JEE MAIN - Mathematics (2019 - 8th April Evening Slot - No. 6)
The number of integral values of m for which the
equation
(1 + m2 )x2 – 2(1 + 3m)x + (1 + 8m) = 0 has no real root is :
(1 + m2 )x2 – 2(1 + 3m)x + (1 + 8m) = 0 has no real root is :
2
infinitely many
1
3
Explanation
(1 + m2
)x2
– 2(1 + 3m)x + (1 + 8m) = 0
Given equation has no real solution,
$$ \therefore $$ Discriminant (D) < 0
$$ \Rightarrow $$ 4(1 + 3m)2 - 4(1 + m2)(1 + 8m) < 0
$$ \Rightarrow $$ 4[9m2 + 6m + 1 - 8m - 1 - 8m3 - m2] < 0
$$ \Rightarrow $$ -8m3 + 8m2 - 2m < 0
$$ \Rightarrow $$ -2m(4m2 - 4m + 1) < 0
$$ \Rightarrow $$ m(2m - 1)2 > 0
$$ \therefore $$ m > 0 and m $$ \ne $$ $${{1 \over 2}}$$
So we can say number of integral values of m are infinitely many.
Given equation has no real solution,
$$ \therefore $$ Discriminant (D) < 0
$$ \Rightarrow $$ 4(1 + 3m)2 - 4(1 + m2)(1 + 8m) < 0
$$ \Rightarrow $$ 4[9m2 + 6m + 1 - 8m - 1 - 8m3 - m2] < 0
$$ \Rightarrow $$ -8m3 + 8m2 - 2m < 0
$$ \Rightarrow $$ -2m(4m2 - 4m + 1) < 0
$$ \Rightarrow $$ m(2m - 1)2 > 0
$$ \therefore $$ m > 0 and m $$ \ne $$ $${{1 \over 2}}$$
So we can say number of integral values of m are infinitely many.
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