Unit 1.1.1 (a) Algebraic Function
1. (a) Define linear function with example.Solution: A linear function is a mathematical function of the form \( f(x)= mx + c \), where its graph is a straight line with slope \( m \) and y-intercept \( c \). Example: \( f(x) = 3x + 4 \)
1. (b) What is the coordinates of vertex of \( f(x) = ax^2 + bx + c, a\neq 0 \) Solution: The vertex of \( f(x) = ax^2 + bx + c \) is \[ (h,k) = \left( \frac{-b}{2a}, \frac{4ac - b^2 }{4a } \right) \] 1. (c) Identify the identity function: \( f(x) = 5 \) and \( f(x) = x \) . Solution: Identity function is \( f(x) = x \)2. (a) Study the following graphs and identitify their nature as identity, constant, quadratic and cubic function.
Solution:
This is a graph of constant function.
This is a graph of identity function.
This is a graph of cubic function.
This is a graph of quadratic function.
This is a graph of cubic function
This is a graph of quadratic function.
3. (a) Draw the graph of \( y = x + 2 \)
Solution:
Given, \( y = x + 2 \)
When \( x = 1, y = 3 \)
When \( x =2, y = 4 \)
When \( x = 3, y = 5 \)
Therefore, passing points are (1,3), (2,4) and (3,5).
3. (b) Draw the graph of \( y = 6 \)
Solution:
Here \( y = 6 \) represents the horizontal line passing through (0,6) .
3. (c) Draw the graph of \( y = x^2 \)
Solution:
Given, \( y = x^2 \)
When, \( x = -3, y = 9 \)
When \( x = -2, y = 4 \)
When \( x = -1, y = 1 \)
When \( x = 0, y = 0 \)
When \( x= 1, y = 1 \)
When \( x = 2, y = 4 \)
When \( x = 3, y = 9 \)
Hence passing points are (\( - \)3,9), (\( - \)2,4), (\( - \)1,1), (0,0), (1,1), (2,4) and (3,9).
3. (d) Draw the graph of \( y = - x^2 \).
Solution:
Given, \( y = - x^2 \)
When \( x =-3, y = - 9 \)
When \( x = -2, y = -4 \)
When \( x = -1, y = -1 \)
When \( x = 1, y = -1 \)
When \( x = 2, y = -4 \)
When \( x = 3, y = -9 \)
Hence, the passing points are (-3,-9), (-2,-4), (-1,-1), (0,0), (1,-1), (2,-4) and (3,-9).
3. (e) Draw graph of \( y = x^3 \)
Solution:
Given, \( y = x^3 \)
Now,
When, \( x = -3, y = -27 \)
When \( x = -2 ,y = - 8 \)
When \( x= -1, y = -1 \)
When \( x = 0 , y = 0 \)
When \( x = 1, y = 1 \)
When \( x = 2, y = 8 \)
When \( x = 3, y = 27 \)
4. Pemba estimates the minimum ideal weight of a woman, in pounds is to multiply her height, in inches by 4 and subtract 130. Let \( y = \) minimum ideal weight and \( x = \) height.
(a) Express \( y \) as a linear function of \( x \).
Solution:
\[ y = 4x - 130 ...... (i) \]
(b) Find the minimum ideal weight of a woman whose height is 62 inches. Solution: Here, \begin{align} y & = 4x - 130 ...... (i) \\ \text{Putting } x & = 62 \text{ in equation (i), we get, } \\ y & = 4 \times 62 - 130 \\ \text{or, } y & = 248 - 130 \\ \therefore y & = 118 \text{ pounds } \end{align}
(c) Draw the graph of height and weight
Solution:
We have, \( y = 4x - 130 \)
When \( x = 0, y = -130 \)
When \( x = 5, y = - 110 \)
When \( x = 20, y = - 50 \)
When \( x = 40, y = 30 \)
All The Best
Unit: 1.1.1 (b) : Trigonometric Function
1. (a) Write the maximum and minimum values of \( f(x) = \sin x \) .Solution: The maximum and minimum values of \( f(x) = \sin x \) are +1 and -1 respectively. 1. (b) Write the maximum and minimum values of \( f(x) = \cos x \) . Solution: The maximum and minimum values of \( f(x) = \cos x \) are +1 and -1 respectively. 1. (c) Write the maximum and minimum values of \( f(x) = \tan x \) . Solution: The maximum and minimum values of \( f(x) = \tan x \) are \( + \infty \) and \( - \infty \) respectively. 2. (a) Write the period of \( f(x) = \sin x \) . Solution: The period of \( f(x) = \sin x \) is \( 2\pi^c \) .
