Question

For all real values of $a_0, a_1, a_2, a_3$ satisfying $a_{0}+\frac{a_{1}}{2}+\frac{a_{2}}{3}+\frac{a_{3}}{4}=0$, the equation $a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}=0$ has a real root in the interval

• A. $[0, 1]$
• B. $[-1, 0]$
• C. $[1, 2]$
• D. $[-2, -1]$

Answer 1

Answer: (A)

$[0, 1]$

Answer 2

Let $f(x)=\frac{a_{3} x^{4}}{4}+\frac{a_{2} x^{3}}{3}+\frac{a_{1} x^{2}}{2}+a_{0} x$
$\therefore f(0)=0, f(1)=\frac{a_{3}}{4}+\frac{a_{2}}{3}+\frac{a_{1}}{2}+a_{0}=0$
$\Rightarrow f(0)=f(1)$
$\Rightarrow f'(x)=0$ has atleast one real root in $[0,1]$
[according to Rolle's theorem]
$\therefore f'(x)=a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}$
Hence, $a_{3} x^{3}+a_{2} x^{2}+a_{1} x+ a_{0}$ must has a real root in the interval $[0,1]$.