f o f(k) = f(f(k))
= 2(2k - 1) -1 = 4k - 3
Thus, f o f(k) = 4k - 3
But, it is given that f o f(k) = 5
Therefore 4k - 3 = 5 ⇒ k = 2

f(x) = 3 x -2, g(x) = 2x + k
f o g = f(g(x)) = f(2x + k) = 3(2x + k) - 2 = 6x + k - 2
Thus, f o g(x) = 6x + 3k - 2
g o f(x) = g(3x - 2) = 2(3x - 2) + k
Thus, g o f(x) = 6x - 4 + k
Given that f o g = g o f
Therefore, 6x + 3k-2 = 6z - 4 + k
6x - 6x + 3k - k = -4 + 2 ⇒ -1

We set f₂(x) = 2x² - 5x + 3 and f₁(x) =\(\sqrt { x } \)
Then, f(x) = \(\\ \sqrt { 2x^{ 2 }-5x+3 } =\sqrt { { f }_{ 2 }(x) } \)
= f₁{f₂(x)} = f₁f₂(x)

Given X = {3, 4, 6, 8}
Relation R = {(x, f(x)) | x \(\in \) X, f(x) = x² + 1}
When x = 3 ⇒ f(x) = f(3)² = 9 + 1 = 10 \(\in \) N
When x = 4 ⇒ f(x) = f(4)² = 16 + 1 = 17 \(\in \) N
When x = 6 ⇒ f(x) = f(6)² = 36 + 1 = 37 \(\in \) N
When x = 8 ⇒ f(x) = f(8)² = 64 + 1 = 65 \(\in \) N
R = {(3, 10), (4, 17), (6, 37), (8, 65)}
Since, all the elements of X are having natural numbers as images, it is a function from X to N.

gff(x) = g[f{f(x)}] (This means "g of f of f of x")
= g[f(3x + 1)] = g[3(x + 1) + 1] = g(9x + 4)
g(9x + 4) = [(9x + 4) + 3] = 9x + 7
fgg(x) = f[g{g(x)}] (This means " f of g of g of x")
= f[g(x + 3)] = f[(x + 3) + 3] = f(x + 6)
f(x + 6) = [3(x + 6) + 1] = 3x + 19
These two quantities being equal, we get 9x + 7 = 3x + 19. Solving this equation we obtain x = 2.

The function f is defined by three values in intervals I, II, III as shown by the side.
For a given value of x = a, find out the interval at which the point a is located, there after find
f(a) using the particular value defined in that interval.
(i) First, we see that, x = 4 lie in the third interval.
Therefore, f(x) = 3x - 2; f(4) = 3(4) = 10
(ii) x = -2 lies in the second interval
Therefore, f(x) = x² - 2; f(-2) = (-2)² - 2 = 2
(iii) From (i), f(4) =10.
To find f(1) first we see that x = 1 lies in the second interval.
Therefore, f(x) = x²-2 ⇒ f(1) = 1² - 2 = -1
So, f(4) + 2f(1) = 10 + 2(-1) = 8
(iv) We know that f(1) = -1 and f(4) = 10
For finding f(-3), we see that x = −3, lies in the first interval.
Therefore, f(x) = 2x + 7; thus, f(-3) = 2(-3) + 7 = 1
Hence, \(\frac { f(1)-3f(4) }{ f(-3) } =\frac { -2-3(10) }{ 1 } \) = - 31

The distance travelled by the particle in time t hours is given by S(t)=\(\frac { { t }^{ 2 }+t }{ 2 } \).
(i) t = 3.5 hours. Therefore, S(3.5)=\(\frac { (3.5)^{ 2 }+3.5 }{ 2 } =\frac { 15.75 }{ 2 } \)=7.875
The distance travelled in 3.5 hours is 7.875 kms.
(ii) t = 8.25 hours. Therefore, S(8.25)=\(\frac { (8.25)^{ 2 }+8.25 }{ 2 } =\frac { 76.3125 }{ 2 } \)=38.15625
The distance travelled in 8.25 hours is 38.16 kms, approximately.