, x = -1

∴ f(-1) = 0

2x^{3} - kx^{2} - 8x + 5

Putting the value of x = -1 in above relation,

f(-1) = 2 (-1)^{3} - k (-1)^{2} - 8 (-1) + 5

or, - 2 - k + 8 + 5 = 0

or, 11 - k = 0

∴ k = 11 _{Ans}

A polynomial was divided by x - 1 and the quotient was x^{2} + 2x + 1 with the remainder 2. What was the original polynomial?

Quotient Q(x) =x^{2} + 2x + 1, Remainder R(x) = 2, Original polynomial = f(x)

We know,

f(x) = (x + 1) Q(x) + R(x)

or, f(x) = (x + 1) (x^{2} + 2x + 1) + 2

or, f(x) = x^{3} + 2x^{2} + x -x^{2} - 2x - 1 + 2

or, f(x) =x^{3} + x^{2}- x + 1 _{Ans}

If x - k is a factor of the polynomial x^{3} - kx^{2} - 2x + k + 4, find the value of k.

Let,

f(k) =x^{3} - kx^{2} - 2x + k + 4

If x - k is a factor of f(k) then f(k) = 0

f(k) = k^{3} - k⋅k^{2} - 2k + k + 4

or, 0 = k^{3} - k^{3} - k + 4

or, 0 = -k + 4

∴ k = 4 _{Ans}

If x + 3 is a factor of x^{3} - (k - 1) x^{2} + kx +54. find the value of k.

x + 3 is a factor of the given expression.

When x = -3, putting the value in the given expression equal to zero.

(-3)^{3} - (k - 1) (-3)^{2} + k(-3) +54 = 0

or, -27 -9k + 9 - 3k + 54 = 0

or, -12k + 36 = 0

or, -12k = -36

or k = \(\frac {-36}{-12}\)

∴ k = 3 _{Ans}

Define the remainder theorem and find the remainder when 3x^{3} - 5x^{2} + 2x - 3 is divided by x - 2 with the help of remainder theorem.

If a number C is substituted for x in the polynomial p(x) of degree n, then P(C) is the remainder that would be obtained by dividing p(x) by x - c.

i.e. P(x) = Q(x)⋅ (x - c) + P(c)

where, Q(x) is a polynomial of degree n -1

f(x) = 3x^{3} - 5x^{2} + 2x - 3 and g(x) = x - 2 = x - (-2)

If f(x)÷ g(x), quotient = Q and remainder (R) = ?

f(2) = 3(2)^{3} - 5(2)^{2} + 2×2 - 3 = 24 - 20 + 4 -3 = 28 - 23 = 5

Remainder (R) = 5 _{Ans}

State remainder theorem and use it to find the remainder when: x^{4} - 3x^{3} -2x^{2} +x +5 is divided by x + 1.

Remainder theorem: The remainder theorem states that if f(x) is divided by (x - a), then f(a) will be the remainder.

x + 1 is a factor ofx^{4} - 3x^{3} -2x^{2} +x +5

f(-1) =(-1)^{4} - 3(-1)^{3} -2(-1)^{2} +(-1) +5 = 1 +3 -2 -1 +5 = 6 _{Ans}

Calculate the value of p if x + 2 is a factor of 3x^{2} + px^{2} - 2x - 8.

x + 2 is a factor of 3x^{2} + px^{2} - 2x - 8, so the given expression is satisfies by x = -2

3x^{2} + px^{2} - 2x - 8 = 0

or, 3(-2)^{2} + p(-2)^{2} - 2(-2) - 8 = 0

or, -24 + 4p +4 -8 = 0

or, 4p = 28

∴ p = \(\frac {28}{4}\) = 7 _{Ans}

Define "Factor Theorem". Given that the function f(x) = 2x^{4} - 3x^{2} + 6x + k. if f(1) = 0, find the value of k.

Factor Theorem: If a polynomial f(x) is divided by (x - a) and remainder R = f(a) = 0 then x - a is a factor of f(x). This theorem is known as the factor theorem.