3. (b) Draw the graph of \( f(x) = \cos x \ \ ( 0^\circ \le x \le 2\pi^c ) \)
Solution:
Given, \( f(x) = \cos x \ \ ( 0^\circ \le x \le 2\pi^c ) \)
Let us take the values of \( x \) differing 90° and corresponding values of \( y \) for \( y = \cos x \) The maximum and minimum values of \( \cos x \) are 1 and -1 respectively.
Now,
When \( x = 0, y = 1 \)
When \( x = \frac{\pi^c}{2} , y = 0 \)
When \( x = \pi^c , y = -1 \)
When \( x = \frac{3\pi^c}{2}, y = 0 \)
When \( x =2\pi^c, y = 1 \)
3. (c) Draw the graph of \( f(x) = \tan x \ \ ( 0^\circ \le x \le \pi^c ) \)
Solution:
Given, \( f(x) = \tan x \ \ ( 0^\circ \le x \le \pi^c ) \)
Now,
When \( x = 0^\circ, y = 0 \)
When \( x = \frac{\pi^c}{6} , y = 0.58 \)
When \( x = \frac{\pi^c}{4} , y = 1 \)
When \( x = \frac{\pi^c}{3} , y = 1.73 \)
When \( x = \frac{\pi^c}{2} , y = \text{ Undefined }\)
When \( x = \frac{2\pi^c}{3} , y = -1.73 \)
When \( x = \frac{3\pi^c}{4} , y = -1 \)
When \( x = \frac{5\pi^c}{6} , y = - 0.58 \)
When \( x = \pi^c , y = 0 \)
3. (d) Draw the graph of \( f(x) = 2\sin x \) \( \left( - \frac{\pi^c}{2} < x < \frac{\pi^c}{2} \right) \)
Solution:
Given, \( f(x) = 2\sin x \) \( \left( - \frac{\pi^c}{2} < x < \frac{\pi^c}{2} \right) \)
Now,
When \( x = -\frac{\pi^c}{2} , y = -2 \)
When \( x = -\frac{-\pi^c}{3} , y = -1.73 \)
When \( x = -\frac{-\pi^c}{4} , y = -1.41 \)
When \( x = -\frac{\pi^c}{6} , y = -1 \)
When \( x = 0 , y = 0 \)
When \( x = \frac{\pi^c}{6} , y = 1 \)
When \( x = \frac{\pi^c}{4} , y = 1.41 \)
When \( x = \frac{\pi^c}{3} , y = 1.73 \)
When \( x = \frac{\pi^c}{2} , y = 2 \)
Graph:
4. Study the topic 'sound' in physics of your science book and find the nature of longitudinal wave. Relate this concept with trigonometric function.
Solution:
The nature of longitudinal wave is like the graph of sine function.All The Best
Unit1.1.2 : Composite Function
1. (a) Let \( f: A\to B \) such that \( f(x) = y\) and \( g:B \to C \) such that \( g(y) = z \). Name the function defined from \( A \) to \( C \).Solution:
The function defined from \( A \) to \( C \) is \(gof(x) \).
1. (b) Write the difference between \( (fog)(x) \) and \( (gof)(x) \) .
Solution:
\( (fog)(x) \) is the composite function of \( g \) and \(f\) and read as \(g\) followed by \(f\).
\( (gof)(x) \) is the composite function of \( f \) and \(g \) and read as \( f \) followed by \( g \).
1. (c) Define composite function of \( f \) and \( g \).
Solution:
If \( f:A \to B \) and \( g:B\to C \) be any two functions then the new function devinded from \( A \to C \) is called composite function of \( f \) anf \( g \). It is denoted by \( gof(x) \).
2. (a) Given \(f \) and \( g \) are two real-valued functions defined as below. Find (i) domain of \( f \) (ii) domain of \( g \) (iii) domain of \( (fog) \) (iv) range of \( (gof)\) in each of the following if exist.
(a) \( f = \{ (3,4),(5,6), (9,10) \} \) and \( g =\{ (4,16),(6,36),(10,100) \} \).
Solution:
(i) Domain of \( f = \{ 3,5,9 \} \)
(ii) Domain of \( g = \{ 4,6,10 \} \)
(iii) Since range of \( g \ \ \cap \) domain of \( f = \phi \) so, \( fog \), does not exists.
