Question

If $x=a$ is a root of multiplicity two of a polynomial equation $f(x)=0$, then

• A. $f^{'}(a)=f^{''}(a)=0$
• B. $f^{''}(a)=f(a)=0$
• C. $f^{'}(a) \neq 0 \neq f^{''}(a)$
• D. $f(a)=f^{'}(a)=0 ; f^{''}(a) \neq 0$

Answer 1

Answer: (D)

$x=a$ is a root of multiplicity two of a polynomial equation $f(x)=0$.
Let $f(x)=(x-a)^{2} g(x)$
$\Rightarrow f^{'}(x)=2(x-a) g(x)+(x-a)^{2} g^{'}(x)$
Now, $f^{''}(x)=2 g(x)+2(x-a) g^{'}(x)+2(x-a) g^{'}(x)+(x-a)^{2} g^{''}(x)$
$\Rightarrow =2 g(x)+4(x-a) g^{'}(x)+(x-a)^{2} g^{''}(x)$
$\Rightarrow f^{'}(a)=2(a-a) g(a)+(a-a)^{2} g^{'}(a)=0$
$\Rightarrow f^{''}(a)=2 g(a)+4(a-a) g^{'}(a)+(a-a)^{2} g^{''}(a)$
$=2 g(a)$
$\therefore f(a)=f^{'}(a)=0, f^{''}(a) \neq 0$