Given,

f(x) = 2x^{4} - 3x^{2} + 6x + k and f(1) = 0

If x = 1 then

2 × 1^{4} - 3 × 1^{2} + 6 × 1 + k = f(1)

or, 2 - 3 + 6 + k = 0

or, 5 + k = 0

∴ k = -5 _{Ans}

What do you mean by constant function? The range of the function f(x) = 7x - 8 is 12, find its domain.

Constant Function: A function f:A → B is called a constant function if there exists an element C∈B such that f(x) = C for all X∈A. A set containing only one element.

Let: Y = f(x) = 7x - 8

Y = 7x -8 [∴ Range = 13]

or, 7x = 13 + 8

or, x = \(\frac {21}{7}\) = 3

∴ Domain = 3 _{Ans}

Define the remainder theorem and find the remainder when 2x^{3} - 7x^{2} + 5x + 4 is divided by (x - 3).

Remainder Theorem: When a polynomial f(x) is divided by a linear polynomial x - a then the remainder R is given by the value f(a) of the polynomial, R = f(a).

x - 3 | 2x^{3} - 7x^{2} + 5x + 4 | 2x^{2} - x + 2 |

2x - + - | ||

- x^{2} + 5x + 4 | ||

- x + - | ||

2x + 4 | ||

2x - 6 - + | ||

10 |

∴ Remainder = 10 _{Ans}

Give the definition of remainder theorem. If a + 1 is a factor of a^{4} - 3a^{3} - 2a^{2} + a + p, find the value of p.

Remainder Theorem: If f(x) is a polynomial of degree n in x and if f(x)is divided by x - a, then the remainder is f(a). This theory is known as Remainder Theorey.

Let: f(x) =a^{4} - 3a^{3} - 2a^{2} + a + p

If a + 1 is factor of f(x) then f(-1) = 0

f(-1) =(-1)^{4} - 3(-1)^{3} - 2(-1)^{2} + (-1) + p

or, 0 = 1 + 3 - 2 - 1 + p

or, p + 1 = 0

∴ p = -1 _{Ans}

If f(2) = 8 and f(x) = 2x^{3} + 3x^{2} + k (1 - \(\frac {3x}{k}\) ), find the value of k.

f(2) = 8

f(2) =2×2^{3} + 3×2^{2} + k (1 - \(\frac {3×2}{k}\) )

or, 8 = 16 + 12 +k (1- \(\frac 6k\))

or, 28 + k \(\frac {k-6}{k}\) = 8

or, k - 6 = 8 - 28

or, k = -20 + 6

∴ k = -14 _{Ans}

Using the remainder theorem, find the remainder when 3x^{3} - 2x^{2} + 4x - 1 is divided by 3x + 2?

f(x) =3x^{3} - 2x^{2} + 4x - 1 and g(x) = 3x + 2

g(x) = x + \(\frac 23\) = x - (- \(\frac 23\))

In f(x)÷ g(x), remainder R = f(a) where a = - \(\frac 23\)

f(-\( \frac 23\)) = 3 (-\( \frac 23)^3\) + 2 (-\( \frac 23)^2\) + 4×(-\(\frac 23\)) - 1

= -\(\frac 89\) + \(\frac 89\) - \(\frac 83 \)- 1

= \(\frac {-8-3}{3}\)

= -\(\frac {11}{3}\)

∴ Remainder R =f(- \(\frac 23\)) =-\(\frac {11}{3}\) _{Ans}

If 2x - 5 is a factor of 6x^{3} - (k + 6)x^{2} + 2kx - 25, find the value of k.

f(x) =6x^{3} - (k + 6)x^{2} + 2kx - 25

If (2x - 5) is a factor of f(x) then f(\(\frac 52\)) = 0

f(\frac 52) = 6(\(\frac 52)^3\) - (k + 6)(\(\frac 52)^2\) + 2k(\(\frac 52\)) - 25

or, 0 = \(\frac {375}{4}\) - \(\frac{25}{4}\) (k + 6) + 5k - 25

or, 0 = \(\frac {375 - 25k - 150 + 20k - 100}{4}\)

or, -5k + 125 = 0

or, -5k = -125

or, k = \(\frac {-125}{-5}\)

∴k = 25 _{Ans}

Define inverse function. Find the inverse function of f = {(2, 5), (3, 6), (-4, 1). (7, 4)}.

