Moderate -0.3 This is a standard C1 question testing routine techniques: discriminant for intersection conditions, completing the square (with coefficient), and finding minimum points. All parts follow textbook methods with no novel problem-solving required, making it slightly easier than average, though the multi-part structure and completing the square with a coefficient of 3 provides some substance.
Find the set of values of \(k\) for which the line \(y = 2x + k\) intersects the curve \(y = 3x^2 + 12x + 13\) at two distinct points. [5]
Express \(3x^2 + 12x + 13\) in the form \(a(x + b)^2 + c\). Hence show that the curve \(y = 3x^2 + 12x + 13\) lies completely above the \(x\)-axis. [5]
Find the value of \(k\) for which the line \(y = 2x + k\) passes through the minimum point of the curve \(y = 3x^2 + 12x + 13\). [2]
Question 3:
3 | (i) | 3x2 + 12x + 13 = 2x + k
3x2 + 10x + 13 k [= 0]
b24ac > 0 oe soi
100 4 × 3 × (13 k) (> 0) oe
k > 14/3 oe | M1
M1
M1
M1
A1
[5] | oe eg M1 for 3x2 + 10x + 13 = k
for rearranging to 0; condone one error in
adding/subtracting; but M0 for 3x2 + 10x +
13 = k or 3x2 + 10x + 13 k = y
may be earned near end with correct
inequality sign used there
for correct substitution ft into b2 4ac, dep
on second M1 earned; brackets / signs must
be correct
accept k > 56/12 or better, isw incorrect
conversion of fraction but not wrong use of
inequalities
if A0, allow B1 for 56/12 oe obtained with
equality or wrong inequality (ie 3rd M1 has
not been earned) | condone 3x2 + 10x + 13 k = y for
this M1
3x2 + 10x + 13 k [= 0] will also earn
the first M1 if a separate statement has
not already done so
allow ‘b2 4ac is positive’ oe;
0 for just ‘discriminant > 0’ unless
implied by later work
can be earned with equality or wrong
inequality, or in formula
M0 for trials of values of k in
b2 −4ac
3 | (ii) | 3(x + 2)2 + 1 www as final answer
y-minimum = 1 [hence curve is above x-axis] | B4
B1
[5] | B1 for a = 3 and B1 for b = 2
and B2 for c = 1 or M1 for 13 3 × their b2
or for 13/3 their b2
or
1
B3 for 3 x22
3
Stating min pt is (2, 1) is sufft
allow ft if their c > 0
B0 for only showing that discriminant is
negative oe; need also to justify that it is all
above not all below x-axis
B0 for stating min point = 1 or ft | condone omission of square symbol;
ignore equating to zero in working or
answer
must be done in this part;
ignore wrong x-coordinate
3 | (iii) | 5 cao | B2
[2] | M1 for substitution of their (2, 1) in
y = 2x + k | allow M1 ft their 3(x + 2)2 + 1;
or use of (2,1) found using calculus;
M0 if they use an incorrect minimum
point inconsistent with their completed
square form
\begin{enumerate}[label=(\roman*)]
\item Find the set of values of $k$ for which the line $y = 2x + k$ intersects the curve $y = 3x^2 + 12x + 13$ at two distinct points. [5]
\item Express $3x^2 + 12x + 13$ in the form $a(x + b)^2 + c$. Hence show that the curve $y = 3x^2 + 12x + 13$ lies completely above the $x$-axis. [5]
\item Find the value of $k$ for which the line $y = 2x + k$ passes through the minimum point of the curve $y = 3x^2 + 12x + 13$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C1 Q3 [12]}}