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4-2.Quadratic Equations and Inequations
normal
The polynomial equation $x^3-3 a x^2+\left(27 a^2+9\right) x+2016=0$ has
A
exactly one real root for any real $a$
B
three real roots for any real $a$
C
three real roots for any $a \geq 0$, and exactly one real root for any $a < 0$
D
three real roots for any $a \leq 0$, and exactly one real root for any $a > 0$
(KVPY-2016)
Solution
(a)
We have,
$x^3-3 a x^2+\left(27 a^2+9\right) x+2016=0$
Let
$f(x) =x^3-3 a x^2+\left(27 a^2+9\right) x+2016$
$f^{\prime}(x) =3 x^2-6 a x+27 a^2+9$
$f^{\prime}(x) =3\left(x^2-2 a x+9 a^2+3\right)$
$f^{\prime}(x) =3\left((x-a)^2+8 a^2+3\right)$
$f^{\prime}(x) > 0, \forall x \in R$
$\therefore(x) \text { is increasing function, } \forall a \in R$.
$f(x)$ has exactly one real root for any real $\varepsilon$.
Standard 11
Mathematics