(iv) Range of \( (gof) \) = range of \( g = \{ 16,36,100 \} \)
(b) \( f = \{(1, 2), (2, 3), (3, 4) \} \) and \( g = \{(2, 3), (3, 4), (4, 5)\}\)
Solution:
(i) Domain of \( f = \{ 1,2,3 \} \)
(ii) Domain of \( g = \{ 2,3,4 \} \)
(iii) As range of \( g \) is not a subset of domain of \(f \) , so \( fog \) can not be defined. Therefore, \( fog \) does not exists.
(iv) Range of \( (gof)\) = Range of \( g = \{ 3,4,5 \} \)
3. Given that \( f(x) = 2x-5 \) and \( g(x) = x^2 -2x+6 \) , calculate:
(a) \( (fog) (x) \) and \( (gof)(x) \)
Solution:
\begin{align} (fog)(x) & = f(g(x)) \\ \ & = f( x^2 - 2x +6 ) \\ \ & = 2(x^2 - 2x +6) - 5 \\ \ & = 2x^2 -4x + 12 - 5 \\ \ & = 2x^2 -4x + 7 \end{align} Also, \begin{align} (gof)(x) & = g(f(x))\\ \ & = g(2x-5) \\ \ & = (2x-5)^2 - 2(2x-5) +6 \\ \ & = (2x)^2 - 2\times 2x \times 5 + 5^2 -4x+10+6 \\ \ & = 4x^2 - 20 x + 25 -4x + 10 +6 \\ \ & = 4x^2 -24x+41 \end{align}
(b) \( (fog)(5) anf (gof)(4) \)
Solution:
\begin{align} (fog)(5) & = f(g(5)) \\ \ & = f( 5^2 - 2\times 5 +6 ) \\ \ & = f(25 -10+6) \\ \ & = f(21) \\ \ & = 2 \times 21 - 5 \\ \ & = 42 - 5 \\ \ & = 37 \end{align} Also, \begin{align} (gof)(4) & = g(f(4)) \\ \ & = g( 2 \times 4 - 5 ) \\ \ & = g(8 - 5 ) \\ \ & = g(3) \\ \ & = 3^2 - 2\times 3 +6 \\ \ & = 9 - 6 +6 \\ \ & = 9 \\ \end{align}
(c) \( (gog)(2) \) and \( (fof)(9) \)
Solution:
\begin{align} (gog)(2) & = g(g(2)) \\ \ & = g( 2^2 - 2\times 2 + 6 ) \\ \ & = g(4 - 4 +6 ) \\ \ & = g(6) \\ \ & = 6^2 -2\times 6 + 6 \\ \ & = 36 - 12 + 6 \\ \ & = 30\\ \end{align} Also, \begin{align} (fof)(9) & = f(f(9)) \\ \ & = f( 2\times 9 - 5 ) \\ \ & = f( 18 - 5 ) \\ \ & = f(13 ) \\ \ & = 2\times 13 - 5 \\ \ & = 26 - 5 \\ \ & = 21 \\ \end{align}
(d) \( (fog)(-4) \) and \( (gof)(-4) \)
Solution:
\begin{align} (fog)(-4) & = f(g(-4)) \\ \ & = f( (-4)^2 - 2(-4) +6 ) \\ \ & = f( 16 +8 +6 ) \\ \ & = f(30) \\ \ & = 2\times 30 - 5 \\ \ & = 60 - 5 \\ \ & = 55 \\ \end{align} Also, \begin{align} (gof)(-4) & = g(f(-4)) \\ \ & = g( 2\times (-4) -5 ) \\ \ & = g(-8 -5 ) \\ \ & = g(-13) \\ \ & = (-13)^2 - 2(-13) +6 \\ \ & = 169 + 26 + 6 \\ \ & = 201 \\ \end{align}
4. (a) If \( f(x) = x, g(x) = x + 1 \) and \( h(x) = 2x – 1 \) then find \( (f(goh)(x)) \) and \( (go(hof)(x)) \).
