Long Questions

Here,
g(x) = 2x - 3
fog(x) = 6x - 11
f^-1(x) = ?

Let, f(x) = mx + c --- (i)
so, f(g(x)) = m(g(x)) + c
or, 6x - 11 = m(2x - 3) + c [fog(x) = f(g(x))]
or, 6x - 11 = 2mx - 3m + c

Using these values at (i),
f(x) = 3x - 2

For f^-1(x),
f(x) = y = 3x - 2
or, y + 2 = 3x
or, $\frac{\text{y + 2}}{3}$ = x
Interchanging the value of 'x' and 'y', we get,
or, $\frac{\text{x + 2}}{3}$ = y
∴ f^-1(x) = $\frac{\text{x + 2}}{3}$

Here,
f(x) = 2x - 3
gof(x) = 6x - 11
g^-1(x) = ?

Let, g(x) = mx + c --- (i)
so, g(f(x)) = m(f(x)) + c
or, 6x - 11 = m(2x - 3) + c [gof(x) = g(f(x))]
or, 6x - 11 = 2mx - 3m + c

Using these values at (i),
g(x) = 3x - 2

For g^-1(x),
g(x) = y = 3x - 2
or, y + 2 = 3x
or, $\frac{\text{y + 2}}{3}$ = x
Interchanging the value of 'x' and 'y', we get,
or, $\frac{\text{x + 2}}{3}$ = y
∴ g^-1(x) = $\frac{\text{x + 2}}{3}$

Here,
f = {x, 5x - 13}
f(x)= 5x - 13
g = {x, $\frac{\text{2x + 7}}{3}$}
g(x) = $\frac{\text{2x + 7}}{3}$
g^-1(x) = fof(x)
x = ?

Now, for g^-1(x),
g(x) = y = $\frac{\text{2x + 7}}{3}$
or, $\frac{\text{3y - 7}}{2}$ = x
Interchanging the value of 'x' and 'y', we get,
or, $\frac{\text{3x - 7}}{2}$ = y
∴ g^-1(x) = $\frac{\text{3x - 7}}{2}$ ---- (A)

Again,
fof(x) = f(5x - 13)
= 5(5x - 13) - 13
= 25x - 65 -13
= 25x - 78 ---- (B)

We have given,
g^-1(x) = fof(x)
or, $\frac{\text{3x - 7}}{2}$ = 25x - 78 (from 'A' and 'B')
or, 3x - 7 = 50x - 156
or, 149 = 47x
∴ x = $\frac{\text{149}}{\text{47}}$

Here,
2.f(x) = kx - 3
f(x) = $\frac{\text{kx - 3}}{2}$
g(x) = $\frac{3}{\text{x + 2}}$
fog^-1(3) = $\frac{-1}{4}$
k = ?

Now, for g^-1(x),
g(x) = y = $\frac{3}{\text{x + 2}}$
or, x + 2 = $\frac{3}{y}$
or, x = $\frac{3}{y}$ - 2
Interchanging the value of 'x' and 'y', we get,
y = $\frac{\text{3 - 2x}}{x}$
∴ g^-1(x) = $\frac{\text{3 - 2x}}{x}$
∴ g^-1(3) = $\frac{\text{3 - 6}}{3}$ = -1

Again, we have given,
fog^-1(3) = $\frac{\text{-1}}{\text{4}}$
or, f(-1) = $\frac{\text{-1}}{\text{4}}$
or, -k - 3 = $\frac{\text{-2}}{\text{4}}$
or, k + 3 = $\frac{\text{1}}{\text{2}}$
∴ k = $\frac{\text{-5}}{\text{2}}$

Solution,
$\begin{aligned} Given, f(x) &\text{ = } 3x - 7 \\ g(x) &\text{ = } \frac{\text{5x + 2}}{3} \\ g^{-1}f(x) &\text{ = } 8 \\ \end{aligned}$
To find, x = ?