Let, f: A→B be a one to one onto function then a function f^{-1} : B→A is called an inverse function of f. i.e.

f = {(2, 5), (3, 6), (-4, 1). (7, 4)}

f^{-1 }={(5, 2), (6, 3), (1, -4). (4, 7)} _{Ans}

If f(x) = 2x^{3} + 3x^{2} - 3x + p and f(2) = 8, find the value of P.

Here,

f(x) = 2x^{3} + 3x^{2} - 3x + p and f(2) = 8

Putting the value of x = 2,

f(2) = 2 × 2^{3} + 3 × 2^{2} - 3 × 2 + p

or, 8 = 16 + 12 - 6 + p

or, p = 8 - 22

∴ p = -14 _{Ans}

If f(x) = 2x^{4} - 3x^{2} + 6x + k and f(1) = 0, then find the value of k.

Here,

f(x) = 2x^{4} - 3x^{2} + 6x + k and f(1) = 0

Putting the value of x = 1

f(1) = 2 × 1^{4} - 3 × 1^{2} + 6 × 1 + k

or, 0 = 2 -3 + 6 + k

or, k + 5 = 0

∴ k = - 5 _{Ans}

Let f(x) = x^{2} and g(x) = 3x, find:

- fog(x)
- gof(x)
- fog(2)
- gof(2)

Here,

f(x) = x^{2} and g(x) = 3x

- fog(x) = f(3x) = (3x)
^{2}=9x^{2}_{Ans} - gof(x) = g(x
^{2}) = 3x^{2}_{Ans} - fog(2) = 9x
^{2}= 9 (2)^{2}= 36_{Ans} - gof(2) = 3x
^{2}= 3 (2)^{2}= 12_{Ans}

If f(x) = 2x - 3 and g(x) = 3x + 4 find:

- gof (x)
- fog
^{-1}(x) - f
^{-1}og(2) - fog (-3)

i. gof(x)

= g(2x - 3) [\(\because\) f(x) = 2x - 3]

= 3(2x - 3) + 4 [\(\because\) g(x) = 3x + 4]

= 6x - 9 + 4

= 6x -5 _{Ans}

Let, y = f(x) = 2x - 3

∴ y = 2x - 3

Interchanging the place of x and y

x = 2y - 3

or, 2y = x + 3

or, y = \(\frac {x+3}{2} \)

i.e. f^{-1}(x) =\(\frac {x+3}{2}\)

Let, y= g(x) = 3x + 4

∴ y = 3x + 4

Interchanging the place of x and y

x = 3y + 4

or, 3y = x - 4

or, y = \(\frac {x-4}{3}\)

i.e. g^{-1}(x) =\(\frac {x-4}{3}\)

ii. fog^{-1}(x)

= f\(\frac {x-4}{3}\) [\(\because\) g^{-1}(x) =\(\frac {x-4}{3}\) ]

= 2 \(\frac {x-4}{3}\) -3

= \(\frac {2x- 8 - 9}{3}\)

= \(\frac {2x - 17}{3} \)_{Ans}

iii. f^{-1}og (2)

= f^{-1}(3× 2 + 4)

=f^{-1}(10)

= \(\frac {10 + 3}{2}\) [\(\because\) \(\frac {x+3}{2}\)]

= \(\frac {13}{2}\) _{Ans}

iv. fog(-3)

= f(3× -3 + 4) [\(\because\) g(x) = 3x + 4]

= f( -9 + 4)

= f(-5)

= 2× (-5) - 3

= -10 - 3

= -13 _{Ans}

If f(x) = 8 - 3x, evaluate f^{-1} (-4) and ff(2).

let, y= f(x) = 8 - 3x

Interchanging the place of x and y

x = 8 - 3y

or, 3y = 8 - x

∴ y = \(\frac {8 - x}{3}\)

i.e. f^{-1}(x) = \(\frac {8 - x}{3}\)

i. f^{-1}(-4) = \(\frac {8 - (-4)}{3}\) = \(\frac {12}{3}\) = 4 _{Ans}

ff(x) = f(8 - 3x) = 8 - 3(8- 3x) = 8 - 24 + 9x = 9x - 16

ff(2) = 9× 2 - 16 = 18 - 16 = 2 _{Ans}

It is given that the function f(x) = 4x + 7 and g(x) = 3x - 5. If fg^{-1}(x) = 15, find the value of x.

