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If exactly one root of the quadratic equation $x^{2} - (a + 1) x + 2 a = 0$ lies in the interval (0, 3), then the set of values a is given by

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(- \infty, 0) \cup (6, \infty)

(- \infty, 0] \cup (6, \infty)

(- \infty, 0] \cup [6, \infty)

(0, 6)

answer is B.

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Detailed Solution

Given quadratic equation,

x^{2} - (a + 1) x + 2 a = 0

The interval = (0, 3)

f (0) \times f (3) < 0

\Rightarrow (2 a) (9 - 3 (a + 1) + 2 a) < 0

\Rightarrow a (6 - a) < 0

\Rightarrow a (a - 6) > 0)

\Rightarrow a < 0

and

a > 6

Discriminant,

D = b^{2} - 4 ac \geq 0

\Rightarrow {(a + 1)}^{2} - 8 a \geq 0

\Rightarrow a^{2} - 6 a + 1 \geq 0

\Rightarrow {(a - 3)}^{2} - 8 \geq 0

\Rightarrow (a - (3 + 2 \sqrt 2)) (a - (3 - 2 \sqrt 2)) \geq 0)

\Rightarrow a \geq (3 + 2 \sqrt{2})

\Rightarrow a \leq (3 - 2 \sqrt{2})

The set of values of a is,

a < 0

and

a > 6

Hence, the set of values, a is

(- \infty, 0] \cup (6, \infty)

.
Option 2 is correct.

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