JEE MAIN - Mathematics (2019 - 9th January Evening Slot - No. 13)
If both the roots of the quadratic equation x2 $$-$$ mx + 4 = 0 are real and distinct and they lie in the interval [1, 5], then m lies in the interval :
($$-$$5, $$-$$4)
(4, 5)
(5, 6)
(3, 4)
Explanation
x2 $$-$$mx + 4 = 0
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Case-I :
D > 0
m2 $$-$$ 16 > 0
$$ \Rightarrow $$ m $$ \in $$ ($$-$$ $$\infty $$, $$-$$ 4) $$ \cup $$ (4, $$\infty $$)
Case-II :
$$ \Rightarrow \,\,1 < {{ - b} \over {2a}} < 5$$
$$ \Rightarrow \,\,1 < {m \over 2} < 5 \Rightarrow \,m \in \left( {2,10} \right)$$
Case-III :
f(1) > 0 and f(5) > 0
1 $$-$$ m + 4 > 0 and 25 $$-$$ 5m + 4 > 0
m < 5 and m < $${{29} \over 5}$$
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)
Case-I :
D > 0
m2 $$-$$ 16 > 0
$$ \Rightarrow $$ m $$ \in $$ ($$-$$ $$\infty $$, $$-$$ 4) $$ \cup $$ (4, $$\infty $$)
Case-II :
$$ \Rightarrow \,\,1 < {{ - b} \over {2a}} < 5$$
$$ \Rightarrow \,\,1 < {m \over 2} < 5 \Rightarrow \,m \in \left( {2,10} \right)$$
Case-III :
f(1) > 0 and f(5) > 0
1 $$-$$ m + 4 > 0 and 25 $$-$$ 5m + 4 > 0
m < 5 and m < $${{29} \over 5}$$
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)
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