Question

If $f\left(x\right)=a{x}^2+bx+c$, where $a,b,c$ are rational nos. and if $f\left(2^{1/3}\right)=0$; then

• A. $f\left(x\right)$ is a constant function
• B. $f\left(x\right)$ is an Even function
• C. $a=b=c=0$
• D. None of the above

Answer 1

Answer: (A), (B), (C)

$f\left(x\right)$ is an Even function
$f\left(x\right)$ is a constant function
$a=b=c=0$

$f\left(x\right)=a{x}^2+bx+c$
$f\left(2^{1/3}\right)=a.2^{2/3}+b.2^{1/3}+c=0$ (given)

Now consider the equation,
$4^{1/3}a+2^{1/3}b+c=0$

If $a=\frac{p_1}{q_1},b=\frac{p_2}{q_2},c=\frac{p_3}{q_3}$, (since $a,b,c$ are rational) .... where $p_1,p_2,p_3,q_1,q_2,q_2ϵI$ and $q_1,q_2,q_3\ne 0$.

So, we get
$4^{1/3}\frac{p_1}{q_1}+2^{1/3}\frac{p_2}{q_2}+\frac{p_3}{q_3}=0$

$\Rightarrow 4^{1/3}{p}_1{q}_2{q}_3+2^{1/3}{p}_2{q}_1{q}_2+p_3{q}_1{Q}_2=0$ (Multiplying by $q_1{q}_2{q}_3)$

This is of the form $A.4^{1/3}+B.2^{1/3}+C=0$, where $A,B,CϵI;(A=p_1{p}_2{q}_2;B=p_2{q}_3{q}_1;C=p_3{q}_1{q}_2)$

Now, If $U+V+W=0$, Then $U^3+V^3+W^3=3UVW$.

Taking $U=A,V=B.2^{1/3},W=C.2^{2/3}$, we get;
$A^3+2B^3+4C^3=\left(3\right)\left(A\right)(B.2^{1/3})(C.2^{2/3})$
$\Rightarrow A^3+2B^3+4C^3=6ABC$

Thus $A$ must be Even.

Let $A=2A_1$.

Thus, $(2A_1{)}^3+2B^3+4C^3=6(2A_1)BC$

$8A_1^3+2B^3+4C^3=12A_{1}BC$

ie $4A_1^3+B^3+2C^3=6A_{1}BC$

Thus $B$ must also be Even.

Let $B=2B_2$

$\Rightarrow$ Thus, $4A_1^3+(2B_2{)}^3+2C^3=6A_1(2B_1\left)\right(C)$
$4A_1^3+8B_1^3+2C^3=12A_1{B}_{1}C$
$\Rightarrow 2A_1^3+4B_1^3+C^3=6A_1{B}_{1}C$
$\Rightarrow C$ is also Even.
Let $C=2C_1$
So, $A,B,C$ are all Even.
Now, If $A,B,C$ are not Equal to zero, then we can Assume that at least one of them is Odd.
This is because, we could have divided by $2$ enough number of times till atleast one of $A,B$, or $C$ becomes odd.
But, we have already proved that $A,B,C$ are all even.
This means, that our assumption that $A,B,C$ are not zeros is incorrect.

Thus, $A=B=C=0$
So $p_1=0,p_2=0,p_3=0$
$\Rightarrow a=0,b=0,c=0$
Thus, $f\left(x\right)=0.x^2+0.x+0=0$
ie $f\left(x\right)=0$
So $f\left(x\right)$ is a constant fnc (having value $0$)
$f\left(x\right)$ is also Even (since $f(-x)=0=f\left(x\right))$

So Option (A),(B) and (C) to be selected.