IBDP Maths SL 2.7 Use of the discriminant AA HL Paper 1- Exam Style Questions- New Syllabus

(a)

For the quadratic equation to have no real roots, its discriminant must be negative.

\( b^2-4ac<0 \)

Here,

\( a=\frac12,\qquad b=k,\qquad c=13. \)

Therefore,

\( k^2-4\left(\frac12\right)(13)<0 \)

\( k^2-26<0 \)

\( k^2<26. \)

Since \(k\) is a positive integer,

\( k<\sqrt{26}\approx5.10. \)

Hence the greatest possible value is

\( \boxed{k=5.} \)

(b)(i)

Using \(k=5\),

\( f(x)=\frac12x^2+5x+13. \)

The axis of symmetry is

\( x=-\frac{b}{2a} =-\frac5{2(\frac12)} =-5. \)

✅ Axis of symmetry:

\( \boxed{x=-5} \)

(b)(ii)

Substitute \(x=-5\) into the function:

\( f(-5) =\frac12(25)+5(-5)+13 =\frac{25}{2}-25+13 =\frac12. \)

Therefore the minimum point is

\( \boxed{\left(-5,\frac12\right)}. \)

The vertex gives the minimum since the coefficient of \(x^2\) is positive, so the parabola opens upwards.

(c)

Differentiate:

\( \frac{dy}{dx}=x+5. \)

At \(x=-3\),

\( \frac{dy}{dx}=2. \)

Hence the normal has gradient

\( -\frac12. \)

The point on the curve is

\( f(-3) =\frac12(9)-15+13 =\frac52. \)

Using point-gradient form,

\( y-\frac52=-\frac12(x+3). \)

Simplifying,

\( \boxed{y=-\frac12x+1.} \)

(d)

The required area is the area between the curve and the line from \(x=-3\) to \(x=0\).

\( A=\int_{-3}^{0} \left[\left(\frac12x^2+5x+13\right)-\left(-\frac12x+1\right)\right]dx \)

\( =\int_{-3}^{0} \left(\frac12x^2+\frac{11}{2}x+12\right)dx. \)

Integrate:

\( A= \left[ \frac{x^3}{6} +\frac{11x^2}{4} +12x \right]_{-3}^{0}. \)

Evaluating the limits,

\( A = 0- \left( -\frac{27}{6} +\frac{99}{4} -36 \right) = \frac{63}{4}. \)

The shaded area is obtained by integrating the difference between the upper curve and the lower straight line.

✅ Shaded area:

\( \boxed{\frac{63}{4}.} \)