First slide

Theory of equations

Question

$Consider f (x) = x^{2} - 3 x + a + \frac{1}{a}, a \in R - {0}, such that f (3) > 0 and f (2) \leq 0 . If α and β are the roots of equation$ $f (x) = 0$ $then the value of α^{2} + β^{2} is equal to$

Moderate

A

greater than 11
B

less than 5
C

5
D

depends upon a and a cannot be determined.

Solution

$\begin{array}{l} f (x) = x^{2} - 3 x + a + \frac{1}{a} \\ \Rightarrow f (3) = 9 - 9 + a + \frac{1}{a} > 0 \\ \Rightarrow a + \frac{1}{a} > 0 \\ \Rightarrow a > 0 \\ f (2) = 4 - 6 + a + \frac{1}{a} \leq 0 \\ \Rightarrow \frac{a^{2} - 2 a + 1}{a} \leq 0 \\ \Rightarrow \frac{(a - 1)^{2}}{a} \leq 0 \\ \Rightarrow a = 1 \\ \Rightarrow a + \frac{1}{a} = 2 \end{array}$
$\begin{array}{l} Therefore, f (x) = x^{2} - 3 x + 2 = 0 has roots 1 and 2 . \\ ∴ α^{2} + β^{2} = 5 \end{array}$

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