Let, y = g(x) = 3x - 5

y = 3x - 5 ---------- (1)

Interchanging the place of xand y in equation (1)

x = 3y - 5

or, 3y = x + 5

∴ y = \(\frac {x + 5}{3}\)

i.e. g^{-1}(x) = \(\frac {x + 5}{3}\)

fg^{-1}(x) = 15

or, f\(\frac {x + 5}{3}\)=15

or, 4\(\frac {x + 5}{3}\) = 15

or, \(\frac {4x + 20 + 21}{3}\) = 15

or, 4x + 41 = 45

or, 4x = 45 - 41 = 4

or, x = \(\frac 44\)

∴ x = 1 _{Ans}

If f(x) = 3x + 5, find ff^{-1}(2) and f^{-1}(3).

Let, y = f(x) = 3x + 5

or, y= 3x + 5

Interchanging the position of x and y,

x = 3y + 5

or, 3y = x - 5

or, y = \(\frac {x - 5}{3}\)

∴ f^{-1}(x) =\(\frac {x - 5}{3}\)

f^{-1}(2) =\(\frac {2 - 5}{3}\) = - \(\frac 33\) = -1

ff^{-1}(2) = f(-1) = 3× (-1) + 5 = - 3 + 5 = 2

f^{-1}f(3) = f^{-1} (3× 3 +5) = f^{-1} (14) = \(\frac {14 - 5}{3}\) = \(\frac 93\) = 3

Hence, ff^{-1}(2) = 2

f^{-1} f(3) = 3 _{Ans}

If f(x) = 3x - 4, g(x) = x + 3, h(x) = - 2x + 1, find ghf(x) and g^{-1}hf(x).

ghf(x)

= gh(3x - 4)

= g [ -2 (3x - 4) + 1]

= g(-6x + 8 + 1)

= g(- 6x + 9)

= -6x + 9 +3

= - 6x + 12

Let,

y = g(x) = x + 3

∴ y = x + 3

Interchanging the place of x and y,

x = y + 3

or, x - 3 = y

i.e y = x - 3 [\(\because\) g^{-1}(x) = x - 3]

g^{-1}hf(x)

= g^{-1}h(3x - 4)

= g^{-1} [ - 2(3x - 4) + 1]

= g^{-1} (-6x + 8 + 1)

= g^{-1} (-6x + 9)

= - 6x + 9 -3

= - 6x + 6

Hence, ghf(x) = - 6x + 12

and, g^{-1}hf(x) = -6x + 6 _{Ans}

If f(x) = 3x + 4; g(x) = 2(x + 1), prove that f_{o}g = g_{o}f and find the value of f^{-1}(2).

Given,

f(x) = 3x + 4

g(x) = 2(x + 1) = 2x + 2

f_{o}g(x) = f(2x + 2) = 3 (2x + 2) + 4 = 6x + 6 + 4 = 6x + 10

g_{o}f(x) = g(3x + 4) = 2 (3x + 4) + 2 = 6x + 8 +2 =6x + 10

Hence, f_{o}g(x) = g_{o}f(x) _{proved}

Let y = 3x + 4 = f(x)

y = 3x + 4

Interchanging the place of x and y

x = 3y + 4

or, 3y = x - 4

or, y = \(\frac {x - 4}{3}\)

f^{-1}(x) = \(\frac {x - 4}{3}\)

∴ f^{-1}(2) = \(\frac {2 - 4}{3}\) = - \(\frac 23\)_{Ans}

If f(x) =\( \frac {1}{x}\), x ≠ 0.Prove that fof^{-1} = f^{-1}of.

Let y = f(x) = \(\frac 1x\)

y = \(\frac 1x\)

Interchanging the place of x and y

x = \(\frac 1y\)

y = \(\frac 1x\)

∴ f^{-1}(x) = \(\frac 1x\)

L.H.S

=fof^{-1} (x)

= f\(\frac 1x\)

=\( \cfrac{1}{

\cfrac{1}{x}}\)

= x

R.H.S

= f^{-1}of(x)

= f^{-1}\(\frac 1x\)

= \(\cfrac{1}{

\cfrac{1}{x}}\)

= x

Hence, L.H.S = R.H.S _{proved}

If f(x) = \(\frac{x}{x - 3}\), find the value of x for which f(x) = f^{-1}(x).

