The figure shows the graphs of \(f(x)=x+3\) and \(g(x)=\sqrt{9-x^2}\). Solve \(f(x)\ge g(x)\).
Solution
The function \(g(x)=\sqrt{9-x^2}\) is defined on \([-3,3]\). From the graph, the line \(f(x)=x+3\) is not below the semicircle at \(x=-3\) and on \([0,3]\). Thus \(x\in\{-3\}\cup[0;3]\). Answer: A.
Shapes
Grid
Shapes can be selected, moved, and resized.
2
Task
Numeric answer
Find the domain of \(y=\sqrt[4]{50-3x}\). Enter the greatest two-digit integer belonging to the domain.
Solution
For an even root we need \(50-3x\ge0\). Hence \(x\le\dfrac{50}{3}=16\dfrac23\). The greatest two-digit integer in the domain is \(16\).
Shapes
Grid
Shapes can be selected, moved, and resized.
3
Task
One correct answer
Solve \(\sqrt{2x+3}\ge-4\).
Solution
The left-hand side is always nonnegative, so the inequality holds for every admissible \(x\). Domain: \(2x+3\ge0\), hence \(x\ge-1.5\). Thus \(x\in[-1.5,+\infty)\). Answer: E.
Shapes
Grid
Shapes can be selected, moved, and resized.
4
Task
One correct answer
How many integer solutions does \(\sqrt[3]{x^3-2x}\gt x+2\) have?
Solution
Since \(t\mapsto t^3\) is increasing, cube both sides: \(x^3-2x\gt(x+2)^3\). This simplifies to \(3x^2+7x+4\lt0\), i.e. \((3x+4)(x+1)\lt0\). Hence \(x\in(-\frac43,-1)\), which contains no integers. Answer: B.
Shapes
Grid
Shapes can be selected, moved, and resized.
5
Task
One correct answer
Solve \(\sqrt{9x-20}\lt x\).
Solution
Domain: \(x\ge\dfrac{20}{9}\). On this domain the right-hand side is positive, so square: \(9x-20\lt x^2\), or \((x-4)(x-5)\gt0\). Thus \(x\lt4\) or \(x\gt5\). Intersecting with the domain gives \([\frac{20}{9},4)\cup(5,+\infty)\). Answer: B.
Shapes
Grid
Shapes can be selected, moved, and resized.
6
Task
One correct answer
Solve \(\sqrt{(x^2-5x+6)^2}\gt2-x\).
Solution
\(\sqrt{(x^2-5x+6)^2}=|x^2-5x+6|\). For \(x\le2\) or \(x\ge3\), the inequality becomes \((x-2)^2\gt0\), so \(x\ne2\). For \(2\lt x\lt3\), resolving the absolute value gives \((x-2)(x-4)\lt0\), which holds throughout that interval. Therefore the solution is \((-\infty,2)\cup(2,+\infty)\). Answer: D.
Shapes
Grid
Shapes can be selected, moved, and resized.
7
Task
One correct answer
Solve \(\sqrt{x^2-3x-4}\lt\sqrt{x^2-5x+6}\).
Solution
Domain: \((x-4)(x+1)\ge0\) and \((x-2)(x-3)\ge0\), so \(x\in(-\infty,-1]\cup[4,+\infty)\). Both sides are nonnegative there, so square: \(x^2-3x-4\lt x^2-5x+6\), giving \(x\lt5\). Hence \((-\infty,-1]\cup[4,5)\). Answer: B.
Shapes
Grid
Shapes can be selected, moved, and resized.
8
Task
One correct answer
Solve \(\dfrac{\sqrt{x-3}}{\sqrt{1-5x}}\le0\).
Solution
For the numerator we need \(x-3\ge0\), so \(x\ge3\). Because the square root is in the denominator, we need strictly \(1-5x\gt0\), so \(x\lt0.2\). These conditions are incompatible, so the domain is empty. Therefore the inequality has no solutions: \(\varnothing\).
Shapes
Grid
Shapes can be selected, moved, and resized.
9
Task
Numeric answer
Solve \(\sqrt[3]{2-x}+\sqrt{x-1}\gt1\). Enter the smallest integer solution.
Solution
Domain: \(x\ge1\). Rearrange and cube both sides, which is valid because the cube function is increasing. Let \(t=\sqrt{x-1}\ge0\). Simplification gives \(t(t-1)(t-3)\gt0\). Hence \(0\lt t\lt1\) or \(t\gt3\). Returning to \(x\): \(1\lt x\lt2\) or \(x\gt10\). The smallest integer solution is \(11\).
Shapes
Grid
Shapes can be selected, moved, and resized.
10
Task
Numeric answer
Find the number of integer solutions of \((x-2)\sqrt{16-x^2}\le0\).
Solution
Domain: \(x\in[-4,4]\). The square root is nonnegative there. The product is nonpositive when \(x\le2\), or when the square root is zero, i.e. at \(x=\pm4\). Thus \(x\in[-4,2]\cup\{4\}\), giving \(8\) integer solutions.
