Question

If $3(a+2c)=4(b+3d)$, then the equation $a{x}^3+b{x}^2+cx+d=0$ will have

• A. no real solution
• B. atleast one real root in $(-1,0)$
• C. atleast one real root in $(0,1)$
• D. none of these

Answer 1

The correct option is B atleast one real root in $(-1,0)$

Let $f\left(x\right)=\frac{a{x}^4}{4}+\frac{b{x}^3}{3}+\frac{c{x}^2}{2}+dx$
Clearly, f(x) is continuous on [-1,0] and differentiable in (-1,0)
Now, $f\left(0\right)=0$
and $f(-1)=\frac{a}{4}-\frac{b}{3}+\frac{c}{2}-d$

$=\frac{a+2c}{4}-\frac{b+3d}{3}$
$=0$
$\Rightarrow f\left(0\right)=f(-1)=0$
Hence, all the condition of Rolle's theorem are satisfied, so there exists at least one root of $f^{\prime }\left(x\right)=0$ in $[-1,0]$
and $f^{\prime }\left(x\right)=a{x}^3+b{x}^2+cx+d$
So, it has at least one root in $[-1,0]$