Book Online Demo

Check Your IQ

Try Test

Courses

Offline Centres

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

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

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

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

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

(0, 6)

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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.

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE