JEE MAIN - Mathematics (2020 - 8th January Morning Slot - No. 1)
The least positive value of 'a' for which the
equation
2x2 + (a – 10)x + $${{33} \over 2}$$ = 2a has real roots is
2x2 + (a – 10)x + $${{33} \over 2}$$ = 2a has real roots is
Answer
8
Explanation
For real roots Discriminate $$ \ge $$ 0.
(a – 10)2 – 4$$\left( {{{33} \over 2} - 2a} \right).2$$ $$ \ge $$ 0
$$ \Rightarrow $$ a2 + 100 – 20a – 132 + 16a $$ \ge $$ 0
$$ \Rightarrow $$ a 2 – 4a – 32 $$ \ge $$ 0
$$ \Rightarrow $$ (a – 8) (a + 4) $$ \ge $$ 0
$$ \Rightarrow $$ a $$ \le $$ -4 $$ \cup $$ a $$ \ge $$ 8
$$ \therefore $$ least positive a = 8
(a – 10)2 – 4$$\left( {{{33} \over 2} - 2a} \right).2$$ $$ \ge $$ 0
$$ \Rightarrow $$ a2 + 100 – 20a – 132 + 16a $$ \ge $$ 0
$$ \Rightarrow $$ a 2 – 4a – 32 $$ \ge $$ 0
$$ \Rightarrow $$ (a – 8) (a + 4) $$ \ge $$ 0
$$ \Rightarrow $$ a $$ \le $$ -4 $$ \cup $$ a $$ \ge $$ 8
$$ \therefore $$ least positive a = 8
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