Now, for g^-1(x),
$\begin{aligned} g(x) = y &\text{ = } \frac{\text{5x + 2}}{3} \\ or, 3y - 2 &\text{ = } 5x \\ or, \frac{\text{3y - 2}}{5} &\text{ = } x \\ \end{aligned}$
Interchanging the value of 'x' and 'y', we get,
y = $\frac{\text{3y - 2}}{5}$
∴ g^-1(x) = $\frac{\text{3x - 2}}{5}$

We have given,
g^-1f(x) = 8
or, g^-1(3x - 7) = 8
or, $\frac{\text{3(3x - 7) - 2}}{5}$ = 8
or, 9x - 21 - 2 = 40
∴ x = $\frac{\text{63}}{\text{9}}$ = 7

Solution,
$\begin{aligned} Given, f(x) &\text{ = } \frac{\text{x + 1}}{2} \\ g(x) &\text{ = } \frac{\text{x - 5}}{2} \\ fog^{-1}(x) &\text{ = } 6 \\ \end{aligned}$
To find, x = ?

Now, for g^-1(x),
$\begin{aligned} g(x) = y &\text{ = } \frac{\text{x - 5}}{2} \\ or, 2y + 5 &\text{ = } x \\ \end{aligned}$
Interchanging the value of 'x' and 'y', we get,
or, 2x + 5 = y
∴ g^-1(x) = 2x + 5

Again, we have given,
fog^-1(x) = 6
or, f(2x + 5) = 6
or, 2x + 5 + 1 = 12
or, 2x = 6
∴ x = 3

Solution,
Given, f(x) = 3x + 4
g(x) = 2(x + 1) = 2x + 2

To prove, fog(x) = gof(x)

Now, for fog(x),
fog(x) = f(2x + 2)
= 3(2x + 2) + 4
= 6x + 6 + 4
= 6x + 10 ---- (A)

Again, for gof(x),
gof(x) = g(3x + 4)
= 2(3x + 4) + 2
= 6x + 8 + 2
= 6x + 10 ---- (B)
So, from (A) and (B), fog(x) = gof(x).

Here,
f(x) = 2x - 7
g(x) = a(let)
fog(x) = 4x + 3
(gof)^-1(x)= ?

Now,
fog(x) = 4x + 3
or, f(a) = 4x + 3
or, 2a - 7 = 4x + 3
or, 2a = 4x + 10
or, a = 2x + 5
∴ g(x) = 2x + 5 --- (A)

Again,
gof(x) = g(2x - 7)
= 2(2x - 7) + 5 [from (A)]
= 4x - 14 + 5
= 4x - 9

Finally, for (gof)^-1(x),
gof(x) = y = 4x - 9
or, $\frac{\text{y + 9}}{4}$ = x
Interchanging the value of 'x' and 'y', we get,
y = $\frac{\text{x + 9}}{4}$
∴ (gof)^-1(x) = $\frac{\text{x + 9}}{4}$

Here,
f(x) = 3x - 1
g(x) = a (let)
fog(x) = 6x + 5
(gof)^-1(x) = ?

Now,
fog(x) = 6x + 5
or, f(a) = 6x + 5
or, 3a - 1 = 6x + 5
or, 3a = 6x + 6
∴ a = g(x) = 2(x + 1)

Again,
gof(x) = g(3x - 1)
= 2(3x - 1) + 2
= 6x - 2 + 2
= 6x

Finally, for (gof)^-1(x),
gof(x) = y = 6x
or, $\frac{y}{6}$ = x
Interchanging the value of 'x' and 'y', we get,
or, $\frac{x}{6}$ = y
∴ (gof)^-1(x) = $\frac{x}{6}$

Solution,
$\begin{aligned} \text{Given, } \text{f(x)} & \text{ = } \text{3x - 7} \\ \text{g(x)} &\text{ = } \frac{\text{4x - 2}}{3} \\ f^{-1}(x) &\text{ = } \text{g(x)} \end{aligned}$
To find, x = ?

Now, for f^-1(x),
$\begin{aligned} f(x) \text{ = } y &\text{ = } \text{3x - 7} \ \text{or, y + 7} & \text{ = } 3x \ \text{or, }\frac{\text{y + 7}}{3} & \text{ = } x \ \end{aligned}$
Interchanging the value of 'x' and 'y', we get,
$\begin{aligned} \frac{\text{x + 7}}{3} & \text{ = } y \ \therefore f^{-1}(x) & \text{ = } \frac{\text{x + 7}}{3} \ \end{aligned}$

Again, we have given,
$\begin{aligned} f^{-1}(x) &\text{ = } \text{g(x)} \\ \text{or, }\frac{\text{x + 7}}{\cancel{3}} &\text{ = } \frac{\text{4x - 2}}{\cancel{3}} \\ \text{or, 9} &\text{ = } \text{3x} \\ \therefore \text{3} &\text{ = } \text{x} \\ \end{aligned}$

Solution,
$\begin{aligned} \text{Given, } \text{h(x)} &\text{ = }\frac{\text{x}}{\text{2 - x}} \\ \text{g(x)} &\text{ = }\text{ax - 5} \\ \text{goh(4)} &\text{ = }\text{-13} \\ \end{aligned}$

To find, a = ?, g^-1(1) = ?

$\begin{aligned} \text{Now, } \text{h(x)} &\text{ = }\frac{\text{x}}{\text{2 - x}} \\ \text{or, h(4)} &\text{ = }\frac{\text{4}}{\text{2 - 4}} \\ &\text{ = }\text{-2} \\ \end{aligned}$

$\begin{aligned} \text{Again, } \text{goh(4)} &\text{ = }\text{-13} \\ \text{or, g(-2)} &\text{ = }\text{-13} \\ \text{or, -2a - 5} &\text{ = }\text{-13} \\ \text{or, -2a} &\text{ = }\text{-13 + 5} \\ \therefore \text{a} &\text{ = }\text{4} \end{aligned}$

Again, for g^-1(x),
$\begin{aligned} \text{g(x)}\text{ = }y &\text{ = }\text{4x - 5} \ \text{or, } \frac{\text{y + 5}}{\text{4}} &\text{ = }\text{x} \ \end{aligned}$
Interchanging the value of 'x' and 'y', we get,
$\begin{aligned} \text{or, y} &\text{ = }\frac{\text{x + 5}}{\text{4}} \ \text{or, } g^{-1}(x) &\text{ = }\frac{\text{x + 5}}{\text{4}} \ \therefore g^{-1}(1) &\text{ = }\frac{\text{6}}{\text{4}}\text{ = }\frac{\text{3}}{\text{2}} \ \end{aligned}$

Solution,
$\begin{aligned} \text{Given, } \text{f(x)} &\text{ = }\text{3x - 4} \\ \text{g(x)} &\text{ = }\text{a (let)} \\ f^{-1}g(x) &\text{ = }\frac{\text{3x + 2}}{\text{3}} \\ \end{aligned}$

$\begin{aligned} \text{To find, } \text{g(x)} &\text{ = }\text{?} \ f^{-1}\left(\frac{1}{2}\right) &\text{ = }\text{?} \ \end{aligned}$

For, f^-1(x),
$\begin{aligned} \text{f(x)}\text{ = }\text{y} &\text{ = }\text{3x - 4} \ \text{or, } \frac{\text{y + 4}}{3} &\text{ = }\text{x} \ \end{aligned}$
Interchanging the value of 'x' and 'y', we get,
$\begin{aligned} \text{or, } \frac{\text{x + 4}}{3} &\text{ = }\text{y} \ \therefore f^{-1}(x) &\text{ = }\frac{\text{x + 4}}{3} \ \end{aligned}$
$\therefore f^{-1}\left(\frac{1}{2}\right)\text{ = }\frac{\text{1 + 8}}{\text{6}} \text{ = }\frac{\text{3}}{\text{2}}$

Again,
$\begin{aligned} f^{-1}g(x) &\text{ = }\frac{\text{3x + 2}}{\text{3}} \\ \text{or, } f^{-1}(a) &\text{ = }\frac{\text{3x + 2}}{\text{3}} \\ \text{or,}\frac{\text{a + 4}}{3} &\text{ = } \frac{\text{3x + 2}}{3} \\ \text{or, a + 4} &\text{ = }\text{3x + 2} \\ \therefore \text{a}\text{ = }\text{g(x)} &\text{ = }\text{3x - 2} \\ \end{aligned}$

Solution,
$\begin{aligned} \text{Given, f(x)} &\text{ = } \text{2x + 3} \\ \text{g(x) } &\text{ = } \frac{\text{x + 5}}{\text{2}} \\ \text{ff(x)} &\text{ = } g^{-1}(x) \\ \end{aligned}$
To find, x = ?

