Question

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

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

Answer 1

Answer: (D)

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