Let y = f(x) =\(\frac {x}{x-3}\)

y = \(\frac {x}{x - 3}\)

Interchanging the place of x and y,

x = \(\frac {y}{y - 3}\)

or, xy- 3x= y

or, xy - y = 3x

or, y (x - 1) = 3x

or, y = \(\frac {3x}{x - 1}\)

i.e., f^{-1}(x) = \(\frac {3x}{x - 1}\)

From the question, f(x) = f^{-1}(x)

\(\frac {x}{x - 3}\) =\(\frac {3x}{x - 1}\)

or, x^{2} - x = 3x^{2} - 9x

or, 3x^{2} - 9x - x^{2} + x = 0

or, 2x^{2} - 8x = 0

or, 2x(x - 4) = 0

Either, x = 0 and x - 4 = 0 or, x = 4

∴x = 0, 4 _{Ans}

It is given that f(x) = x^{2} - 2x and g(x) = 2x + 3. If fg^{-1}(x) = 3, calculate the value of x.

Here,

f(x) = x^{2} - 2x and g(x) = 2x + 3

Let: y = g(x) = 2x + 3

Interchanging the place of x and y

x = 2y + 3

or, 2y = x - 3

or, y = \(\frac {x-3}{2}\)

∴ g^{-1}(x) =\(\frac {x-3}{2}\)

fg^{-1}(x) = 3

f\(\frac {x-3}{2}\) = 3

or, \((\frac {x-3}{2})^2\) - 2 \(\frac {x-3}{2}\) = 3

or, \(\frac {x^2 - 6x + 9}{4}\) -\(\frac {x-3}{2}\) = 3

or, \(\frac {x^2 - 6x + 9 - 4x + 12}{4}\) = 3

or, x^{2} - 10x + 21 =12

or, x^{2} -10x + 21 - 12 = 0

or, x^{2} - 10x + 9 = 0

or, x^{2} - 9x - x + 9 = 0

or, x(x - 9) -1 (x - 9)= 0

or, (x - 9) (x - 1) = 0

Either, x - 9 = 0 i.e. x = 9

or, x - 1 = 0 i.e. x = 1

∴ x = 9 and 1 _{Ans}

If f(x) = 1 + 2x and g(x) = \(\frac {1}{1-x}\), find the value of g^{-1 }\(\frac 12\) and fg (-1).

Let: y =g(x) = \(\frac {1}{1-x}\)

y = \(\frac {1}{1-x}\)

Interchanging the place of x and y

x =\(\frac {1}{1-y}\)

or, 1 - y = \(\frac 1x\)

or, y = 1 - \(\frac 1x\) = \(\frac {x - 1}{x}\)

g^{-1}(x) = \(\frac {x - 1}{x}\)

g^{-1}\(\frac 12\)

= \(\cfrac{(\frac{1}{2})-1}{

\cfrac{1}{2}}\)

= \(\frac {1-2}{2}\)× \(\frac 21\)

= -1 _{Ans}

Again, fg(x) = f\(\frac {1}{1-x}\) = 1 + \(\frac {2×1}{1-x}\)

fg(x) = f\(\frac {1}{1-x}\) = 1 + \(\frac {2×1}{1-x}\)

= f\(\frac {1}{1-x}\) = 1 + \(\frac {2×1}{1-x}\)

= 1 + \(\frac {2×1}{1-x}\)

= \(\frac {1 - x + 2}{1 - x}\)

= \(\frac {3 - x}{1 - x}\)

fg(-1)

= \(\frac {3 - (-1)}{1 - (-1)}\)

= \(\frac {3 + 1}{1 + 1}\)

= \(\frac 42\)

= 2 _{Ans}

If function f(x) = \(\frac {2x + 3}{x + 2}\) and g(x) = x - 2, find the values of f^{-1}(x), f^{-1}(1), fg (x) and fg(1).