Solution:
\begin{align} \text{Given}, & \\ \ & f(x) = x \\ \ & g(x) = x+1 \\ \ & h(x) = 2x - 1 \end{align} Now, \begin{align} (goh)(x) & = g(h(x)) \\ \ & = g(2x-1) \\ \ & = 2x-1+1 \\ \ & = 2x \\ \therefore (goh)(x) & = 2x \\ \text{Also} & \\ (f(goh)(x)) & = f( (gof)(x) ) \\ \ & = f(2x) \\ \ & = 2x \\ \therefore (f(goh)(x)) & = 2x \\ \end{align} Also, \begin{align} (hof)(x) & = h(f(x)) \\ \ & = h(x) \\ \ & = 2x-1 \\ \therefore (hof)(x) & = 2x-1 \\ \text{Now} & \\ (go(hof)(x)) & = g(hof(x)) \\ \ & = g( 2x-1) \\ \ & = 2x-1+1 \\ \ & = 2x \\ (go(hof)(x)) & = 2x \\ \end{align}
4. (b) Given that \( f(x) = 2x -3, g(x) = x^3 + 2 \) and \( h(x) = x^2 -2x+3 \), find \( (f_o(g_oh)(x)) \) and \( ((h_of)_og(x)) \). (Taking composition of two functions as a single function).
Solution:
\begin{align} \text{Given}, & \\ f(x) & = 2x -3 \\ g(x) & = x^3 + 2 \\ h(x) & = x^2 -2x+3 \\ \end{align} \begin{align} (g_oh)(x) & = g(h(x)) \\ \ & = g(x^2 - 2x +3 ) \\ \ & = (x^2 - 2x + 3)^3 + 2 \\ \end{align} \begin{align} (f_o(g_oh)(x)) & = f(g_oh(x)) \\ \ & = f ( (x^2 - 2x +3 )^3 + 2 ) \\ \ & = 2 ((x^2 -2x+3)^3 +2) -3 \\ \ & = 2(x^2 -2x+3)^3 + 4 - 3 \\ \ & = 2(x^2 -2x+3)^3 + 1 \\ \therefore(f_o(g_oh)(x)) & = 2(x^2 -2x+3)^3 +1 \\ \end{align} \begin{align} (h_of)(x) & = h (f(x)) \\ \ & = h( 2x-3) \\ \ & = (2x-3)^2 - 2(2x-3)+3 \\ \ & = 4x^2 - 12x+9 - 4x+6 + 3 \\ \ & = 4x^2 - 16x + 18 \\ \therefore (h_o f)(x) & = 4x^2 - 16x + 18 \\ \end{align} \begin{align} ((h_of)_og(x)) & = (h_o f)(g(x)) \\ \ & = (h_o f)( x^3 +2 ) \\ \ & = 4(x^3+2)^2-16(x^3 + 2) + 18 \\ \therefore ((h_of)_og(x)) & = 4(x^3+2)^2-16(x^3 + 2) + 18 \\ \end{align}
5. If \( f \) and \( g \) are linear functions, what can you say about the domain of \((f_o g) \) and \( (g_o f) \) ? Explain.
Solution:
The domain of \((f_o g) \) = Doman of \( g = - \infty < x < \infty \).
The domain of \( (g_o f) \) = Domain of \( f = - \infty < x < \infty \).
6. Dolma determines the domain of \( f_o g \) by examining only the formula for \( (f_o g)(x). \) Is her approach valid? Why or why not?
Solution:
If Dolma determines the domain of \( fog \) by examining only the formula of \( (fog)(x) \) then her approach may not be valid. We know the domain of \( fog(x) \) is same as the domain of \( g(x). \) If formula of \( g(x) \) contains radical sign or fraction having zero values in the denominator for some values of \( x \) then the domain of \( fog(x) \) cannot be emamined just by the formula of \( (fog)(x). \) For example,
(i)Let’s define the functions \( f \) an \( g \) by \( f (x) = \frac{1 }{x} \) and \( g (x) = \frac{1}{x^2} \) then \( (fog)(x) = x^2 \) which is defined at \( x = 0 \) but \( g (x) \) is not defined at \( x = 0. \) That means 0 is not the element in the domain of \( g \) and hence 0 is not the element in the domain of \( fog \). (ii)Let’s define the functions \( f \) an \( g \) by \( f (x) = \frac{1}{x-2} \) and \( g (x) = \frac{1}{ x-3 } \) then \( (fog)(x) = \frac{ x-3}{4-x } \) which is defined at \( x = 3 \) but \( g (x) \) is not defined at \( x = 3.\) That means 3 is not the element in the domain of g and hence 3 is not the element in the domain of \( (fog)(x). \) From the above counter examples, we can claim that her approach is not valid. Her approach is valid if and only if both \( f(x) \) and \( g(x) \) and are linear functions.
Alternative: Let us start from an example.
