f: [-5,9] ⟶ R
(i) f(-3) + f(2)
= [6(-3) + 1 ] + [ 5(2)² - 1]
= ( -18 + 1) + ( 20 - 1)
 = -17 + 19 = 2.
(ii) f(7) - f(1)
= [ 3(7) - 4 ] - [6(1) + 1 ]
= (21 - 4) - (6 + 1)
=17 - 7 = 10
(iii) 2 f(4) + f(8)
= 2 [ 5(4)² - 1] + [3(8) - 4]
= 2[80 - 1] + [ 24 - 4]
= 158 + 20 = 178
(iv) \(\frac { 2f(-2)-f(6) }{ f(4)+f(-2) } \)
f(-2) = 6x + 1 = 6(-2) + 1 = -11
f(6) = 3x - 4 = 3(6) - 4 = 14
f(4) = 5x² - 1 = 5(4²) - 1 = 79
f(-2) = 6x + 1 = 6(-2) + 1 =-11
\(\frac { 2f(-2)-f(6) }{ f(4)+f(-2) } =\frac { 2(-11)-14 }{ 79+(-11) } =\frac { -22-14 }{ 68 } \)
= \(\frac { -36 }{ 68 } =\frac { -9 }{ 17 } \)

Given the function f: x ⟶ x² - 5x + 6.
i) f(-1) = (-1)² - 5(-1) + 6 = 1 + 5 + 6 = 12
ii) f(2a) = (2a)² - 5(2a) + 6 = 4a² - 10a + 6
iii) f(2) = 2² - 5(2) + 6 = 4 - 10 + 6 = 0
iv) f (x - 1)² - 5(x - 1) + 6
= x²- 2x + 1 - 5x + 5 + 6
= x²-7x + 12

L.C.M.= 3 x 2 x 2 x 2 x 5 x 3 = 360
Greatest number of 6 digit is 999999
L.C.M. of 24,15 and 36 = 360

On dividing 999999 by 360 remainder obtained is 279.
Greatest number of 6 digit, divisible by 24,15 and 36 = 999999 - 279 = 999720
Hence the required number is = 999720

Given 113400 = \({ p }_{ 1 }^{ { x }_{ 1 } }\times { p }_{ 2 }^{ { x }_{ 2 } }\times { p }_{ 3 }^{ { x }_{ 3 } }\times { p }_{ 4 }^{ { x }_{ 4 } }\)

The number 113400 can be factorized as 113400 = 2³ x 3⁴ x 5² x 7¹ where 2,3,5,7 are primes in ascending order.
Comparing 
\({ p }_{ 1 }^{ { x }_{ 1 } }\times { p }_{ 2 }^{ { x }_{ 2 } }\times { p }_{ 3 }^{ { x }_{ 3 } }\times { p }_{ 4 }^{ { x }_{ 4 } }\) = 2³ x 3⁴ x 5² x 7¹
P₁ = 2; P₂ = 3; P₃ = 5; P₄ = 7
x₁ = 3; x₂ = 4; x₃ = 2; x₄ = 1.

252525 = 3¹ x 5² x 7¹ x 13¹ x 37¹
363636 = 2² x 3³ x 7¹ x 13¹ x 37¹
H.C.F. = 3¹ x 7¹ x 13¹ x 37¹
= 3 x 3367
= 10101

To find HCF of three given numbers, first we have to find HCF of the first two numbers.
To find HCF of 396 and 504
Using Euclid’s division algorithm we get 504 = 396  x  1 + 108
The remainder is 108 \(\neq \) 0
Again applying Euclid’s division algorithm 396  = 108  x 3 + 72 
The remainder is 72 \(\neq \) 0
Again applying Euclid’s division algorithm 108  = 72 x 1 + 36 
The remainder is 36 \(\neq \) 0
Again applying Euclid division algorithm 72 = 36 x 2 + 0
Here the remainder is zero. Therefore HCF of 396 , 504 = 36, To find the HCF of 636 and 36
Using Euclid’s division algorithm we get 636 = 36 x 17 + 24
The remainder is 24 \(\neq \) 0
Again applying Euclid's division algorithm 36 = 24 x 1 + 12
The remainder is 12 \(\neq \) 0
Again applying Euclid's division algorithm 24 = 12 x 2 + 0
Here the remainder is zero. Therefore HCF of 636,36 = 12
Therefore Highest Common Factor of 396, 504 and 636 is 12.