Every Inequalities question from the WMA11 papers in our bank, 2020–2026, sorted from easiest to hardest. Try each one first, then open the worked solution.
Level is estimated from marks: 1–6 easy, 7–8 medium, 9+ hard.
$4x^2-3x+7\ge4x+9$ gives $4x^2-7x-2\ge0$.
$4x^2-7x-2=(4x+1)(x-2)$.
The roots are $x=-\dfrac14$ and $x=2$. Since the quadratic opens upwards, it is non-negative outside the roots.
$x\le-\dfrac14$ or $x\ge2$.
$\dfrac4z<8$ gives $\dfrac4z-8<0$, so $\dfrac{4-8z}{z}<0$.
The critical values are $z=0$ and $z=\dfrac12$.
A sign check gives $z<0$ or $z>\dfrac12$.
$x+5=x(x-3)$ gives $x^2-4x-5=0$, so $(x-5)(x+1)=0$.
The inequality $x+5\le y\le x(x-3)$ is possible when $x\le-1$ or $x\ge5$. With $x\ge-2$, the shaded x-ranges are $-2\le x\le-1$ and $x\ge5$.
Draw the vertical boundary $x=-2$ and shade the two regions between the line and the parabola: from $x=-2$ to $x=-1$, and to the right of $x=5$.
$(i)\ z<0\text{ or }z>\dfrac12.\quad (ii)$ shade the region above $y=x+5$, below $y=x(x-3)$, with $x\ge-2$.
The bottom horizontal length is 5, from $x=0$ to the intercept of $y=10-2x$ at $x=5$.
The top horizontal length is $4-a$, because $y=2$ meets $y=10-2x$ at $x=4$.
Area $R_1=\dfrac12(5+4-a)(2)=9-a=\dfrac{27}{4}$, so $a=\dfrac94$.
The line $x=a$ passes through the intersection of $y=2$ and $y=kx$, so $2=ka$. Hence $k=\dfrac{2}{9/4}=\dfrac89$.
Region $R_2$ is to the right of $x=a$, above $y=kx$, and below $y=10-2x$.
$(a)\ a=\dfrac94,\ k=\dfrac89.\quad (b)\ x\ge\dfrac94,\ y\ge\dfrac89x,\ y\le10-2x$.
$k(2x-1)=x^2+2x+11$, so $x^2+(2-2k)x+(11+k)=0$.
For the line not to cross or touch the curve, the quadratic in $x$ must have no real roots, so its discriminant is less than zero.
$(2-2k)^2-4(1)(11+k)<0$. Dividing by $4$ gives $(1-k)^2-(11+k)<0$, so $k^2-3k-10<0$.
$(k-5)(k+2)<0$, so $k$ lies between the two roots: $-2<k<5$.
$-2<k<5$.
$-2x^2+12x+7=-2(x^2-6x)+7$.
$x^2-6x=(x-3)^2-9$, so $-2x^2+12x+7=-2\big((x-3)^2-9\big)+7=-2(x-3)^2+25$.
Because the coefficient of $(x-3)^2$ is negative, the curve has its maximum when $(x-3)^2=0$. Hence $x=3$, and the maximum value is $y=25$, so $P=(3,25)$.
The line through $O=(0,0)$ and $P=(3,25)$ has gradient $\dfrac{25}{3}$, so $l$ has equation $y=\dfrac{25}{3}x$.
Points in $R$ are below the curve, above the line, and on or to the right of the $y$-axis. Therefore $y\leq-2x^2+12x+7$, $y\geq\dfrac{25}{3}x$, and $x\geq0$.
$(a)\ -2(x-3)^2+25.\qquad (b)\ P=(3,25).\qquad (c)\ y\leq-2x^2+12x+7,\ y\geq\dfrac{25}{3}x,\ x\geq0.$
$y-5x=75$, so $y=5x+75$.
$5x+75=2x^2+x-21$, so $2x^2-4x-96=0$.
$x^2-2x-48=0=(x-8)(x+6)$, so $x=-6$ or $x=8$.
Using $y=5x+75$, $x=-6$ gives $y=45$ and $x=8$ gives $y=115$.
From the diagram, $P=(-6,45)$ and $Q=(8,115)$.
The shaded region is above the x-axis and on the left part of the diagram, so $y\ge0$ and $x\le0$.
The region is below both the line and the curve, so $y\le5x+75$ and $y\le2x^2+x-21$.
$(a)\ P=(-6,45),\ Q=(8,115).\quad (b)\ y\ge0,\ y\le5x+75,\ y\le2x^2+x-21,\ x\le0$.
Elite Mathematics · Questions classified by topic, solutions checked by Dr Eslam
Elite Mathematics