Question

The equation $3x^2+4ax+b=0$ has at least one root in (0, 1) if

• A. $4a+b+3=0$
• B. $2a+b+1=0$
• C. $b=0,a=-\frac{4}{3}$
• D. none of these

Answer 1

Answer: (B)

Let $f(x)=x^3+2a{x}^2+bx$ in $[0,1]$. since $f(x)$ is a polynomial
$\therefore f(x)$ is continuous in $[0,1]$ and derivable in $(0,2)$. Also
$f(0)=0$, $f(1)=1+2a+b$
$f(0)=f(1)$ gives $0=1+2a+b$
i.e., $2a+b+1=0$
$\therefore f(x)$ satisfies all the conditions of Rolle’s Theorem if $2a+b+1=0$
$\therefore f^{\prime }(x)=0$ has at least one root in $(0,1)$
i. e., $3x^2+4ax+b=0$ has at least one root in $(0,1)$
Hence reqd. condition is $2a+b+1=0$