Class 12th

To understand the phenomenon behind this, have a close look at the formula of the first order reaction.

t1/2 = 0.693/k

Here, we can see that the half life is only dependent on k (rate constant) since there is no [A]' in the formula. Hence, we can easily conclude that the half life of a first order reaction is independent of the initial concentration [A]'.

Kindly consider the following Image

f' (c) = 1 + lnc = e/ (e-1)
lnc = e/ (e-1) - 1 = (e - (e-1)/ (e-1) = 1/ (e-1)
c = e^ (1/ (e-1)

log (k? /k? ) = (Ea / 2.303R) [1/T? - 1/T? ]
log (3.555) = (Ea / (2.303R) [1/303 - 1/313]
1.268 * 8.314 * 303 * 313 = 10Ea
So, Ea = 100 kJ

According to IUPAC convention for naming of elements with atomic number more than 100, different digits are written in order and at the end ium is added. For digits following naming is used.

0 -nil

1-un

2-bi

3-tri

and so on.

y = (2/π x - 1)cosec x
dy/dx = (2/π)cosec x - (2/π x - 1)cosec x cot x
⇒ dy/dx + (2/π x - 1)cosec x cot x = 2/π cosec x
⇒ dy/dx + ycot x = 2/π cosec x
This is a linear differential equation. The integrating factor P (x) is the coefficient of y.
⇒ P (x) = cot x

Refractive index (n) is a dimensionless quantity that has no units. Since both numerator and denominator are in meters per seconds (m/s), both units cancel each other. This makes refractive index simply a number. 

f (x) = sin²x (λ + sinx)
f' (x) = 2sinxcosx (λ + sinx) + sin²x (cosx) = sinxcosx (2λ + 3sinx)
For extrema, f' (x) = 0
sinx = 0, cosx = 0, or sinx = -2λ/3
For more than 2 points of extrema in the interval, sinx = -2λ/3 must have solutions other than where sinx=0 or cosx=0.
-1 < -2/3 < 1 and -2/3 0
This gives λ ∈ (-3/2, 3/2) - {0}