Given: The piecewise function is defined as:
\(f(x) = \begin{cases} n(1 - 2nx), & \text{if } 0 \leq x < \frac{1}{2n} \\ 2n(2n - 1), & \text{if } \frac{1}{2n} \leq x < \frac{3}{4n} \\ 4n(1 - nx), & \text{if } \frac{3}{4n} \leq x < \frac{1}{n} \\ \frac{n}{n - 1}(nx - 1), & \text{if } \frac{1}{n} \leq x \leq 1 \end{cases}\)
The function \( f(x) \) is defined over the interval \( x \in [0, 1] \).
We need to determine the intervals where \( f(x) \) is increasing or decreasing.
We can calculate the derivative of \( f(x) \) for each piece and find the intervals where the derivative is positive (increasing) or negative (decreasing).
By examining the given intervals and the derivatives, we can summarize the behavior of \( f(x) \):
The correct answer is 8.

Let $ P(x_1, y_1) $ and $ Q(x_2, y_2) $ be two distinct points on the ellipse $$ \frac{x^2}{9} + \frac{y^2}{4} = 1 $$ such that $ y_1 > 0 $, and $ y_2 > 0 $. Let $ C $ denote the circle $ x^2 + y^2 = 9 $, and $ M $ be the point $ (3, 0) $. Suppose the line $ x = x_1 $ intersects $ C $ at $ R $, and the line $ x = x_2 $ intersects $ C $ at $ S $, such that the $ y $-coordinates of $ R $ and $ S $ are positive. Let $ \angle ROM = \frac{\pi}{6} $ and $ \angle SOM = \frac{\pi}{3} $, where $ O $ denotes the origin $ (0, 0) $. Let $ |XY| $ denote the length of the line segment $ XY $. Then which of the following statements is (are) TRUE?
Let f : (0,1) → R be the function defined as f(x) = √n if x ∈ [\(\frac{1}{n+1},\frac{1}{n}\)] where n ∈ N. Let g : (0,1) → R be a function such that \(\int_{x^2}^{x}\sqrt{\frac{1-t}{t}}dt<g(x)<2\sqrt x\) for all x ∈ (0,1).
Then \(\lim_{x\rightarrow0}f(x)g(x)\)
Let n ≥ 2 be a natural number and f:[0,1)]\(\rightarrow\) R be the function defined by
\(f(x) \begin{cases} n(1-2nx)\ \ \ \ \ \ \ \ \ \text{if}\ 0\le x\le\frac{1}{2n} \\ 2n((2nx-1)\ \ \ \ \ \text{if}\ \frac{1}{2n}\le x\le \frac{3}{4n}\\4n(1-nx)\ \ \ \ \ \ \ \ \ \text{if}\ \frac{3}{4n}\le x\le\frac{1}{n} \\ \frac{n}{n-1}(nx-1)\ \ \ \ \ \ \ \text{if}\ \frac{1}{n}\le x\le1 \end{cases}\)
If n is such that the area of the region bounded by the curves x = 0, x = 1, y = 0 and y = f(x) is 4, then the maximum value of the function f is
Monocyclic compounds $ P, Q, R $ and $ S $ are the major products formed in the reaction sequences given below.
The product having the highest number of unsaturated carbon atom(s) is:
For the reaction sequence given below, the correct statement(s) is(are): 
Consider a reaction $ A + R \rightarrow Product $. The rate of this reaction is measured to be $ k[A][R] $. At the start of the reaction, the concentration of $ R $, $[R]_0$, is 10-times the concentration of $ A $, $[A]_0$. The reaction can be considered to be a pseudo first order reaction with assumption that $ k[R] = k' $ is constant. Due to this assumption, the relative error (in %) in the rate when this reaction is 40% complete, is ____. [$k$ and $k'$ represent corresponding rate constants]
Given below is the list of the different methods of integration that are useful in simplifying integration problems:
If f(x) and g(x) are two functions and their product is to be integrated, then the formula to integrate f(x).g(x) using by parts method is:
∫f(x).g(x) dx = f(x) ∫g(x) dx − ∫(f′(x) [ ∫g(x) dx)]dx + C
Here f(x) is the first function and g(x) is the second function.
The formula to integrate rational functions of the form f(x)/g(x) is:
∫[f(x)/g(x)]dx = ∫[p(x)/q(x)]dx + ∫[r(x)/s(x)]dx
where
f(x)/g(x) = p(x)/q(x) + r(x)/s(x) and
g(x) = q(x).s(x)
Hence the formula for integration using the substitution method becomes:
∫g(f(x)) dx = ∫g(u)/h(u) du
This method of integration is used when the integration is of the form ∫g'(f(x)) f'(x) dx. In this case, the integral is given by,
∫g'(f(x)) f'(x) dx = g(f(x)) + C