Now, for g^-1(x),
$\begin{aligned} \text{g(x) } \text{ = } y &\text{ = } \frac{\text{x + 5}}{\text{2}} \\ \text{or, 2y} &\text{ = } \text{x + 5} \\ \text{or, 2y - 5} &\text{ = } \text{x} \\ \end{aligned}$
Interchanging the value of 'x' and 'y', we get,
$\begin{aligned} \text{or, 2x - 5} &\text{ = } \text{y} \\ \therefore g^{-1}(x) &\text{ = } \text{ 2x - 5} \\ \end{aligned}$

Again, we have given,
$\text{ff(x)} \text{ = } g^{-1}(x)$
$\text{or, f(2x + 3)} \text{ = } \text{2x - 5}$
$\text{or, 4x + 6 + 3} \text{ = } \text{2x - 5}$
$\text{or, 4x + 9 + 5} \text{ = } \text{2x}$
$\text{or, -2x} \text{ = } \text{14}$
$\therefore \text{x} \text{ = } \text{-7}$

Solution,
$\begin{aligned} \text{Given, f(x)} &\text{ = } \text{2x + 1} \\ \text{g(x)} &\text{ = } \frac{\text{3x + 1}}{\text{2}} \\ \text{ff(x)} &\text{ = } g^{-1}(x) \\ \end{aligned}$
To find, x = ?

Now, for g^-1(x),
$\begin{aligned} \text{g(x)} \text{ = } y &\text{ = } \frac{\text{3x + 1}}{2} \\ \text{or, } \frac{\text{2y - 1}}{\text{3}} &\text{ = } \text{x} \\ \end{aligned}$
Interchanging the value of 'x' and 'y', we get,
$\begin{aligned} \text{or, y} &\text{ = } \frac{\text{2x - 1}}{\text{3}} \\ \therefore g^{-1}(x) &\text{ = } \frac{\text{2x - 1}}{\text{3}} \end{aligned}$

Again, we have given,
$\text{ff(x)} \text{ = } g^{-1}(x)$ $\text{or, f(2x + 1)} \text{ = } \frac{\text{2x - 1}}{\text{3}}$
$\text{or, 2(2x + 1) + 1} \text{ = } \frac{\text{2x - 1}}{\text{3}}$
$\text{or, 4x + 3} \text{ = } \frac{\text{2x - 1}}{\text{3}}$
$\text{12x + 9} \text{ = } \text{2x - 1}$
$\text{12x + 10} \text{ = } \text{2x}$
$\text{10x} \text{ = } \text{-10}$
$\therefore \text{x} \text{ = } \text{-1}$

Solution,
$\begin{aligned} \text{Given, f(x)} &\text{ = } \frac{\text{x}}{\text{3 + 2x}} \\ \text{f(x)} &\text{ = } f^{-1}(x) \\ \end{aligned}$
To find, x = ?

Now, for f^-1(x),
$\text{f(x)} \text{ = } \text{y} \text{ = } \frac{\text{x}}{\text{3 + 2x}}$
$\text{or, 3y + 2xy} \text{ = } \text{x}$ $\text{or, 3y} \text{ = } \text{x - 2xy}$ $\text{or, 3y} \text{ = } \text{x(1 - 2y)}$ $\text{or, x} \text{ = } \frac{\text{3y}}{\text{1 - 2y}}$
Interchanging the value of 'x' and 'y', we get,
$\begin{aligned} \text{or, y} &\text{ = } \frac{\text{3x}}{\text{1 - 2x}} \\ \therefore f^{-1}(x) &\text{ = } \frac{\text{3x}}{\text{1 - 2x}} \\ \end{aligned}$

Again, we have given,
$\begin{aligned} \text{f(x)} &\text{ = } f^{-1}(x) \ \text{or, } \frac{\text{x}}{\text{3 + 2x}} &\text{ = } \frac{\text{3x}}{\text{1 - 2x}} \ \text{or, x - 2}x^2 &\text{ = } \text{9x + 6}x^2 \ \text{or, 8x + 8}x^2 &\text{ = } \text{0} \ \text{or, 8x(1 + x)} &\text{ = } \text{0} \ \end{aligned}$
Either, x = 0
Or, x = -1
∴ x = -1, 0

Solution,
$\begin{aligned} \text{Given, 3.f(x) } &\text{ = } \text{4x + 5} \\ \text{f(x) } &\text{ = } \frac{\text{4x + 5}}{\text{3}} \\ \text{g(x)} &\text{ = } \text{5x - 4} \\ \end{aligned}$
To find, f^-1og^-1(1) = ?