\( f(x) = \frac{1}{x} \ \ x\neq 0 \)
\( g(x) = \frac{1}{x-1} \ \ x\neq 1 \)
Then, \begin{align} fog(x) & = f(g(x)) \\ & = f \left( \frac{1}{x-1} \right) \\ & = x - 1 \\ & \therefore fog(x) & = x - 1 \end{align}
Now if we examine domain of \( fog \) just by considering its formula for \( fog(x) \), certainly it will be whole number \( R \), also range will again be whole number \( R \). But as range for \( f \) and \( g \) excludes certain points in \( R \), so will \( fog \). Hence it will not be valid to examine domain of a composite function just by taking its direct formula into account. In example above, domain for \( fog \) must be certainly \( R - \{ 0, 1 \} \).
7. Write yourself any two real-valued function. Find their composition.
Solution:
Let \( f(x) = 2x - 1 \text{ and } g(x) = 3 - 5x = \) Now, \begin{align} fog(x) & = f(g(x)) \\ & = f( 3-5x) \\ & = 2(3-5x) - 1 \\ & = 6 - 10x - 1 \\ & = 5 - 10x \\ \therefore fog(x) & = 5 - 10x \end{align} Also, \begin{align} gof(x) & = g(f(x)) \\ & = g(2x-1) \\ & = 3-5(2x-1) \\ & = 3 - 10x+5 \\ & = 8 -10x \\ \therefore gof(x) & = 8 - 10x \end{align}
8. A stone is thrown into a pond, creating a circular ripple that spreads over the pond in such a way that the radius is increasing at the rate of 3 ft/sec.
(a) Find a function \( r(t) \) for the radius in terms of \( t. \)
Solution:
\( r(t) = 3t \)
(b) Find a function A(r) for the area of the ripple in terms of the radius r.
Solution:
\( A(r) = \pi r^2 \)
(c) Find \( (A_o r)(t).\) Explain the meaning of this function.
Solution:
\begin{align} ( A_o r ) (t) & = A(r(t)) \\ \ & = A ( 3t ) \\ \ & = \pi (3t)^2 \\ \ & = \pi (9 t^2 ) \\ \ & = 9 \pi t^2 \\ \therefore ( A_o r ) (t) & = 9 \pi t^2 \end{align} This function represents the area of circular ripple after \( t \) seconds.
All The Best
Unit 1.1.3: Inverse Function
1. (a) Define inverse of function, \( f:R\to R \)
Solution:
If \( f:A\to B \) is one to one onto function then the new function defined from B ot A is called inverse function. It is denoted by \( f^{-1} : B\to A \) .
(b) What is the relation between composition of a function and its inverse.
Solution:
The relation between composition of a function and its inverse is an identity function. \( \therefore fof^{-1} (x ) =f^{-1}of(x) = x \)
2. Represent the following functions in mapping diagram and find their inverse.
(a) \( f = \{ (1,2),(2,3),(4,5) \} \)
\( f^{-1} = \{(2,1), (3,2), (5,4) \} \)
(b) \( g = \{ (1,4),(2,5),(3,6) \} \)
Solution:
\( g^{-1} = \{ (4,1), (5,2), (6,3) \} \)
(c) \( h = \{ (-2,4),(-3,9),(-6,36) \} \)
Solution:
\( h^{-1} = \{ (4,-2), (9,-3), (36, -6) \} \)
3. If \( f \) is the real - valued function, find
(a) \( f^{-1}(x) \) (b) \( f^{-1}(6) \) (c) \( f^{-1}\left( \frac{1}{4} \right) \) (d) \( f^{-1}(-2) \) in each of the following:
(i) \( f(x) = 3x+1 \)
Solution:
