Question

Assertion :The number of minimum possible complex roots of $x^6-3x^5+4x^3+3x^2+4=0$ is two. Reason: Let $f\left(x\right)=x^6-3x^5+4x^3+3x^2+4\Rightarrow f\left(x\right)$ has two changes in sign.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Let $f\left(x\right)=x^6-3x^5+4x^3+3x^2+4$
Here, degree of $f\left(x\right)$ is $6$. So $f\left(x\right)$ has $6$ roots
And $f(-x)=x^6-3x^5-4x^3+3x^2+4$
$\because f\left(x\right)$ has two changes in sign and $f(-x)$ has also, two changes in sign hence the number of maximum real roots is $4$.
Hence the equation $f\left(x\right)=0$ has at least two $(6-4=2)$ complex roots
Both statement $1$ and Statement $2$ are individually true but statement $2$ is not the correct explanation of statement $1$.