d²y/dx² = e^-2x

d(dy/dx) = e^-2x.dx

Integrate both the sides :

dy/dx = -e^-2x/2 + c

dy = (-e^-2x/2 + c)dx

Again integrate both the sides ;

y = e^-2x/4 + cx + d where c and d are the constants of integration.

In case of formation our desired reaction is N₂ + 2H₂  ->  N₂H₄

In such type of questions multiply all the reaction with some variables say p,q,r,s.

After multiplication add those reactions, you'll find a single reaction equation. Equate its coefficients to our desired reaction.

In case of formation I got p = -1/4, q = 3/4, r = 1/4, s = -1/4

so enthalpy of formation = -1010/4 + (3/4)(317) + 413/4 - 286/4 = 17 kJ

Similarly you can do it for combustion reaction.

Let x and y be the dimensions of the whole paper.

Given xy = 1536

so area of printed part, A = (x-9)(y-6) = xy + 54 - 6x - 9y = 1536 + 54 - 6x - 9y = 1590 - 6x - 9y

From above y = 1536/x

so A = 1590 - 6x - 9(1536/x)

We have to maximize this area. At maximum dA/dx = 0

-6 + (9)(1536)/x² = 0

x = 48 and y = 1536/48 = 32

d²A/dx² = -(2)(9)(1536)/x³ < 0 is satisfying the condition of maximum.

so dimensions of the page for maximum printed area are 48cm x 32cm

Suppose foci are S (ae,0) and S' (-ae,0) and extremity of the minor axis be P(0,b).

slope of PS = -b/ae  and  slope of PS' = b/ae

since they are perpendicular the product of their slopes must be -1.

(-b/ae)(b/ae) = -1

e² = b²/a²

Also we know for ellipse : b²/a² = 1 - e²

hence e² = 1 - e²

e = 1/root(2)

Amines (including α-amino acids) react with ninhydrin to give a coloured product.

so correct option is d.