Let: y = f(x) = \(\frac {2x + 3}{x + 2}\)

Interchanging the position of x and y

x= \(\frac {2y + 3}{y + 2}\)

or, xy+ 2x - 2y = 3

or, y(x - 2) = 3 - 2x

∴y= \(\frac {3 - 2x}{x - 2}\)

- f-1(x) = \(\frac {3 - 2x}{x - 2}\)
- f-1(1) = \(\frac {3 - 2(1)}{1 - 2}\) = -\(\frac 11\) = -1
- fg(x) = f(x - 2) =\(\frac {2(x-2) + 3}{(x-2) + 2}\) =\(\frac {2x - 4 + 3}{x}\) = \(\frac {2x-1}{x}\)
- fg(1) = \(\frac {2×1-1}{1}\) = \(\frac {2-1}{1}\)
_{Ans}

If f(x) = \(\frac {3x + 11}{x - 3}\), x ≠ 3 and g(x) - \(\frac {x - 3}{2}\) are given functions. Find f^{-1}(x) and if f(x) = g^{-1}(x) find the value of x.

Let, y= f(x) =\(\frac {3x + 11}{x - 3}\)

Interchanging the value of x and y,

x =\(\frac {3y + 11}{y - 3}\)

or, xy - 3x = 3y + 11

or, xy - 3y = 3x + 11

or, y(x - 3) = 3x + 11

or, y = \(\frac {3x + 11}{x - 3}\)

∴ f^{-1}(x) = \(\frac {3x + 11}{x - 3}\) _{Ans}

Again,

Let: y = g(x) = \(\frac {x - 3}{2}\)

y =\(\frac {x - 3}{2}\)

Interchanging the place of x and y,

x =\(\frac {y - 3}{2}\)

or, 2x = y - 3

or, y = 2x + 3

∴ g^{-1}(x) = 2x + 3

From the given question,

f(x) = g^{-1}(x)

or,\(\frac {3x + 11}{x - 3}\) = 2x + 3

or, 3x + 11 = 2x^{2} + 3x - 6x - 9

or, 2x^{2} - 3x - 9 - 3x - 11 = 0

or, 2x^{2} - 6x - 20 = 0

or, 2(x^{2} - 3x - 10) = 0

or, x^{2} - 5x + 2x - 10 = 0

or, x(x - 5) + 2(x - 5) = 0

or, (x - 5) (x + 2) = 0

Either, x - 5 = 0∴ x = 5

Or, x + 2 = 0∴ x = -2

Hence, x = 5 or -2 _{Ans}

^{-1}(x) = 15, find the value of x.

Here,

Let: y = g(x) = 3x - 5

y = 3x - 5 ------------------(1)

Interchanging the position of x and y in equation (1),

x = 3y - 5

or, 3y = x + 5

or, y = \(\frac {x + 5}{3}\)

∴ g^{-1}(x) = \(\frac {x + 5}{3}\)

fg^{-1}(x) = 15

or, f \(\frac {x + 5}{3}\) = 15

or, 4 \(\frac {x + 5}{3}\) + 7 = 15

or, \(\frac {4x + 20}{3}\) + 7 = 15

or, \(\frac {4x + 20 + 21}{3}\) = 15

or, 4x + 41 = 45

or, 4x = 45 - 41

or, 4x = 4

or, x = \(\frac 44\)

∴ x = 1 _{Ans}

Solve the equation:

x^{3} - 4x^{2} + x + 6 = 0

x^{3} - 4x^{2} + x + 6 = 0

(x - 2) is a factor ofx^{3} - 4x^{2} + x + 6

or, x^{3} - 2x^{2} - 2x^{2} + 4x - 3x + 6 = 0

or, x^{2}(x - 2) - 2x(x - 2) - 3(x - 2) = 0

or, (x - 2) (x^{2} - 2x - 3) = 0

or, (x - 2) (x^{2} - 3x + x - 3) = 0

or, (x - 2) [x(x - 3) + 1(x - 3)] = 0

or, (x - 2) (x - 3) (x + 1) = 0

Either, x - 2 = 0 ∴ x = 2

Or, x - 3 = 0 ∴ x = 3

or, x + 1 = 0 ∴ x = - 1

Hence, x = 2, 3, -1 Ans

Rough:

x = 2

x^{3} - 4x^{2} + x + 6

= 2^{3} - 4⋅ 2^{2} + 2 +6

= 16 - 16

= 0

Solve:

