JEE MAIN - Mathematics (2021 - 20th July Morning Shift - No. 15)
The probability of selecting integers a$$\in$$[$$-$$ 5, 30] such that x2 + 2(a + 4)x $$-$$ 5a + 64 > 0, for all x$$\in$$R, is :
$${7 \over {36}}$$
$${2 \over {9}}$$
$${1 \over {6}}$$
$${1 \over {4}}$$
Explanation
D < 0
$$\Rightarrow$$ 4(a + 4)2 $$-$$ 4($$-$$5a + 64) < 0
$$\Rightarrow$$ a2 + 16 + 8a + 5a $$-$$ 64 < 0
$$\Rightarrow$$ a2 + 13a $$-$$ 48 < 0
$$\Rightarrow$$ (a + 16) (a $$-$$ 3) < 0
$$\Rightarrow$$ a $$\in$$ ($$-$$16, 3)
$$\therefore$$ Possible a : {$$-$$5, $$-$$4, ............., 3}
$$\therefore$$ Required probability = $${8 \over {36}}$$ = $${2 \over {9}}$$
$$\Rightarrow$$ 4(a + 4)2 $$-$$ 4($$-$$5a + 64) < 0
$$\Rightarrow$$ a2 + 16 + 8a + 5a $$-$$ 64 < 0
$$\Rightarrow$$ a2 + 13a $$-$$ 48 < 0
$$\Rightarrow$$ (a + 16) (a $$-$$ 3) < 0
$$\Rightarrow$$ a $$\in$$ ($$-$$16, 3)
$$\therefore$$ Possible a : {$$-$$5, $$-$$4, ............., 3}
$$\therefore$$ Required probability = $${8 \over {36}}$$ = $${2 \over {9}}$$
Comments (0)
