If one of the roots of ${x^2} + f\left( a \right)x + a = 0 $ is equal to the third power of the other for all real a, then
$a.{\text{ }}$ 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
384.6k+ views
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.
Recently Updated Pages
Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Find the values of other five trigonometric ratios class 10 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Trending doubts
Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is the past tense of read class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

What is pollution? How many types of pollution? Define it

Give 10 examples for herbs , shrubs , climbers , creepers

Which state has the longest coastline in India A Tamil class 10 social science CBSE

Write an application to the principal requesting five class 10 english CBSE
