At an intersection, \(cx^2+(3-2c)x-(c+3)=0\). If \(c\ne0\), its discriminant is \((3-2c)^2+4c(c+3)=8c^2+9>0\), so two real intersections exist. If \(c=0\), the equation is \(3x-3=0\), which has an intersection at \(x=1\). Therefore \(\boxed{\text{B}}\) is correct.

Equate the curve and line: \[(2k-3)x^2-(k+3)x-(k-6)=0.\] For tangency, the quadratic must have a repeated root. Thus \[(k+3)^2+4(2k-3)(k-6)=0\] \[9k^2-54k+81=9(k-3)^2=0.\] Therefore \[\boxed{k=3}.\] With this value the difference of curve and line is \(3(x-1)^2\), confirming tangency at \((1,-1)\). The exceptional value \(k=3/2\) gives a distinct straight line and cannot satisfy the tangency condition.

Equate the equations: \[x^2+(k-3)x+6-k=0.\] Two distinct real intersections require a positive discriminant: \[(k-3)^2-4(6-k)=k^2-2k-15=(k+3)(k-5)>0.\] Therefore \[\boxed{k<-3\quad\text{or}\quad k>5}.\] Equality gives tangency and is excluded.

Intersections satisfy \(3x^2-(m+4)x+5-m=0\). Two distinct real roots require \[\Delta=(m+4)^2-12(5-m)=m^2+20m-44=(m+22)(m-2)>0.\] Hence \(\boxed{m<-22\text{ or }m>2}\).

Eliminating \(y\) gives \(2x^2+(2m-6)x+m-4=0\). Its discriminant is \[\Delta=(2m-6)^2-8(m-4)=4(m^2-8m+17)=4((m-4)^2+1)>0\] for every real \(m\). The quadratic therefore has two distinct real roots, so the line has two distinct intersections with the curve.

Intersections satisfy \(2x^2-mx+8=0\). No meeting means no real root, so \(\Delta=m^2-64<0\). Hence \(\boxed{-8<m<8}\). The endpoints are excluded because \(\Delta=0\) gives a point of tangency.

Intersections satisfy \(3x^2+(2-m)x+3=0\). Two distinct real roots require \[\Delta=(2-m)^2-36=(m+4)(m-8)>0.\] Hence \(\boxed{m<-4\text{ or }m>8}\).

(a) Substitution gives \(mx^2+cx-16=0\). For tangency the discriminant is zero: \(c^2+64m=0\). Thus \(\boxed{m=-c^2/64}\); tangency also requires \(c\ne0\), since \(c=0\) would give \(m=0\) and no meeting.

(b) With \(m=-4\), two roots require \(c^2-256>0\), giving \(\boxed{c<-16\text{ or }c>16}\).

Intersection gives \(x^2-(k+3)x+2k+2=0\). Its discriminant is \((k+3)^2-4(2k+2)=(k-1)^2\geq0\) for every real \(k\). Thus at least one real intersection always exists, proving the statement. At \(k=1\) there is one repeated intersection; otherwise there are two.

(a) At an intersection, \[4x^2-(k+1)x+\frac12k^2+a=0.\] Since the roots are 0 and \(3/4\), this is \(4x(x-3/4)=4x^2-3x\). Comparing coefficients gives \[k+1=3,\qquad\frac12k^2+a=0,\] so \(k=2,a=-2\).

(b) The intersection equation is now \[4x^2-(k+1)x+\frac12k^2-\frac72=0.\] Tangency requires zero discriminant: \[(k+1)^2-16\left(\frac12k^2-\frac72\right)=0\quad\Longrightarrow\quad7k^2-2k-57=0.\] Thus \[(k-3)(7k+19)=0,\qquad k=3\text{ or }-\frac{19}{7}.\]

At an intersection, \[x^2+(2c-4)x+4-c=0.\] Two distinct points require a positive discriminant: \[(2c-4)^2-4(4-c)=4c^2-12c=4c(c-3)>0.\] Therefore \[c<0\text{ or }c>3.\] The boundary values give tangency, so they are excluded.

