Question

The roots of the equation $a{x}^3+b{x}^2+cx+d=0$, are ${Î\pm }_1,{Î\pm }_2,{Î\pm }_3$ and roots of $g(z)=a{x}^3+\frac{f^{′′}(y)z^2}{2!}+\frac{f^{′}(y)}{1!}z+f(y)=0$ are ${β}_1,{β}_2,{β}_3$ then ${Î\pm }_1−{β}_1$ equals

• A. ${Î\pm }_2−{β}_2$
• B. ${Î\pm }_3−{β}_3$
• C. $y$
• D. All of these

Answer 1

Answer: (D)

Let $f(x)=a{x}^3+b{x}^2+cx+d…(1)$
$g(z)=a{z}^3+\frac{f^{′′}(y)}{2!}{z}^2+\frac{f^{′}(y)}{1!}z+f(y)=0…(2)$
Now, $f^{′′}(y)=6ay+2b$
$f^{′}(y)=3a{y}^2+2by+c$
$f(y)=a{y}^3+b{y}^2+cy+d$ Putting in (2), we get
$g(z)=a{z}^3+\frac{6ay+2b}{2!}{z}^2+(3a{y}^2+2by+c)z+a{y}^3{+}^{′}b{y}^2+cy+d$
$=a{z}^3+(3ay+b)z^2+(3a{y}^2+2by+c)z+a{y}^3+b{y}^2+cy+d$
$=a{z}^3+3ay{z}^2+b{z}^2+3a{y}^{2}z+2byz+cz+a{y}^3+b{y}^2+cy+d$
collecting all a, b. c terms, we get
$=a(z^3+y^3+3z^{2}y+3z{y}^2)+b(z^2+y^2+2zy)+(cz+cy+d)$
$=a(z+y{)}^3+b(z+y{)}^2+c(z+y)+d$
$g(z)=f(z+y)$, which is obtained by replacing x by $z+y$ in equation $(1)$, say $z+y=x$
$∴z=x+y$
Hence, which means the roots of (1) are connected by a quantity y
$∴$ If ${Î\pm }_1,{Î\pm }_2,{Î\pm }_3$ are roots of (1) and
${β}_1,{β}_2,{β}_3$ are roots of (2)
${Î\pm }_1−y={β}_1,{Î\pm }_2−y={β}_2,{Î\pm }_3−y=B_3$
$⇒{Î\pm }_1−{β}_1=y,{Î\pm }_2−{β}_2=y,{Î\pm }_3−y=B_3$
$⇒{Î\pm }_1−{β}_1=y,{Î\pm }_2−{β}_2=y,{Î\pm }_3−{β}_3=y$
$∴{Î\pm }_1−{β}_1={Î\pm }_2−{β}_2={Î\pm }_3−{β}_3=y$