\begin{align} (a) \ \ \text{Let } & y = f(x) \\ \text{or, } & y = 3x+1 \\ \text{or, } & y - 1 = 3x \\ \text{or, } & \frac{y-1}{3} = x \\ \text{or, } & x = \frac{y-1}{3} \\ \text{or, } & f^{-1}(y) = \frac{y-1}{3} \\ \therefore \ \ & f^{-1}(x) = \frac{x-1}{3} \end{align} \begin{align} (b) \ \ \text{Now } & f^{-1}(x) = \frac{x-1}{3} \\ \text{or, } & f^{-1}(6) = \frac{6-1}{3} \\ \therefore \ \ & f^{-1}(6) = \frac{5}{3} \\ \end{align} \begin{align} (c) \ \ \text{Now } & f^{-1}(x) = \frac{x-1}{3} \\ \text{or, } & f^{-1}(\frac{1}{4}) = \frac{\frac{1}{4}-1}{3} \\ \text{or, } & f^{-1}(\frac{1}{4}) = \frac{\frac{1-4}{4}}{3} \\ \text{or, } & f^{-1}(\frac{1}{4}) = \frac{-3}{12} \\ \therefore \ \ & f^{-1}(\frac{1}{4}) = \frac{-1}{4} \\ \end{align} \begin{align} (d) \ \ \text{Now } & f^{-1}(x) = \frac{x-1}{3} \\ \text{or, } & f^{-1}(-2) = \frac{-2-1}{3} \\ \text{or, } & f^{-1}(-2) = \frac{-3}{3} \\ \therefore \ \ & f^{-1}(-2) = - 1 \\ \end{align}
(ii) \( f(x) = 2x - 5 \)
Solution: \begin{align} (a) \ \ \text{Let } & y = f(x) \\ \text{or, } & y = 2x-5 \end{align} Interchanging the the role of x and y, we get \begin{align} \text{or, } & x=2y-5 \\ \text{or, } & x+5 = 2y \\ \text{or, } & \frac{x+5}{2} = y \\ \text{or, } & y = \frac{x+5}{2} \\ \therefore \ \ & f^{-1}(x) = \frac{x+5}{2} \end{align} \begin{align} (b) \ \ \text{Now } & f^{-1}(x) = \frac{x+5}{2} \\ \text{or, } & f^{-1}(6) = \frac{6+5}{2} \\ \therefore \ \ & f^{-1}(6) = \frac{11}{2} \\ \end{align} \begin{align} (c) \ \ \text{Now } & f^{-1}(x) = \frac{x+5}{2} \\ \text{or, } & f^{-1}(\frac{1}{4}) = \frac{\frac{1}{4}+5}{2} \\ \text{or, } & f^{-1}(\frac{1}{4}) = \frac{\frac{1+20}{4}}{2} \\ \text{or, } & f^{-1}(\frac{1}{4}) = \frac{21}{8} \\ \therefore \ \ & f^{-1}(\frac{1}{4}) = \frac{21}{8} \\ \end{align} \begin{align} (d) \ \ \text{Now } & f^{-1}(x) = \frac{x+5}{2} \\ \text{or, } & f^{-1}(-2) = \frac{-2+5}{2} \\ \text{or, } & f^{-1}(-2) = \frac{3}{2} \\ \therefore \ \ & f^{-1}(-2) = \frac{3}{2} \\ \end{align}
(iii) \( f(x) = \frac{x+1}{2} \)
Solution: \begin{align} (a) \ \ \text{Let } & y = f(x) \\ \text{or, } & y = \frac{x+1}{2} \\ \text{or, } & 2y=x+1 \\ \text{or, } & 2y-1=x \\ \text{or, } & x = 2y - 1 \\ \text{or, } & f^{-1}(y) = 2y-1 \\ \therefore \ \ & f^{-1}(x) = 2x-1 \\ \end{align} \begin{align} (b) \ \ \text{Now } & f^{-1}(x) = 2x-1 \\ \text{or, } & f^{-1}(6) = 2\times 6 - 1 \\ \text{or, } & f^{-1}(6) = 12 - 1 \\ \therefore \ \ & f^{-1}(6) = 11 \\ \end{align} \begin{align} (c) \ \ \text{Now } & f^{-1}(x) = 2x - 1 \\ \text{or, } & f^{-1}(\frac{1}{4}) = 2\times \frac{1}{4} - 1 \\ \text{or, } & f^{-1}(\frac{1}{4}) = \frac{1}{2} - 1 \\ \text{or, } & f^{-1}(\frac{1}{4}) = \frac{1-2}{2} \\ \therefore \ \ & f^{-1}(\frac{1}{4}) = \frac{-1}{2} \\ \end{align} \begin{align} (d) \ \ \text{Now } & f^{-1}(x) = 2x - 1 \\ \text{or, } & f^{-1}(-2) = 2(-2) - 1 \\ \text{or, } & f^{-1}(-2) = -4 - 1 \\ \therefore \ \ & f^{-1}(-2) = -5 \\ \end{align}
