QJEEM19Ma91c5: JEE Main 2019 Mathematics - Class 11 JEEM19Ma91c5 Solution: The number of integral values of m for which the equation (1 +…

(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.