Function and Polynomials
FUNCTION
Let A and B be two nonempty 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 preimage 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.
Composite Function
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.
Inverse Function
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 oneone 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 viceversa. 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 manyone 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).
Simple Algebraic Functions
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}
$$f(x)= a_ox^n +a_1x^n1+……………+ a_n1 x+a_n $$
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 xaxis 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 yintercept = 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
Trigonometric functions
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 codomain 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 
POLYNOMIALS
Simple Operations on Polynomials
Multiplication of 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*}(x)\times f(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 \(f(x)=x^23x+2\) and \(g(x)=x^3X^2+2x+4\),find \(f(x).g(x)\).
Solution:
Horizantal method:
\begin{align*} f(x)\times g(x)&=(x^23x+2)(x^3x^2+4) \\ &=x^2(x^3x^2+2x+4)3x(x^3x^2+2x+4)+2(x^3x^2+2x+4) \\ &=x^5x^4+2x^3+4x^23x^4+3x^36x^26x^212x+2x^32x^2+4x+8 \\ &=x^5+(13)x^4+(2+3+1)x^3+(462)x^2 (12+4)x+8 \\ &=x^54x^4+7x^34x^28x+8\end{align*}
Vertical Method:
\begin{align*}\frac{\frac{x^3x^2+2x+4\\\times\:x^23x+2}{2x^32x^2+4x+8\\3x^4+3x^36x^212x\\x^5x^4+2x^34x^2\\}}{x^54x^4+7x^34x^28x+8}\\ \end{align*}
Division of Polynomials
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.
Remainder Theorem
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
Factor theorem
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 xa, 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
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)
Factorization of a polynomial
Factor theorem and the synthetic division are very useful to find the factors of a polynomial.
Polynomial Equation
Let f(x) = a_{n}x^{n} + a_{n1}x^{n1} + ..... + 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).
Ifα is a real number such that f(α) = 0, thenα is called a root of the polynomial equation f(x) = 0.
1. 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 and domain of f is the range of g, then 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).
2.if \(f= {(1, 2),(3, 4),(5, 6)}\), then inverse of f i,e., \(f^1\)={(2, 1),(4, 3), (6, 5)}.
asdTThe composite function of f and g is denoted by gf and read as 'f followed by g'.
The composite function of g and f is denoted by fg and read as 'g followed by f'.
The composite function is a function of a function f^{2 }means fof or ff.
If the divisor is x+2 , then a = 2
If the divisor is 2x3 , then a = \(\frac{3}{2}\)
If the divisor is 2x+3 , then a = \(\frac{3}{2}\)
TThe
TT
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}
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}
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}
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}
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}
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}
Here, f(x) = 3x^{2} + x  28 and g(x) = 6x
f(x)⋅ g(x)
= (3x^{2} + x  28) (6x)
= 18x^{2} + 6x  168 3x^{3}x^{2} + 28x
= 3x^{3} + 17x^{2}+ 34x  168 _{Ans}
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}
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
x1 = 2y
2y = x1
y = \(\frac {x1}{2}\)
∴ f^{1} = \(\frac{x1}{2}\) _{Ans}
f(4x15) = 8x27
Or,
f(4x15) = 2(4x15) + 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}
g(2x5) = 8x19
g(2x5) = 4(2x5) +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}
f(x+3) = 3x+5
f(x+3) = 3(x+3) 4
f(x) = 3x4
Let y = f(x) = 3x4
Y = 3x4
Interchanging the place of x and y in the above relation
x = 3y4
Or, 3y = x+4
y = \(\frac {x+4}{3}\)
f^{1} (x) = \(\frac {x+4}{3}\) _{Ans}
fog(x) = 6x2
Or, f[g(x)] = 6x2
Or, f(2x) = 6x2
Or, f(2⋅\( \frac x2\) = 6 \(\frac x2\)2
f(x) = 3x2 _{Ans}
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}
Function f = {(x, 2x+1) : x∈R}
Let, y = 2x + 1
Interchanging the place of x and y
x = 2y + 1
or, x1 = 2y
or, y = \(\frac {x1}{2}\)
∴ f^{1}= {(x, \(\frac {x1}2\)) :x∈R } _{Ans}
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}
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}
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}
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}
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) = (xa) Q(x) + R = (xa) Q(x) + 0
f(x) = (xa) Q(x)
∴(xa) is a factor of f(x). _{proved}
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}
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}
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^{3} + x^{2}    
2x^{2}  4x + 2  
2x^{2} + 2x    
6x + 2  
6x  6 + +  
8 
∴ Quotient Q(x) = x^{2} + 2x  6
Remainder R(x) = 8 _{Ans}
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}
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}
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}
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 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}
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}
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}
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}
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}
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}
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}
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}
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}
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}
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}
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}
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^{3}  6x^{2}  445  +   
 x^{2} + 5x + 4  
 x^{2} + 3x +   
2x + 4  
2x  6  +  
10 
∴ Remainder = 10 _{Ans}
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}
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 \(\frac {k6}{k}\) = 8
or, k  6 = 8  28
or, k = 20 + 6
∴ k = 14 _{Ans}
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
= \(\frac {83}{3}\)
= \(\frac {11}{3}\)
∴ Remainder R =f( \(\frac 23\)) =\(\frac {11}{3}\) _{Ans}
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}
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}
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}
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}
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}
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 = \(\frac {x4}{3}\)
i.e. g^{1}(x) =\(\frac {x4}{3}\)
ii. fog^{1}(x)
= f\(\frac {x4}{3}\) [\(\because\) g^{1}(x) =\(\frac {x4}{3}\) ]
= 2 \(\frac {x4}{3}\) 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}
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}
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}
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}
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}
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}
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}
Let y = f(x) =\(\frac {x}{x3}\)
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}
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 = \(\frac {x3}{2}\)
∴ g^{1}(x) =\(\frac {x3}{2}\)
fg^{1}(x) = 3
f\(\frac {x3}{2}\) = 3
or, \((\frac {x3}{2})^2\)  2 \(\frac {x3}{2}\) = 3
or, \(\frac {x^2  6x + 9}{4}\) \(\frac {x3}{2}\) = 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}
Let: y =g(x) = \(\frac {1}{1x}\)
y = \(\frac {1}{1x}\)
Interchanging the place of x and y
x =\(\frac {1}{1y}\)
or, 1  y = \(\frac 1x\)
or, y = 1  \(\frac 1x\) = \(\frac {x  1}{x}\)
g^{1}(x) = \(\frac {x  1}{x}\)
g^{1}\(\frac 12\)
= \(\cfrac{(\frac{1}{2})1}{
\cfrac{1}{2}}\)
= \(\frac {12}{2}\)× \(\frac 21\)
= 1 _{Ans}
Again, fg(x) = f\(\frac {1}{1x}\) = 1 + \(\frac {2×1}{1x}\)
fg(x) = f\(\frac {1}{1x}\) = 1 + \(\frac {2×1}{1x}\)
= f\(\frac {1}{1x}\) = 1 + \(\frac {2×1}{1x}\)
= 1 + \(\frac {2×1}{1x}\)
= \(\frac {1  x + 2}{1  x}\)
= \(\frac {3  x}{1  x}\)
fg(1)
= \(\frac {3  (1)}{1  (1)}\)
= \(\frac {3 + 1}{1 + 1}\)
= \(\frac 42\)
= 2 _{Ans}
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= \(\frac {3  2x}{x  2}\)
 f1(x) = \(\frac {3  2x}{x  2}\)
 f1(1) = \(\frac {3  2(1)}{1  2}\) = \(\frac 11\) = 1
 fg(x) = f(x  2) =\(\frac {2(x2) + 3}{(x2) + 2}\) =\(\frac {2x  4 + 3}{x}\) = \(\frac {2x1}{x}\)
 fg(1) = \(\frac {2×11}{1}\) = \(\frac {21}{1}\)_{Ans}
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}
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}
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
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
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
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}
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}
(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}
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}
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
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}