(iv) \( f = \left\{ \left(x,\frac{x-2}{x+2}\right), x\neq -2 \right\} \)
Solution:
\begin{align} (a) \ \ \text{Given, } & \\ f & = \left\{ \left(x,\frac{x-2}{x+2}\right), x\neq -2 \right\} \\ \text{or, } & f(x) = \frac{x-2}{x+2}, x \neq - 2 \\ \text{Now, } & \\ \text{Let } & y = f(x) \\ \text{or, } & y = \frac{x-2}{x+2} \\ \text{or, } & xy+2y =x-2 \\ \text{or, } & 2y + 2 = x - xy \\ \text{or, } & 2y + 2 = x(1-y) \\ \text{or, } & \frac{2y + 2}{1-y} = x \\ \text{or, } & x = \frac{2y + 2}{1-y} \\ \text{or, } & f^{-1}(y) = \frac{2y + 2}{1-y} \\ \therefore \ \ & f^{-1}(x) = \frac{2x + 2}{1-x} \\ \end{align}
\begin{align}(b) \ \ \text{Now } & f^{-1}(x) = \frac{2x + 2}{1-x} \\ \text{or, } & f^{-1}(6) = \frac{2\times 6 + 2}{1- 6 } \\ \text{or, } & f^{-1}(6) = \frac{12 + 2}{-5} \\ \text{or, } & f^{-1}(6) = \frac{14}{-5} \\ \therefore \ \ & f^{-1}(6) = - \frac{14}{ 5} \\ \end{align} \begin{align} (c) \ \ \text{Now } & f^{-1}(x) = \frac{2x + 2}{1-x} \\ \text{or, } & f^{-1}(\dfrac{1}{4}) = \dfrac{2\times \dfrac{1}{4} + 2}{1-\dfrac{1}{4}} \\ \text{or, } & f^{-1}(\dfrac{1}{4}) = \dfrac{\dfrac{1}{2} + 2}{\dfrac{4-1}{4}} \\ \text{or, } & f^{-1}(\dfrac{1}{4}) = \dfrac{\dfrac{1+4}{2}}{\dfrac{4-1}{4}} \\ \text{or, } & f^{-1}(\dfrac{1}{4}) = \dfrac{\dfrac{5}{2}}{\dfrac{3}{4}} \\ \text{or, } & f^{-1}(\frac{1}{4}) =\frac{5}{2}\times \frac{4}{3} \\ \therefore \ \ & f^{-1}(\frac{1}{4}) = \frac{10}{3} \\ \end{align} \begin{align} (d) \ \ \text{Now } & f^{-1}(x) = \frac{2x + 2}{1-x} \\ \text{or, } & f^{-1}(-2) = \frac{2(-2) + 2}{1-(-2)} \\ \text{or, } & f^{-1}(-2) = \frac{-4 + 2}{1+2} \\ \therefore \ \ & f^{-1}(-2) = \frac{-2}{3} \\ \end{align}
4. If \( f(x) = x + 1, g(x) = 2x, \) find
(a) \( (fog^{-1})(x) \) (b) \( (gof^{-1}) (x) \). \( f \) and \( g \) are real-valued functions.
Solution:
\begin{align} (a) \ \ \text{Let } & y_1 = g(x) \\ \text{or, } & y_1 = 2x \\ \text{or, } & \frac{y_1}{2} = x \\ \text{or, } & x = \frac{y_1}{2} \\ \text{or, } & f^{-1}(y_1) = \frac{y_1}{2} \\ \therefore \ \ & f^{-1}(x) =\frac{x}{2} \\ \text{Now, } & \\ (fog^{-1})(x) & = f(g^{-1}(x)) \\ \ & = f \left( \frac{x}{2} \right) \\ \ & = \frac{x}{2} + 1 \\ \ & = \frac{x+2}{2} \\ \therefore (fog^{-1})(x) & = \frac{x+2}{2} \\ \end{align} \begin{align} (b) \ \ \text{Let } & y_2 = f(x) \\ \text{or, } & y_2 = x+1 \\ \text{or, } & y_2 - 1 = x \\ \text{or, } & x = y_2 - 1 \\ \text{or, } & f^{-1}(y_2) = y_2 - 1 \\ \therefore \ \ & f^{-1}(x) = x - 1 \\ \text{Now, } & \\ (gof^{-1})(x) & = g(f^{-1}(x)) \\ \ & = g( x - 1 ) \\ \ & = 2(x-1) \\ \therefore (fog^{-1})(x) & = 2(x-1) \\ \end{align}
5. (a) If \( f(x) = 3x - 7, g(x) = \frac{x+2}{5} \) and \( (g^{-1}of)(x) = f(x),\) find the value of \(x \), \ \ \( f \text{ and } g \) are real-valued functions.
