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 $f\left(x\right)=\frac{a{x}^4}{4}+\frac{b{x}^3}{3}+\frac{c{x}^2}{2}+dx$
which is continuous and differentiable, and
$f\left(0\right)=0,f(-1)=\frac{a}{4}-\frac{b}{3}+\frac{c}{2}-d=\frac{1}{4}\times (a+2c)-\frac{1}{3}\times (b+3d)=0[\because 3(a+2c)=4(b+3d\left)\right]$
From Rolle's theorem: there exists at least one root of $f^{\prime }\left(x\right)=0$ in $(-1,0)$
$\therefore a{x}^3+b{x}^2+cx+d=0$ have atleast one root in $(-1,0)$