For g^-1(x),
$\begin{aligned} \text{g(x)} = y &\text{ = } \text{5x - 4} \\ \text{or, y + 4} &\text{ = } \text{5x} \\ \text{or, } \frac{\text{y + 4}}{\text{5}} &\text{ = } \text{x} \\ \end{aligned}$
Interchanging the value of 'x' and 'y', we get,
$\begin{aligned} \text{or, y} &\text{ = } \frac{\text{x + 4}}{\text{5}} \\ \therefore g^{-1}(x) &\text{ = } \frac{\text{x + 4}}{\text{5}} \\ \therefore g^{-1}(x) &\text{ = } \text{1} \end{aligned}$

For f^-1(x),
$\begin{aligned} \text{f(x) } = \text{y} &\text{ = } \frac{\text{4x + 5}}{\text{3}} \\ \text{or, } \frac{\text{3y - 5}}{\text{4}} &\text{ = } \text{x} \\ \end{aligned}$
Interchanging the value of 'x' and 'y', we get,
$\begin{aligned} \text{or, y} &\text{ = } \frac{\text{3x - 5}}{\text{4}} \\ \therefore f^{-1}(x) &\text{ = } \frac{\text{3x - 5}}{\text{4}} \\ \end{aligned}$

Finally,
$\begin{aligned} f^{-1}g^{-1}(1) &\text{ = } f^{-1}(1) \ &\text{ = } \frac{\text{3(1) - 5}}{\text{4}} \ &\text{ = } \frac{\text{-2}}{\text{4}} \ \therefore f^{-1}g^{-1}(1) &\text{ = } \frac{\text{-1}}{\text{2}} \ \end{aligned}$

Solution,
$\begin{aligned} \text{Given, f(x) } &\text{ = } x^2 - \text{2x} \\ \text{g(x)} &\text{ = } \text{2x + 3} \\ fog^{-1}(x) &\text{ = } \text{3} \\ \end{aligned}$
To find, (x) = ?

Now, for g^-1(x),
$\begin{aligned} \text{g(x)} = \text{y} &\text{ = } \text{2x + 3} \\ \text{or, } \frac{\text{y - 3}}{2} &\text{ = } \text{x} \\ \end{aligned}$
Interchanging the value of 'x' and 'y', we get,
$\begin{aligned} \text{or, } \frac{\text{x - 3}}{2} &\text{ = } \text{y} \\ \therefore g^{-1}(x) &\text{ = } \frac{\text{x - 3}}{2} \\ \end{aligned}$

Again, we've given,
$fog^{-1}(x) \text{ = } \text{3}$
$\text{or, } \text{f}\left(\frac{\text{x - 3}}{\text{2}}\right) \text{ = } \text{3}$
$\text{or, } \left(\frac{\text{x - 3}}{\text{2}}\right)^{2} - \cancel{2}\left(\frac{\text{x - 3}}{\cancel{2}}\right) \text{ = } \text{3}$
$\text{or, } \frac{x^2 \text{- 6x + 9}}{4} \text{- x + } \cancel{3} \text{ = } \cancel{3}$
$\text{or, } x^2 \text{- 6x + 9} \text{ = } \text{4x}$
$\text{or, } x^2 \text{- 6x + 9 - 4x} \text{ = } \text{0}$
$\text{or, } x^2 \text{- 10x + 9} \text{ = } \text{0}$
$\text{or, } x^2 \text{- 9x - x + 9} \text{ = } \text{0}$
$\text{or, } \text{x(x - 9) -1(x + 9)} \text{ = } \text{0}$
$\text{or, } \text{(x - 9)(x - 1)} \text{ = } \text{0}$
$\therefore \text{x} \text{ = } \text{1, 9}$