Since x + 1/x = 2

Hence, x = 1

So,

x ¹²³+x^-123 = (1) ¹²³+(1)^-123

= 1 + 1 = 2

Sum of the roots of equation = coefficient of x / coefficient of x² = -7/3

Product of the roots of equation = Constant term/ coefficient of x² = -187/3

Required difference = -7/3-(-187/3) 

= -7/3+187/3

180/3 = 60

The minimum value of the above expression is 

=2√(Coefficient of Sin² x coefficient of cosec²x)

= 2√(9 x 36)         

=2×3×6

= 36

Hence option A

Let the distance from school to home be x km

It is given that,               

Student goes to his school at the speed of 14 km/hr and come back at the speed of 29 km/hr, total time taken by him is 4.3 hours

x/14 + x/29 = 4.3

Then, x = 40. 6 km

Hence option B 

Let the total work is 144 units (LCM of 36, 48 and 24)

Amount of work done by A in one day = 144/36 = 4 units

Amount of work done by B in one day = 144/ 48 = 3 units

Amount of work done by the A, B and C together = 144/ 18 = 8 units

Amount of work done by C in one day = 8 – 3 – 4 = 1 unit

Time taken by C to do whole work alone = 144/ 1 = 144 days

Time taken by C to do 87.5% of the work alone = 87.5% of 144 = 126 Days

Hence option A         

We know that 8.33% is equivalent to 1/12

So if we consider the original length of rectangle is 12

Then, after an increment of 8.33%

The length of rectangle will be 12+1 = 13

Now,                                 

37.5% is equivalent to 3/8

So if the original breadth of rectangle is 8

The breadth after increment = 8+3 = 11

So now,

Original area = 12x8 = 96 square units

Area after increment = 13x11 = 143 square units

% increment in area = (143 – 96/143) x 100

(47/143) x 100 = 48.96%

Hence option A

Let the numbers are   a and b,

a – b = 4              

ab = 32

So now, we know that,

(a + b)² = (a – b)²+4ab

= (4)²+4x32

= 16+128

(a + b)² = 144

a + b = 12

So the numbers are 8 and 4

Product of their square is = 64 x 16 = 1024

Hence option D

Let the numbers are 23a and 23b

According to question,

23a x 23b = 23x1679

ab= 73                         

Pairs of the numbers is (1, 73) only

When x= 0 then y = 39

And,

When y= 0 then x = 46

Hence,

Area of triangle = ½(46 x 39) = 23 x 39 = 897 square units

Number of Diagonals in a hexagon = n (n – 2)/2

= 6(6 – 3)/2 = 3 x 3 = 9

Number of Diagonals in an octagon = = n (n – 2)/2

= 8(8 – 3)/2 = 20

Required Product = 9 x 20 = 180

Hence option D