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Showing 2 questions
Find least integer $k$ such that
$(k-2)x^2 + k + 8x + 4 > 0$ for all $x\in\mathbb{R}$.
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2
3
4
Solution
For quadratic $ax^2+bx+c>0$ for all $x$:
$a>0$ → $k-2>0$ → $k>2$
Discriminant $<0$
$D=b^2-4ac=8^2-4(k-2)(k+4)$
Compute:
$D=64-4(k^2+2k-8)=64-4k^2-8k+32$
$D=96-4k^2-8k<0$
Divide by $-4$:
$k^2+2k-24>0$
$(k+?)(k+?)$ → roots $4$ and $-6$
So $k>4$ or $k<-6$
Combine with $k>2$ ⇒ $k>4$
Least integer = $5$
If $x < -1$ and
$2^{|x+1|} - 2^x = |2x - 1| + 1$
then the value of $x$ is:
1
2
3
4
Solution
Since $x < -1$, then
$|x+1| = -(x+1)$ and $|2x - 1| = -(2x - 1)$ because $2x - 1 < -3 < 0$.
Equation:
$2^{-(x+1)} - 2^x = -(2x - 1) + 1$
Right side:
$-2x + 1 + 1 = -2x + 2$
Left side:
$2^{-(x+1)} = \dfrac{1}{2^{x+1}}$
Try $x = -2$:
LHS:
$2^{ -(-2+1) } - 2^{-2} = 2^{1} - \dfrac{1}{4} = 2 - 0.25 = 1.75$
RHS:
$-2(-2) + 2 = 4 + 2 = 6$ → not equal
Try $x = -2$ again carefully:
Wait — check systematically.
Better approach: substitute $x=-2$ in original equation:
Left side:
$2^{| -2 + 1 |} - 2^{-2} = 2^{1} - \dfrac{1}{4} = \dfrac{7}{4}$
Right side:
$|2(-2) - 1| + 1 = | -5 | + 1 = 6$
Not equal → reject.
Try $x = -1.5$ type pattern?
Better substitute only options (valid since only one option has $x< -1$):
Only option with $x < -1$ is (1) −2.
But we tested −2 and it doesn't satisfy?
Check original equation carefully:
Original:
$2^{|x+1|} - 2^x = |2x - 1| + 1$
Try $x=-2$ again:
Left:
$2^{|-2+1|} - 2^{-2} = 2^{1} - \dfrac{1}{4} = \dfrac{7}{4}$
Right:
$|2(-2)-1| + 1 = |-5| +1 = 6$
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