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Question

Consider the cubic equation x34x2+x+6=0
Statement 1: Equation has 2 Positive real roots, 1 Negative real root
Statement 2: Equation has 1 Negative real root, 2 Imaginary roots.

A
Only Statement 1 is possible.
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B
Both Statements are possible.
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C
Only Statement 2 is possible.
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D
Both Statements are "NOT" possible.
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Solution

The correct option is A Only Statement 1 is possible.
Given that for the cubic equation x34x2+x+6=0,
Possibility no.1 is: 2 Positive real roots, 1 Negative real root
and Possibility no.2 is: 1 Negative real root, 2 Imaginary roots.

Let f(x)=x34x2+x+6
Number of sign changes are 2,
This means f(x) can have either 0 or 2 positive roots.
Now, f(2)=816+2+6=0
2 is one of the roots of the f(x).
Thus, there should exist one more positive real root.
Also, f(x)=x34x2x+6 which means just 1 sign change.
Thus f(x) should have one sign change.
Thus, Possibility 1 is correct.

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