Solution:
\begin{align} (a) \ \ \text{Let } & y = \frac{x+2}{5} \\ \text{or, } & 5y = x + 2 \\ \text{or, } & 5y - 2 = x \\ \text{or, } & x = 5y - 2 \\ \text{or, } & f^{-1}(y ) = 5y - 2 \\ \therefore \ \ & f^{-1}(x) = 5x - 2 \\ \text{Now, } & \\ \ &(g^{-1}of)(x) = f(x), \\ \text{or, } & g^{-1} ( f(x) ) = 3x - 7 \\ \text{or, } & g^{-1} ( 3x - 7 ) = 3x - 7 \\ \text{or, } & 5(3x-7) - 2 = 3x - 7 \\ \text{or, } & 15x -35 - 2 = 3x - 7 \\ \text{or, } & 15x -35 - 2 = 3x - 7 \\ \text{or, } & 15x - 37 = 3x - 7 \\ \text{or, } & 15x - 3x = 37 - 7 \\ \text{or, } & 12 x = 30 \\ \text{or, } & x = \frac{30}{12} \\ \text{or, } & x = \frac{5}{2} \\ \end{align}
5. (b) \( f \) is real-valued function defined as \( f(x) = 3x+a\). If \( (fof)(6) = 10 \) then find the values of \( a \) and \( f^{-1}(4) \).
Solution:
\begin{align} \text{Given,} & \\ \ & f(x) = 3x + a \\ \text{Now,} & \\ \ & (fof)(6) = 10 \\ \text{or, } & f(f(6)) = 10 \\ \text{or, } & f( 3\times 6 + a ) = 10 \\ \text{or, } & f( 18 + a ) = 10 \\ \text{or, } & 3(18+a) + a = 10 \\ \text{or, } & 54+3a + a = 10 \\ \text{or, } & 4a = 10-54 \\ \text{or, } & 4a = -44 \\ \text{or, } & a = \frac{-44}{4} \\ \text{or, } & a = - 11 \\ \therefore \ & f(x) = 3x - 11 \\ \text{Now,} & \\ \text{Let } & y = f(x) \\ \text{or, } & y = 3x -11 \\ \text{or, } & y +11 = 3x \\ \text{or, } & \frac{y +11}{3} = x \\ \text{or, } & x = \frac{y+11}{3} \\ \text{or, } & f^{-1}(y) = \frac{y+11}{3} \\ \text{or, } & f^{-1}(x) = \frac{x+11}{3} \\ \text{or, } & f^{-1}(4) = \frac{4+11}{3} \\ \text{or, } & f^{-1}(4) = \frac{15}{3} \\ \therefore \ \ & f^{-1}(4) = 5 \\ \end{align}
6. Write the formula of volume and surface area of sphere in terms of radius. Find the functional relation and write their inverse.
Solution:
\begin{align} \text{Let radius of sphere be } x \\ \text{Then, volume of sphere } = \frac{4}{3} \pi x^3 \\ \therefore \ f(x) = \frac{4}{3} \pi x^3 \\ \text{To find inverse }\\ \text{Let } y = f(x) \\ \text{or, } y = \frac{4}{3} \pi x^3 \\ \text{or, } 3y = 4 \pi x^3 \\ \text{or, } \frac{3y}{4 \pi } = x^3 \\ \text{or, } \sqrt[3]{\frac{3y}{4 \pi } } = x \\ \text{or, } x = \sqrt[3]{\frac{3y}{4 \pi } } \\ \text{or, } f^{-1}(y) = \sqrt[3]{\frac{3y}{4 \pi } } \\ \therefore \ \ f^{-1}(x) = \sqrt[3]{\frac{3x}{4 \pi } } \\ \end{align} \begin{align} \text{Let radius of sphere be } x \\ \text{Then, surface area of sphere } = 4 \pi x^2 \\ \therefore \ f(x) = 4 \pi x^2 \\ \text{To find inverse }\\ \text{Let } y = f(x) \\ \text{or, } y = 4 \pi x^2 \\ \text{or, } \frac{y}{4 \pi } = x^2 \\ \text{or, } \pm \sqrt{\frac{y}{4 \pi } } = x \\ \text{or, } x = \pm \sqrt{\frac{y}{4 \pi } } \\ \text{or, } f^{-1}(y) = \pm \sqrt{\frac{y}{4 \pi } } \\ \therefore \ \ f^{-1}(x) = \pm \sqrt{\frac{x}{4 \pi } } \end{align}
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