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)

Let $f(x)=\frac{a_3{x}^4}{4}+\frac{a_2{x}^3}{3}+\frac{a_1{x}^2}{2}+a_{0}x$
$\therefore f(0)=0,f(1)=\frac{a_3}{4}+\frac{a_2}{3}+\frac{a_1}{2}+a_0=0$
$\Rightarrow f(0)=f(1)$
$\Rightarrow f^{\prime }(x)=0$ has atleast one real root in $[0,1]$
[according to Rolle's theorem]
$\therefore f^{\prime }(x)=a_3{x}^3+a_2{x}^2+a_{1}x+a_0$
Hence, $a_3{x}^3+a_2{x}^2+a_{1}x+a_0$ must has a real root in the interval $[0,1]$.