求 kk 的取值范围,使得对于所有实数 xx,恒有 kx2+8x+k<6kx^2 + 8x + k < 6。 已知曲线 y=8x2+mx+m−6y = 8x^2 + mx + m - 6,其中 mm 是常数,求 使 x 轴为该曲线切线的 mm 的可能值。 使该曲线具有正 y 轴截距的 mm 的取值范围。 证明对于所有实数 cc,方程 2x2−cx+c2+6=02x^2 - cx + c^2 + 6 = 0 均无实根。

先不看答案,让 AI 诊断你卡在哪里?

它不会直接告诉你答案,而是问你卡在哪一步看看 AI 是怎么陪你一步步弄懂这道题的

完整解法

5(a)

1. Rearrange the inequality into standard quadratic form and state the conditions for it to be always negative

kx2+8x+k<6kx^2 + 8x + k < 6
kx2+8x+(k−6)<0kx^2 + 8x + (k - 6) < 0
For the quadratic expression to be always negative for all real values of xx, we must have:

  1. Coefficient of x2<0  ⟹  k<0x^2 < 0 \implies k < 0
  2. Discriminant <0< 0

2. Set up and simplify the discriminant inequality

b2−4ac<0b^2 - 4ac < 0
(8)2−4(k)(k−6)<0(8)^2 - 4(k)(k - 6) < 0
64−4k2+24k<064 - 4k^2 + 24k < 0
Divide by −4-4 (and flip the inequality sign):
k2−6k−16>0k^2 - 6k - 16 > 0

3. Solve the quadratic inequality for $k$

(k−8)(k+2)>0(k - 8)(k + 2) > 0
This gives the ranges:
k<−2ork>8k < -2 \quad \text{or} \quad k > 8

4. Combine the conditions to find the final range of $k$

Since we established earlier that k<0k < 0, we must reject k>8k > 8.
Taking the intersection of k<0k < 0 and (k<−2 or k>8)(k < -2 \text{ or } k > 8):
k<−2k < -2

答案

k<−2k < -2

5(b)(i)

1. State the condition for the x-axis to be a tangent to the curve

For the x-axis (y=0y = 0) to be a tangent to the curve, the equation 8x2+mx+m−6=08x^2 + mx + m - 6 = 0 must have two equal real roots, so the discriminant b2−4ac=0b^2 - 4ac = 0.

2. Form the equation in terms of $m$

m2−4(8)(m−6)=0m^2 - 4(8)(m - 6) = 0
m2−32m+192=0m^2 - 32m + 192 = 0

3. Solve the quadratic equation for $m$

(m−8)(m−24)=0(m - 8)(m - 24) = 0
m=8 or m=24m = 8 \text{ or } m = 24

答案

m=8m = 8 or m=24m = 24

5(b)(ii)

1. Find the expression for the y-intercept

When x=0x = 0,
y=8(0)2+m(0)+m−6y = 8(0)^2 + m(0) + m - 6
y=m−6y = m - 6

2. Set up and solve the inequality for a positive y-intercept

For a positive y-intercept, y>0y > 0:
m−6>0m - 6 > 0
m>6m > 6

答案

m>6m > 6

5(c)

1. Formulate the discriminant of the quadratic equation in $x$

For the equation 2x2−cx+(c2+6)=02x^2 - cx + (c^2 + 6) = 0, the discriminant is:
Δ=b2−4ac\Delta = b^2 - 4ac
Δ=(−c)2−4(2)(c2+6)\Delta = (-c)^2 - 4(2)(c^2 + 6)

2. Simplify the discriminant and show it is always negative

Δ=c2−8c2−48\Delta = c^2 - 8c^2 - 48
Δ=−7c2−48\Delta = -7c^2 - 48
Since c2⩾0c^2 \geqslant 0 for all real values of cc, −7c2⩽0-7c^2 \leqslant 0.
Therefore, −7c2−48⩽−48<0-7c^2 - 48 \leqslant -48 < 0.
Since Δ<0\Delta < 0 for all real cc, the equation has no real roots.

答案

Since the discriminant is always negative, the equation has no real roots.

AI 分析,可能出错

用 AI 批改,圈出写错的那一步,并自动将你的薄弱点存入错题本,形成你的错因库。

同一知识点的其他题

  1. 1已知一条曲线的方程为 y=2x−33x+4y = \frac{2x - 3}{3x + 4},其中 x≠−43x \ne -\frac{4}{3}。 曲线在点 PP(其中 x>0x > 0)处的法线平行于直线 y=−2517x+2y = -\frac{25}{17}x + 2。…
  2. 2如果抛物线y=(k−4)x2+ky=(k-4)x^{2}+k的图都在xx轴上方,那么kk的取值范围是 ___.
  3. 3解方程 2x+1+2−x=32^{x+1}+2^{-x}=3。 已知曲线方程为 y=2x2−4ax+by=2x^2-4ax+b,其中 aa 和 bb 为非零常数。解释为什么当 b>2a2b>2a^2 时,有 y>0y>0。
  4. 4曲线的方程为 y=x3+kx2+kx+8y = x^3 + kx^2 + kx + 8。求满足下列条件的 kk 的取值集合: 该曲线为单调递增函数, 该曲线恰好有 1 个驻点。
  5. 5若函数y=x3+x2+mx+1y=x^{3}+x^{2}+mx+1是RR上的单调函数,则实数mm的取值范围是()() A. (13,+∞)( \dfrac{1}{3} ,+∞) B. (−∞,13](-∞, \dfrac{1}{3} ] C. [13,+∞)[ \dfrac{1}{3} ,+∞)…