Functions

$3-2^x-2^{1-x}\ge 0put{2}^x=t$

$3-t-\frac{2}{t}\ge 0\Rightarrow t^2-3t+2\le 0$

$\left(t-1\right)\left(t-2\right)\le 0\Rightarrow t\in \left[1,2\right]$

? P? = 6!/2! = 360

f (x) = 2e²? / (e²? +e? ) and f (1-x) = 2e²? / (e²? +e¹? )
∴ f (x) + f (1-x) = 1/2
i.e. f (x) + f (1-x) = 2
∴ f (1/100) + f (2/100) + . + f (99/100)
Σ? f (x/100) + f (1-x/100) + f (1/2)
= 49 x 2 + 1 = 99

lim? (x→7) (18- [1-x])/ ( [x-3a])
exist & a∈I.
= lim? (x→7) (17- [-x])/ ( [x]-3a)
exist
RHL = lim? (x→7? ) (17- [-x])/ ( [x]-3a) = 25/ (7-3a) [a ≠ 7/3]
LHL = lim? (x→7? ) (17- [-x])/ ( [x]-3a) = 24/ (6-3a) [a ≠ 2]
LHL = RHL
25/ (7-3a) = 8/ (2-a)
∴ a = -6

A function f (x) is continuous at x=1, so lim (x→1? ) f (x) = lim (x→1? ) f (x) = f (1).
Assuming a piecewise function like f (x) = { -x, x<1; ax+b, x1 } (structure inferred from derivative).
Continuity at x=1: f (1) = 1. a (1)+b = 1 => a+b=1.

The function is differentiable at x=1. The derivative of f (x) at x=1 from the left is -1. The derivative from the right is a.
So, a = -1. (The image has 2a = -1, which would imply a function like -x and ax²+b). Let's assume f' (x) = 2a for x>1.
2a = -1 => a = -1/2.

From a+b=1, b = 1 - a = 1 - (-1/2) = 3/2.
So, a = -1/2 and b = 3/2.

Given the function f (x) = cosec? ¹ (x) / √ {x - [x]} where [x] is the greatest integer function.

The domain of cosec? ¹ (x) is (-∞, -1] U [1, ∞).
For the denominator to be defined, x - [x] ≠ 0, which means {x} ≠ 0 (the fractional part of x is not zero). This implies that x cannot be an integer (x ∉ I).

Combining these conditions, the domain is all non-integer numbers except for those in the interval (-1, 1).

Kindly consider the following Image 

19.00
Only '2' in range → 1 function
one element out of 1, 3, 4 is in range with '2'
number of ways = ³C? * (3!/(2!1!)) * 2! = 18
(Select one from 1, 3, 4 and distribute among a, b, c)
Total function = 1 + 18 = 19

$y=lo{g}_{10}x+lo{g}_{10}{x}^{1/3}+lo{g}_{10}{x}^{1/9}+........upto\infty terms$

$=lo{g}_{10}x\left(1+\frac{1}{3}+\frac{1}{9}+......\right)$   

y = log₁₀ x * $\frac{1}{1-\frac{1}{3}}=\frac{3}{2}lo{g}_{10}x$  

Now, $\frac{2+4+6+....+2y}{3+6+9+....+3y}=\frac{4}{lo{g}_{10}x}$

$\Rightarrow \frac{2*y\frac{\left(y+1\right)}{2}}{3y\frac{\left(y+1\right)}{2}}=\frac{4}{lo{g}_{10}x}solo{g}_{10}x=6$

$\Rightarrow y=\frac{3}{2}*6=9$

So, (x, y) = (10⁶, 9)

$f\left(x\right)=\frac{1-x}{1+x},0<x<1$ (on simplification)

$f\text{'}\left(x\right)=\frac{-2}{{\left(1+x\right)}^2}$

${\left(1-x\right)}^{2}f\text{'}\left(x\right)=-2{\left(f\left(x\right)\right)}^2$