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$a.{\text{ }}$ The domain of the real value function $f$ is the set of non-negative real numbers.

$

b.{\text{ }}f\left( x \right) = - {x^{\dfrac{1}{4}}}\left( {1 + {x^{\dfrac{1}{2}}}} \right) \\

c.{\text{ }}f\left( x \right) = - {x^{\dfrac{1}{4}}} + {x^{\dfrac{3}{2}}} \\

d.{\text{ None of these}} \\

$

Answer

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Hint: - Use sum of the roots $ = \dfrac{{ - {\text{Coefficient of }}x}}{{{\text{Coefficient of }}{x^2}}}$, and product of roots $ = \dfrac{{{\text{constant term}}}}{{{\text{Coefficient of }}{x^2}}}$

Given quadratic equation ${x^2} + f\left( a \right)x + a = 0$

Let $\alpha $and $\beta $be the roots of the equation.

As you know sum of the roots $ = \dfrac{{ - {\text{Coefficient of }}x}}{{{\text{Coefficient of }}{x^2}}}$

And product of roots $ = \dfrac{{{\text{constant term}}}}{{{\text{Coefficient of }}{x^2}}}$

$

\Rightarrow \alpha + \beta = \dfrac{{ - f\left( a \right)}}{1} = - f\left( a \right).........\left( 1 \right), \\

\alpha \beta = \dfrac{a}{1} = a...............\left( 2 \right) \\

$

Now it is given that one root is the cube of other

$ \Rightarrow {\alpha ^3} = \beta $

From equation 2, considering real values of $\alpha $ we get

$

{\alpha ^3}\alpha = a \\

\Rightarrow {\alpha ^4} = a \\

\Rightarrow \alpha = \pm {a^{\dfrac{1}{4}}} \\

$

For positive value of $\alpha $

$ \Rightarrow \alpha = + {a^{\dfrac{1}{4}}}.............\left( 3 \right)$

Therefore from equation 1

$

\alpha + \beta = - f\left( a \right) \\

\Rightarrow \alpha + {\alpha ^3} = - f\left( a \right) \\

$

Now, from equation 3

\[

\Rightarrow {a^{\dfrac{1}{4}}} + {a^{\dfrac{3}{4}}} = - f\left( a \right) \\

\Rightarrow f\left( a \right) = - \left( {{a^{\dfrac{1}{4}}} + {a^{\dfrac{3}{4}}}} \right) \\

\Rightarrow f\left( a \right) = - {a^{\dfrac{1}{4}}}\left( {1 + {a^{\dfrac{1}{2}}}} \right) \\

\]

Now, in place of $a$ substitute $x$ in the above equation.

\[ \Rightarrow f\left( x \right) = - {x^{\dfrac{1}{4}}}\left( {1 + {x^{\dfrac{1}{2}}}} \right)\]

Hence, option (b) is correct.

Note: - In such types of questions the key concept we have to remember is that for a quadratic equation always remember the sum of roots and product of roots which is stated above, then simplify according to the given condition we will get the required answer.

Given quadratic equation ${x^2} + f\left( a \right)x + a = 0$

Let $\alpha $and $\beta $be the roots of the equation.

As you know sum of the roots $ = \dfrac{{ - {\text{Coefficient of }}x}}{{{\text{Coefficient of }}{x^2}}}$

And product of roots $ = \dfrac{{{\text{constant term}}}}{{{\text{Coefficient of }}{x^2}}}$

$

\Rightarrow \alpha + \beta = \dfrac{{ - f\left( a \right)}}{1} = - f\left( a \right).........\left( 1 \right), \\

\alpha \beta = \dfrac{a}{1} = a...............\left( 2 \right) \\

$

Now it is given that one root is the cube of other

$ \Rightarrow {\alpha ^3} = \beta $

From equation 2, considering real values of $\alpha $ we get

$

{\alpha ^3}\alpha = a \\

\Rightarrow {\alpha ^4} = a \\

\Rightarrow \alpha = \pm {a^{\dfrac{1}{4}}} \\

$

For positive value of $\alpha $

$ \Rightarrow \alpha = + {a^{\dfrac{1}{4}}}.............\left( 3 \right)$

Therefore from equation 1

$

\alpha + \beta = - f\left( a \right) \\

\Rightarrow \alpha + {\alpha ^3} = - f\left( a \right) \\

$

Now, from equation 3

\[

\Rightarrow {a^{\dfrac{1}{4}}} + {a^{\dfrac{3}{4}}} = - f\left( a \right) \\

\Rightarrow f\left( a \right) = - \left( {{a^{\dfrac{1}{4}}} + {a^{\dfrac{3}{4}}}} \right) \\

\Rightarrow f\left( a \right) = - {a^{\dfrac{1}{4}}}\left( {1 + {a^{\dfrac{1}{2}}}} \right) \\

\]

Now, in place of $a$ substitute $x$ in the above equation.

\[ \Rightarrow f\left( x \right) = - {x^{\dfrac{1}{4}}}\left( {1 + {x^{\dfrac{1}{2}}}} \right)\]

Hence, option (b) is correct.

Note: - In such types of questions the key concept we have to remember is that for a quadratic equation always remember the sum of roots and product of roots which is stated above, then simplify according to the given condition we will get the required answer.

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