Question

Assertion :If $a,b,c\in R$ and $2a+3b+6c=0$, then the equation $a{x}^2+bx+c=0$ has at least one real root in $(0,1)$. Reason: If $f\left(x\right)$ is a polynomial which assumes both positive and negative values, then it has at least one real root.

• 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

$a,b,cϵR,2a+3b+6c=0$

$6(\frac{a}{3}+\frac{b}{2}+c)=0$

$\therefore \frac{a}{3}+\frac{b}{2}+c=0$

Let $f\left(x\right)=\left(\frac{a}{3}\right)x^3+\frac{b}{2}\left(x^2\right)+cx$

$f\left(1\right)=\frac{a}{3}+\frac{b}{2}+c=0$

$f\left(0\right)=0$

$\therefore f\left(x\right)$ has $0,1$ as roots

$\therefore f^{\prime }\left(x\right)=0$ has root lies in $(0,1)$

$\therefore 3\left(\frac{a}{3}\right)x^2+2\left(\frac{b}{2}\right)x+c=0$ has root lies in $(0,1)$

$\therefore a{x}^2+bx+c=0$ has one real root in $(0,1)$

$\therefore$ If $f\left(x\right)$ has both positive and negative values

Then there must be a'x' which is real such that $f\left(x\right)=0$ i.e., $f\left(x\right)$ must cross x axis at some 'x'

$\therefore f\left(x\right)$ must have atleast one real root if it has both positive and negative values.

Both assertion and reason are correct but assertion cannot be explained by reason correctly.