The real function f(x) = arccosec(x) / √(x - [x]), where [x] denotes the greatest integer less than or equal to x, is defined for all x belonging to:

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, x≥1 } (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

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

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

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

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

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

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

So, (x, y) = (10⁶, 9)

$f\left(x\right)=\{\begin{array}{l}x-\left[x\right]if\left[x\right]isodd\\ 1+\left[x\right]-x,if\left[x\right]iseven\end{array}$

Graph of f (x)

So

$=2{\pi }^2{\displaystyle \underset{0}{\overset{1}{\int }}\left(1-x\right)cos\pi xdx}$

=4