2x^{3} + 3x^{2} - 11x - 6 = 0

x - 2 is a factor of given expression

or, 2x^{3} - 4x^{2} + 7x^{2} - 14x + 3x - 6 = 0

or, 2x^{2} (x - 2) + 7x (x - 2) + 3 (x - 2) = 0

or, (x - 2) (2x^{2} + 7x + 3) = 0

or, (x - 2) (2x^{2} + 6x + x +3) = 0

or, (x - 2) [2x (x + 3) + 1 (x + 3)] = 0

or, (x - 2) (x + 3) (2x + 1) = 0

Either, x + 3 = 0∴ x = -3

Or, x - 2 = 0∴ x = 2

Or, 2x + 1 = 0∴ x = -\(\frac 12\)

∴ x = 2 , - 3 , -\(\frac12\) _{Ans}

Rough:

If x = 2

2x^{3} + 3x^{2} - 11x - 6

= 2(2)^{3} + 3(2)^{2} - 11(2) - 6

= 16 + 12 - 22 - 6

= 0

Solve:

3x^{3} - 13x^{2} + 16 = 0

x = -1 is correct.

x + 1 is a factor of above equation.

or, 3x^{3} + 3x^{2} - 16x^{2} - 16x + 16x + 16 = 0

or, 3x^{2}(x + 1) - 16x(x + 1) + 16(x + 1) = 0

or, (x + 1) (3x^{2} - 16x + 16) = 0

or, (x + 1) (3x^{2} - 12x - 4x + 16) = 0

or, (x + 1) [3x (x - 4) - 4 (x - 4)] = 0

or, (x + 1) (x - 4) (3x - 4) = 0

Either, x + 1 = 0 ∴ x = - 1

Or, x - 4 = 0 ∴ x = 4

Or, 3x - 4 = 0 ∴ x = \(\frac 43\)

∴ x = -1, 4 and \(\frac 43\) _{Ans}

Rough:

Put x = -1

or, 3(-1)^{3} - 13(-1)^{2} + 16 = 0

or, -3 -13 + 16 = 0

or, - 16 + 16 = 0

∴ 0 = 0

Solve:

x^{3} - 19x - 30 = 0

x + 2 is a factor of given equation.

x^{3} + 2x^{2} - 2x^{2} - 4x - 15x - 30 = 0

or, x^{2}(x + 2) - 2x(x + 2) - 15(x + 2) = 0

or, (x + 2) (x^{2} - 2x - 15) = 0

or, (x + 2) (x^{2} - 5x + 3x - 15) = 0

or, (x + 2) [x(x - 5) + 3(x - 5)] = 0

or, (x + 2) (x - 5) (x + 3) = 0

Either, x + 2 = 0 ∴ x = -2

Or, x - 5 = 0 ∴ x = 5

Or, x + 3 = 0 ∴ x = -3

∴ x = -2, -3, 5 _{Ans}

Solve:

x^{3} - 3x - 2 = 0

x^{3} - 3x - 2 = 0

x + 1 is a factor of given equation,

or, x^{3} + x^{2} - x^{2} - x - 2x - 2 = 0

or, x^{2} (x + 1) - x (x + 1) - 2 (x + 1) = 0

or, (x + 1) (x^{2} - x - 2) = 0

or, (x + 1) [x(x - 2) + 1 -(x - 2)] = 0

or, (x + 1) (x - 2) (x + 1) = 0

Either, x + 1 = 0 ∴ x = -1

Or, x - 2 = 0 ∴ x = 2

∴ x = -1 and 2 _{Ans}

Solve:

