Question

If family of straight lines $ax+by+c=0$ always passes through a fixed point $(\frac{3}{2},1),$ then equation $36a{x}^2+8bx+2c=0$ has

• A. at least one root in $[0,1]$
• B. atleast one root in $[\frac{-1}{2},\frac{1}{2}]$
• C. atleast one root in $[-1,2]$
• D. atleast one root in $[0,\frac{1}{2}]$

Answer 1

Answer: (A), (B), (C), (D)

atleast one root in $[0,\frac{1}{2}]$

$ax+by+c=0$ passes through $(\frac{3}{2},1)$
$\therefore 3a+2b+2c=0$
$\Rightarrow$ $\frac{3}{2}a+b+c=0$
Let $f\left(x\right)=12a{x}^3+4b{x}^2+2cx+d$
such that $f^{\prime }\left(x\right)=36a{x}^2+8bx+2c$
$f\left(x\right)$ is continuous and differentiable
$f\left(0\right)=d$,
$f\left(\frac{1}{2}\right)=\frac{3}{2}a+b+c+d=d$
$f\left(0\right)=f\left(\frac{1}{2}\right)$
From above we see that $f\left(x\right)$ satisfies all the conditions of Rolle's theorem
So, according to Rolle's theorem there is at least one root in $[0,\frac{1}{2}]$