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Chapter 3: Problem 30

If \(f(x)\) is a real-valued function discontinuous at all integral points lying in \([0, n]\) and if \((f(x))^{2}=1 \forall x \in[0, n]\), then number of functions \(f(x)\) are (a) \(2^{n+1}\) (b) \(6 \times 3^{n}\) (c) \(2 \times 3^{n-1}\) (d) \(3^{n+1}\)

Short Answer

Expert verified
The number of such functions is \(2^{n+1}\) (option a).

Step by step solution

01

Understanding function properties

The function \(f(x)\) is discontinuous at all integral points in the interval \([0, n]\). It also satisfies \((f(x))^2 = 1\) for all \(x \in [0, n]\). This implies that for each value of \(x\), \(f(x)\) can either be \(+1\) or \(-1\), since these are the only numbers whose square equals 1.
02

Possible values at non-integral points

Since \((f(x))^2 = 1\), without specific restrictions on non-integral points, \(f(x)\) can still take values \(+1\) or \(-1\) at any non-integral point in \([0, n]\). This increases the number of choices we have for defining \(f(x)\) over this range.
03

Handling discontinuity at integral points

\(f(x)\) must be discontinuous at all integer points \(x=0,1,2,...,n\). For each integer point \(x=k\), the limit from the left \(f(k^-)\) and the limit from the right \(f(k^+)\) must be different, meaning \(f(k^-)\) and \(f(k^+)\) must be opposite (one is \(+1\) and the other is \(-1\)).
04

Calculating the number of functions

There are \(n+1\) integral points (from 0 to \(n\)), hence there are \(n+1\) pairs \((f(k^-), f(k^+))\) with 2 choices (\(+1, -1\) or \(-1, +1\)) each. Each non-integral point in the real number interval has 2 choices (\(+1\) or \(-1\)), but we only consider the constraints of integer discontinuity for our calculation. Thus, there are \(2^{n+1}\) valid configurations for the function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Discontinuity
A function is said to be discontinuous at a particular point if there is a failure in its smoothness, so to speak. In mathematical terms, a function calculates differently when approaching from the left versus the right, creating a kind of 'jump'.

Discontinuity can occur for various reasons, such as:
  • A jump discontinuity, where the function suddenly hops from one value to another at a specific point.
  • An infinite discontinuity, where the function reaches infinity.
  • An oscillating discontinuity, where small variations cause the function value to bounce unpredictably.
For the function discussed in the exercise, it is discontinuous at each integer in the interval \([0, n]\). This means:
  • The left-hand limit \(f(k^-)\) and right-hand limit \(f(k^+)\) at any integer \(k\) in \([0, n]\) are not equal.
  • The function 'jumps' between values at each of these integers.
This characteristic of the function is key in calculating the total number of possible function configurations. By forcing discontinuity at certain points, we limit and define how the function behaves.
Integral Points Discontinuity
Integral points, in the context of the exercise, refer to the whole numbers within the interval \([0, n]\). Here, each of these integral points causes a planned discontinuity for the function. This means for any integer \(k\) inside this interval:
  • The value approaching from the left \(f(k^-)\) and the value from the right \(f(k^+)\) must be different.
  • This results in dynamically switching behavior exactly at these points, described as a 'jump'.
For each of these integer points, the function can exhibit the two behaviors:
  • Jump from \(+1\) to \(-1\)
  • Jump from \(-1\) to \(+1\)
These two possibilities at each integer point mean that for \(n+1\) integral points, the function has, in total, \(2^{n+1}\) possible configurations. This calculation stems from the nature of its discontinuity at each integral point.
Binary Function Values
Given that \((f(x))^2 = 1\) for all \(x\) within the interval \([0, n]\), the function outputs can only be \(f(x) = +1\) or \(f(x) = -1\). These are the only real numbers satisfying the condition of squaring to one.
  • This property implies a binary nature of the function's values at each point on the interval, allowing it to either be \(+1\) or \(-1\).
  • Such a binary function is much easier to work with when calculating for various configurations since each point only has two possible outcomes.
A binary function is attractive when considering calculus problems as well, especially where discontinuity is involved. The binary nature simplifies constraint handling and broadens the scope of possible solutions. This is why, despite the seemingly restricted nature of the function, it allows a vast number of configurations determined by these permissible jumps and maintains discontinuity at integral points.

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Most popular questions from this chapter

Let \(f: R \rightarrow R\) be a function defined by \(f(x)=\\{|\cos x|\\}\), where \(\\{x\\}\) represents fractional part of \(x\). Let \(S\) be the set containing all real values \(x\) lying in the interval \([0,2 \pi]\) for which \(f(x) \neq|\cos x|\). Then, number of elements in the set \(S\) is (a) 0 (b) 1 (c) 3 (d) infinite

Match the statements of Column I with values of Column II. Column I (A) \(\sqrt{\sin (\cos x)}\), has domain (B) \((\sqrt{\cos (\sin x)})^{-1}\), has domain (C) \(\tan (\pi \sin x)\), has domain (D) \(\ln (\tan x)\), has domain Column II (p) \(\quad x \in R\) (q) \(R-\left\\{n \pi \pm \frac{\pi}{6}\right\\}\) (r) \(\quad x \in\left(n \pi, n \pi+\frac{\pi}{2}\right)\) (s) \(\quad x \in\left[2 n \pi-\frac{\pi}{2}, 2 n \pi+\frac{\pi}{2}\right]\)

A quadratic polynomial maps from \([-2,3]\) onto \([0,3]\) and touches \(x\)-axis at \(x=3\), then the polynomial is (a) \(\frac{3}{16}\left(x^{2}-6 x+16\right)\) (b) \(\frac{3}{25}\left(x^{2}-6 x+9\right)\) (c) \(\frac{3}{25}\left(x^{2}-6 x+16\right)\) (d) \(\frac{3}{16}\left(x^{2}-6 x+9\right)\)

Let \(f: R \rightarrow R\) and \(g: R \rightarrow R\) be two one-one and onto functions, such that they are the mirror images of each other about the line \(y=a\). If \(h(x)=f(x)+g(x)\), then \(h(x)\) is (a) one-one and onto (b) only one-one and not onto (c) only onto but not one-one (d) None of these

The range of values of \(a\) so that all the roots of the equation \(2 x^{3}-3 x^{2}-12 x+a=0\) are real and distinct belongs to (a) \((7,20)\) (b) \((-7,20)\) (c) \((-20,7)\) (d) \((-7,7)\)
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