QJEEM19Ma1238: JEE Main 2019 Mathematics - Class 11 JEEM19Ma1238 Solution: If both the roots of the quadratic equation x 2 − mx + 4 = 0…

x² $$-$$mx + 4 = 0

Case-I :

D > 0

m² $$-$$ 16 > 0

$$\Rightarrow$$  m $$\in$$ ($$-$$ $$\mathrm{\infty }$$, $$-$$ 4) $$\cup$$ (4, $$\mathrm{\infty }$$)

Case-II :

$$\Rightarrow 1<\frac{-b}{2a}<5$$

$$\Rightarrow 1<\frac{m}{2}<5\Rightarrow m\in \left(2,10\right)$$

Case-III :

f(1) > 0   and f(5) > 0

1 $$-$$ m + 4 > 0   and 25 $$-$$ 5m + 4 > 0

m < 5   and m < $$\frac{29}{5}$$

Case-IV :

Let one root is x = 1

1 $$-$$ m + 4 = 0

m = 5

Now equation x² $$-$$ 5x + 4 = 0

(x $$-$$ 1) (x $$-$$ 4) = 0

x = 1 i.e. m = 5 is also included

hence m $$\in$$ (4, 5]

So given option is (4, 5)