(x + 1) (x + 2) (x + 3) (x + 4) - 8 = 0

(x + 1) (x + 2) (x + 3) (x + 4) - 8 = 0

or, (x^{2} + x + 4x + 4) ( x^{2} + 2x + 3x + 6) - 8 = 0

or, (x^{2} + 5x + 4) (x^{2} + 5x + 6) - 8 = 0

Let: x^{2} + 5x = k

(k + 4) (k + 6) - 8 = 0

or, k^{2} + 6k + 4k + 24 - 8 = 0

or, k^{2} + 10k + 16 = 0

or, k^{2} + 8k + 2k + 16 = 0

or, k(x + 8) + 2(x + 8) = 0

or, (k + 2) (x + 8) = 0

Putting the value of k

(x^{2} + 5x + 8) (x^{2} + 5x + 2) = 0

Either, x^{2} + 5x + 8 = 0 --------------(1)

Or, x^{2} + 5x + 2 = 0 ------------------(2)

Taking eq^{n}(1)

x =\(\frac {-5 ± (\sqrt{5^{2}-4×1×8})}{2×1}\) =\(\frac {-5 ± \sqrt{-7}}{2×1}\) (Impossible)

Taking eq^{n}(2)

x =\(\frac {-5 ± (\sqrt{5^{2}-4×1×2})}{2×1}\) =\(\frac {-5 ± \sqrt{17}}{2×1}\) _{Ans}

Divide the polynomials using synthetic method.

f(x) = x^{4} - x^{3} - 3x^{2} - 2x + 5, where x = 2, -1, 3

Here,

f(x) = x^{4} - x^{3} - 3x^{2} - 2x + 5

When x = 2 then,

2 | 1 | -1 | -3 | -2 | 5 |

2 | 2 | -2 | -8 | ||

1 | 1 | -1 | -4 | -3 |

∴ Quotient = x^{3} + x^{2} - x - 4 and f(x) = -3 _{Ans}

When x = -1 then,

-1 | 1 | -1 | -3 | -2 | 5 |

-1 | 2 | 1 | 1 | ||

1 | -2 | -1 | -1 | 6 |

∴ Quotient = x^{3} - 2x^{2} - x - 1 and f(2) = 6 _{Ans}

When x = 3 then,

3 | 1 | -1 | -3 | -2 | 5 |

3 | 6 | 9 | 21 | ||

1 | 2 | 3 | 7 | 26 |

∴ Quotient = x^{3} + 2x^{2} + 3x + 7 and f(3) = 26 _{Ans}

Solve:

x^{3} - x^{2} - 14x + 24 = 0

x - 2 = 0

x - 2 is a factor of given expression

or, x^{3} - x^{2} - 14x + 24 = 0

or, x^{3} - 2x^{2} + x^{2} - 2x - 12x + 24 = 0

or, x^{2}(x - 2) + x(x- 2) - 12(x - 2) = 0

or, (x - 2) (x^{2} + x -12) = 0

or, (x - 2) (x^{2} + 4x - 3x - 12) = 0

or, (x - 2) [x(x + 4) - 3(x + 4)] = 0

or, (x - 2) (x + 4) (x -3) = 0

Either, x - 2 = 0 ∴x = 2

Or, x + 4 = 0 ∴ x = - 4

Or, x - 3 = 0 ∴ x = 3

∴ x = 2, 3, -4 Ans

Rough:

x = 2

i.e 2^{3} - 2^{2} - 14× 2 + 24

= 8 - 4 -28 + 24

= 0

Solve:

2x^{3} - 3x^{2} - 3x + 2 = 0

2x^{3} - 3x^{2} - 3x + 2 = 0

or, 2x^{3} + 2 - 3x^{2} - 3x = 0

or, 2(x^{3} + 13) - 3x (x + 1) = 0

or, 2(x + 1) (x^{2} - x + 1) - 3x (x + 1) = 0

or, (x + 1)[2x^{2} - 2x + 2 - 3x] = 0

or, (x + 1) (2x^{2} - 5x + 2) = 0

or, (x + 1) (2x^{2} - 4x - x + 2) = 0

or, (x + 1) [2x(x - 2) - 1(x - 2)] = 0

or, (x + 1) (x - 2) (2x - 1) = 0

Either, x + 1 = 0 ∴x = -1

Or, x - 2 = 0 ∴x = 2

Or, 2x - 1 = 0 ∴ x = \(\frac 12\)

∴ x = -1, 2 , \(\frac12\) _{Ans}

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