Write the range and inverse function of the function.
f = {(2, (frac{1}{2})} , (3, (frac{1}{3})) , (4 , (frac{1}{4}));i:1;s:190:
;i:3;s:190:
;i:4;s:190:
;i:2;s:190:

b. If g(x) = 4x2, find the value of g(^1) (1) .
(frac{1}{2})
(frac{1}{4})
(frac{4}{1})
(frac{2}{1})

If f (^1) (x) = 2x3 then find f (x).
(frac{2 + 3}{x})
(frac{x + 3}{2})
(frac{x + 2}{3})
(frac{3 + x}{2})

If the function f = {(1,2) , (2,3), (3,4)} and g = {(2,a) , (4,c), (3,b)} , find it in ordered pair form .
gof = {(1,2), (a,c) , (3,b)}
gof = {(2,a), (1,b) , (3,c)}
gof = {(1,c), (3,a) , (2,b)}
gof = {(1,a), (2,b) , (3,c)}

If f (x) = 3x + 2 and g(x) = 5x + 1 then find gof(x).
6x + 1
3x + 6
1x + 6
6x + 3

If f (x = 2x3 and g (x) = x (^2) + 1 , find the value of fog (3) .
12
17
15
71

If f (x) = 3x , g (x) = x +2 and fog (x) = 18 , find the value of x.
15
4
3
9

The range of the function f (x) = 4x5 is {1 , 7 }, find its domain.
{1,3}
{1,7}
{1,4}
{3,1}

What will be the preimage of 4 in the function f(x) = (sqrt x)
16
2
8
40

If f (x) = 83x , evaluate f (^1) (4) and fof (2) .
4,2
2,8
4,8
2,4

If f (x) = (frac{x}{2x  3} ) and f (x) = f (^1) (x) , find the value of x .
2,0
3,0
2,3
0,2

If f (x) = (frac{14  3x}{4x}) and f (^1) (x) = 2 , find the value of x and prove that fof (^1) (A) is an identity function .
11
1
17
3

If f (x) = 3x  4 and f (^1) g(x) = (frac{3x+2}{3}) , find g (x) and f (^1) ( (frac{1}{2}) ) .
2x  2 , (frac{2}{2})
3x  2 , (frac{3}{2})
2x  3 , (frac{2}{3})
1x  2 , (frac{3}{3})

If f (x) = 2x  3 and f (^1) g(x) = (frac{x+11}{4}) , find g (x) and g (^1) ( (frac{1}{2}) )
f(x) = (frac{x+5}{3}) , 2
f(x) = (frac{x+3}{2}) , 3
f(x) = (frac{x+5}{5}) , 3
f(x) = (frac{x+5}{2}) , 5

If f (x) = 3x  1 and fog(x) = 6x+5 , then find gof (^1) .
(frac{3}{6})
(frac{x}{6})
(frac{1}{6})
(frac{6}{x})

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Smarikamany errors are there in exercise.....please fix it 
Jan 29, 2017 
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prakriti bhattraiif f=[{2,3},{1,2},{3,4}] and g=[{3,4},{4,5},{2,3}] find gf{1} and fg{2} 
Jan 06, 2017 
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krisuI didn't understand the question 
Dec 30, 2016 
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