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
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
Approach Solution - 1
Given the function \( f(x) \) defined as: \[ f(x) = \begin{cases} n(1 - 2nx) & \text{if } 0 \leq x \leq \frac{1}{2n} \\ 2n((2nx - 1)) & \text{if } \frac{1}{2n} \leq x \leq \frac{3}{4n} \\ 4n(1 - nx) & \text{if } \frac{3}{4n} \leq x \leq \frac{1}{n} \\ \frac{n}{n - 1}(nx - 1) & \text{if } \frac{1}{n} \leq x \leq 1 \end{cases} \]
The function \( f(x) \) is defined on the interval \( x \in [0, 1] \).
Analysis of the increasing and decreasing intervals: - \( f(x) \) is decreasing in \( [0, \frac{1}{2n}] \) - Increasing in \( [\frac{1}{2n}, \frac{3}{4n}] \) - Decreasing in \( [\frac{3}{4n}, \frac{1}{n}] \) - Increasing in \( [\frac{1}{n}, 1] \)
The graph of \( f(x) \) is as follows:
The function \( f(x) \) lies in the range \( [0, n] \), and the area under the curve is given as 4.
Solving for \( n \): \[ \text{Area} = 4 \quad \Rightarrow \quad n = 8 \]
Therefore, the maximum value of \( f(x) \) is \( f(x)_{\text{max}} = n = 8 \).
Final Answer:
The correct value of \( n \) is \( \boxed{8} \).
Approach Solution -2
Given :
\(f(x) \begin{cases} n(1-2nx)\ \ \ \ \ \ \ \ \ \text{if}\ 0\le x\le\frac{1}{2n} \\ 2n((2nx-1)\ \ \ \ \ \text{if}\ \frac{1}{2n}\le x\le \frac{3}{4n}\\4n(1-nx)\ \ \ \ \ \ \ \ \ \text{if}\ \frac{3}{4n}\le x\le\frac{1}{n} \\ \frac{n}{n-1}(nx-1)\ \ \ \ \ \ \ \text{if}\ \frac{1}{n}\le x\le1 \end{cases}\)
x ∈ [0, 1]
f(x) is decreasing in \([0,\frac{1}{2n}]\)
Let's see the increase and decrease :
increasing in \([\frac{1}{2n},\frac{3}{4n}]\)
decreasing in \([\frac{3}{4n},\frac{1}{n}]\)
increasing in \([\frac{1}{n},1]\)
The graph is as follows :
f(x) ∈ [0, n]
Area = 4
⇒ n = 8
f(x)max = n = 8
So, the correct answer is 8.
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Concepts Used:
Methods of Integration
Given below is the list of the different methods of integration that are useful in simplifying integration problems:
Integration by Parts:
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.
Method of Integration Using Partial Fractions:
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)
Integration by Substitution Method
Hence the formula for integration using the substitution method becomes:
∫g(f(x)) dx = ∫g(u)/h(u) du
Integration by Decomposition
Reverse Chain Rule
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