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The least integral value of a for which the graphs $y=2 a x+1$ and $y=(a-6) x^2-2$ do not intersect is
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The correct answer is:
$-5$
$(a-6) x^2-2=2 a x+1$
$(a-6) x^2-2 a x-3=0$
Since the graphs do not intersect,
$\therefore \mathrm{D} \lt 0$
$\Rightarrow 4 a^2+4 \times 3(a-6) \lt 0$
$\Rightarrow a^2+3 a-18 \lt 0$
$\Rightarrow(\mathrm{a}+6)(\mathrm{a}-3) \lt 0$
$\Rightarrow \mathrm{a} \in(-6,3)$
Hence, the least integer value of a is -5
$(a-6) x^2-2 a x-3=0$
Since the graphs do not intersect,
$\therefore \mathrm{D} \lt 0$
$\Rightarrow 4 a^2+4 \times 3(a-6) \lt 0$
$\Rightarrow a^2+3 a-18 \lt 0$
$\Rightarrow(\mathrm{a}+6)(\mathrm{a}-3) \lt 0$
$\Rightarrow \mathrm{a} \in(-6,3)$
Hence, the least integer value of a is -5
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