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I was solving some exam questions but some options are confusing me . I don't know how to ask this confusion so i tried to put it in a form of question .

Imagine there was a multiple choice question about polynomial f(x) and solving it we get the following result that

f(x) has at least 3 real roots and at most 4 imaginary roots

But options are like this

Polynomial f(x) has

(A) At least 3 real roots

(B) At least 2 real roots

(C) At most 4 imaginary roots

(D) At most 6 imaginary roots

So logically which options are correct or i should choose in exam ? Is there something like some options are more correct then others or in any way of thinking answer is same ?

RKK
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    This depends upon your exam, but I feel like I know which exam this is. I don't get the idea of "more correct", (perhaps more appropriate would be more appropriate), but in multiple choice exams you're just asked to select the correct statements, so in this case, one would select all the options, they're all correct. (and logically correct, at that). Any number that is "at least $3$", is also "at least $2$", for example. – Sarvesh Ravichandran Iyer Jun 17 '22 at 07:28
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    Any combination of answers that excludes either A or C is a weaker assertion than any combination of answers that includes both A and C. – ryang Jun 17 '22 at 14:15
  • Thanks Sir , i got it . All options are right . – RKK Jun 19 '22 at 06:28

1 Answers1

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Given the information that "$f(x)$ has at least 3 real roots and at most 4 imaginary roots" for a polynomial $f(x)$, the logically certain option is only (A).

The assertion is a conjunction of two propositions (symbolically, we have $P\wedge Q$):

  1. $f(x)$ has at least 3 real roots.

  2. $f(x)$ has at most 4 imaginary roots.

Our test is: In case that a third proposition compatible with the given options is adjoined, the truth-value of the resultant compound should not be false. Symbolically, we shall have $P\wedge Q\wedge R$.

By (1), we understand that an analysis of the polynomial secures 3 real roots, and leaves the possibility of having more open (see the question as an example).

By (2), we understand that the polynomial can have at most 4 imaginary roots, and leaves the possibility of having nought imaginary roots open.

Suppose the third proposition is

  1. $f(x)$ has 6 imaginary roots.

This is allowed by (D), but not by (2). Similarly, (1) controverts (B).

In the given instance, 'at least' is a stronger quantifier than 'at most', because 'at most' allows the number of imaginary roots to be rendered to nought, whereas 'at least' does not. To illustrate the point, suppose

$$f(x) = ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+g$$

Unconditionally, $f(x)$ has at most 4 imaginary roots and at least one real root. Conditionally, it may have 2 imaginary roots (hence, 3 real roots) or nought.

However, if the condition has been given that $f(x)$ has at least 3 real roots, it cannot have 4 imaginary roots.

In other words, 'at least' operates as the independent quantifier, 'at most' operates as the dependent quantifier. To better grasp what is at stake, switch their roles, and begin the argument with 'at most' as the independent quantifier (as analogous to the order of the standard quantifiers, e.g., $\exists x\forall y\ldots$). 'At least' has existential import, 'at most' does not have.

Consider the following propositions as candidates for the third proposition with respect to the preceding example:

  • $f(x)$ has 4 imaginary roots (and one real root).
  • $f(x)$ has 5 real roots (and no imaginary roots).

The former one is controverted by (1), while the latter one is compatible with both (1) and (2).

In logic, we work with the surest options.

Tankut Beygu
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