At an intersection, \[kx^2+(2-k)x+(2-k)=0.\] For \(k\ne0\), no real intersection requires a negative discriminant: \[D=(2-k)^2-4k(2-k)=5k^2-12k+4=(5k-2)(k-2)<0.\] The roots are \(2/5\) and 2 and the quadratic opens upwards. Hence \[\boxed{\frac25<k<2}.\] If \(k=0\), the equations reduce to the intersecting lines \(y=2x\) and \(y=-2\); it adds no value to the answer.

Intersections satisfy \(cx^2-4x+c-3=0\). The case \(c=0\) gives two distinct lines, not a tangent to the curve. Otherwise tangency requires \(16-4c(c-3)=0\), or \((c-4)(c+1)=0\). For \(c=4\), \((2x-1)^2=0\), so \(P=(1/2,-1)\). For \(c=-1\), \((x+2)^2=0\), so \(P=(-2,-11)\). Hence \(\boxed{c=4,\ P=(1/2,-1);\quad c=-1,\ P=(-2,-11)}\).

Equating the line and curve gives \[x^2-(m+4)x+9=0.\] Tangency requires discriminant zero: \[(m+4)^2-36=0,\qquad m+4=\pm6.\] Thus \(m=2\) or \(m=-10\). The repeated root is \(x=(m+4)/2\). For \(m=2\), \(x=3\) and \(y=0\); for \(m=-10\), \(x=-3\) and \(y=24\). Hence \[\boxed{m=2\text{ at }(3,0),\qquad m=-10\text{ at }(-3,24)}.\]

(a) Equate the expressions: \[3x^2-18x+15=3(x-1)(x-5)=0.\] Thus \(x=1,5\). Substituting into the line gives \(\boxed{(1,8),(5,32)}\).

(b) The inequality is \(3(x-1)(x-5)\le0\), so \(\boxed{1\le x\le5}\).

(a) Substitute \(y=11x+3\): \[2x^2-2x(11x+3)+2=0\Rightarrow10x^2+3x-1=(5x-1)(2x+1)=0.\] Hence \[(-1/2,-5/2),\quad(1/5,26/5).\]

(b) Substituting \(y=4x+3\) gives \[(2-4k)x^2-3kx+2=0.\] Its discriminant is \[9k^2-8(2-4k)=9k^2+32k-16=(k+4)(9k-4).\] It is negative for \(-4<k<4/9\). The degenerate case \(k=1/2\) gives a linear intersection, outside this interval. Thus the answer is \(-4<k<4/9\).

Equating gives \[2x^2+(k-3)x+8=0.\] No real intersection requires \[(k-3)^2-64<0\Rightarrow|k-3|<8\Rightarrow-5<k<11.\]

Substitute y=−3x−k to obtain \(10x^2+3kx-40=0\). Its discriminant \(9k^2+1600>0\) for every real k. Thus two real x-values, and corresponding y-values, always exist.

(a) Substitution in the curve gives \(25k^2-40k+12=0=(5k-2)(5k-6)\). Restriction gives k=2/5. From the line, \(1/2=1+p\), so p=−1/2. Equating the curve and line gives \(4x^2-60x+125=0\), roots 5/2,25/2. The other y-value is \(9/2\): point (25/2,9/2). (b) General intersection quadratic is \(\frac12k^2x^2-3kx+2-p=0\). Its discriminant \(k^2(5+2p)\) must be negative. Since k≠0, p<−5/2.

(a) Substitute y=x+5: \(5x^2+30x+75-k=0\). The discriminant is zero, so \(900-20(75-k)=0\) and k=30. (b) \(5(x+3)^2=0\), giving x=−3,y=2. Thus P=(−3,2).

(a) At intersection, \(2x^2+(k-2)x+k-4=0\). Tangency requires \((k-2)^2-8(k-4)=0\), which simplifies to \((k-6)^2=0\). Hence \(\boxed{k=6}\).

(b) With \(k=2\), \[y=2x^2+2x+1=\boxed{2(x+1/2)^2+1/2}.\] The vertex is \(\boxed{(-1/2,1/2)}\).