Question

Assertion :If a, b, c $\in R$ and $a+b+c=0$, then the quadratic equation $3a{x}^2+2bx+c=0$ has at least one real root in (0, 1). Reason: Between any two roots of a polynomial f(x) there is a root of its derivative f'(x)

• 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 A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Let $f\left(x\right)=a{x}^3+b{x}^2+cx$
$f\left(0\right)=a{0}^3+b{0}^2+c0=0$
$f\left(1\right)=a{1}^3+b{1}^2+c1=a+b+c=0$ $\left(given\right)$
Now,
$f\left(x\right)$ is continous on $[0,1]$
$f\left(x\right)$ is differentiable on $(0,1)$
$f\left(0\right)=f\left(1\right)=0$
$f^{\prime }\left(x\right)=3a{x}^2+2bx+c$
According to Rolle’s Theorem if $f:[a,b]\to R$ is continuous on $[a,b]$ and
differentiable on $(a,b)$, such that $f\left(a\right)=f\left(b\right)$, where $a$ and $b$ are some real numbers.
Then there exists atleast one $x\in (a,b)$ such that $f\text{'}\left(x\right)=0$.
Hence,
$3a{x}^2+2bx+c=0$ for atleast one $x\in (0,1)$