Topic 15 — tasks — Mathematik

1

Task One correct answer

Solve \(6^x\lt\dfrac1{36}\).

А \((-2;+\infty)\) Б \((-\infty;\dfrac16)\) В \((-\infty;-2)\) Г \((\dfrac16;+\infty)\) Д \((-\infty;\dfrac12)\)

Solution

\(\dfrac1{36}=6^{-2}\). Since \(6\gt1\), the function \(6^x\) is increasing. Thus \(6^x\lt6^{-2}\) gives \(x\lt-2\). Answer: C.

2

Task One correct answer

Solve \(2\cdot(0.3)^x\lt0.18\).

А \((-\infty;2)\) Б \((2;+\infty)\) В \((-\infty;0{,}3)\) Г \((0{,}3;+\infty)\) Д \((0;2)\)

Solution

Divide by \(2\): \((0.3)^x\lt0.09=(0.3)^2\). Since \(0.3\lt1\), the exponential function is decreasing, so \(x\gt2\). Answer: B.

3

Task Numeric answer

Solve \(3\cdot9^x-2\cdot15^x-5^{2x+1}\gt0\). If there are integer solutions, enter the greatest one. If solutions exist but no greatest integer solution can be specified, enter \(50\). If there are no solutions, enter \(100\).

Your answer

Solution

Divide by \(25^x\gt0\): \(3\left(\dfrac35\right)^{2x}-2\left(\dfrac35\right)^x-5\gt0\). Let \(t=\left(\dfrac35\right)^x\gt0\). Then \(3t^2-2t-5\gt0\), so \(t\gt\dfrac53\). Since \(\dfrac53=\left(\dfrac35\right)^{-1}\) and the base is less than \(1\), we get \(x\lt-1\). The greatest integer solution is \(-2\).

4

Task Numeric answer

Solve the system \(\begin{cases}(0.5)^{1-2x}\gt(0.5)^{8+x},\\\dfrac4{x-5}\lt0.\end{cases}\) Enter the number of all integer solutions. If there are infinitely many, enter \(100\).

Your answer

Solution

Since \(0.5\lt1\), the first inequality is equivalent to \(1-2x\lt8+x\), so \(x\gt-\dfrac73\). The second inequality gives \(x\lt5\). Hence \(-\dfrac73\lt x\lt5\), containing \(7\) integers.

5

Task One correct answer

Solve \(\left(\dfrac15\right)^x\le\dfrac1{25}\).

А \((-\infty;5]\) Б \((-\infty;2]\) В \((0;2]\) Г \([2;+\infty)\) Д \([5;+\infty)\)

Solution

\(\dfrac1{25}=\left(\dfrac15\right)^2\). Since the base is less than \(1\), the function is decreasing, so \(x\ge2\). Answer: D.

6

Task One correct answer

Solve \(\left(\dfrac\pi4\right)^x\lt\left(\dfrac4\pi\right)^3\).

А \((-3;+\infty)\) Б \((3;+\infty)\) В \((-\infty;3)\) Г \((-\infty;-3)\) Д \((-\infty;\dfrac13]\)

Solution

\(\left(\dfrac4\pi\right)^3=\left(\dfrac\pi4\right)^{-3}\). Since \(0\lt\dfrac\pi4\lt1\), the exponential function is decreasing, so \(x\gt-3\). Answer: A.

7

Task One correct answer

Using the graphs shown, solve \(2^x\gt-x+3\).

А \((-\infty;2)\) Б \((1;+\infty)\) В \((0;1)\) Г \((-\infty;1)\) Д \((2;+\infty)\)

Solution

The graphs \(y=2^x\) and \(y=-x+3\) intersect at \(x=1\). To the right of this point, \(y=2^x\) lies above the line. Hence \(x\gt1\). Answer: B.

8

Task One correct answer

Solve \(\left(\dfrac13\right)^x\gt1\).

А \((-\infty;0)\) Б \((-\infty;1)\) В \((0;+\infty)\) Г \((1;+\infty)\) Д \((3;+\infty)\)

Solution

\(1=\left(\dfrac13\right)^0\). Since the base is less than \(1\), the function is decreasing, so \(x\lt0\). Answer: A.

9

Task One correct answer

Solve \(\left(\dfrac37\right)^{x-5}\gt\dfrac37\).

А \((-\infty;5)\) Б \((-\infty;6)\) В \((0;5)\) Г \((5;+\infty)\) Д \((6;+\infty)\)

Solution

\(\dfrac37=\left(\dfrac37\right)^1\). Since the base is less than \(1\), we need \(x-5\lt1\), hence \(x\lt6\). Answer: B.

10

Task One correct answer

Which of the following numbers satisfies \(5\le3^x\le15\)?

А \(5\) Б \(4\) В \(3\) Г \(2\) Д \(1\)

Solution

For \(x=2\), \(3^2=9\) and \(5\le9\le15\). Thus \(2\) satisfies the inequality. Answer: D.

11

Task One correct answer

Solve \(2^{4x-5}\ge2\).

А \([1{,}5;+\infty)\) Б \([1{,}25;+\infty)\) В \([-1;+\infty)\) Г \((-\infty;-1]\) Д \([\dfrac23;+\infty)\)

Solution

\(2=2^1\). Since the base \(2\gt1\), we get \(4x-5\ge1\). Hence \(x\ge\dfrac32=1.5\). Answer: A.

12

Task Numeric answer

Solve \(\dfrac{10^x-16\cdot5^x}{x+2}\ge0\). Enter the sum of all integer solutions on \([-3,7]\).

Your answer

Solution

Factor out \(5^x\gt0\). The sign is therefore determined by \(\dfrac{2^x-16}{x+2}\). The critical points are \(x=-2\) and \(x=4\). The interval method gives \((-\infty,-2)\cup[4,+\infty)\). On \([-3,7]\), the integer solutions are \(-3,4,5,6,7\), whose sum is \(19\).

13

Task Numeric answer

Solve \(\left(\dfrac12\right)^{x^2-x}\gt8^{x-5}\). Enter the sum of all integer solutions. If there are infinitely many integer solutions, enter \(100\).

Your answer

Solution

Write both sides with base \(2\): \(2^{-x^2+x}\gt2^{3x-15}\). Hence \(-x^2+x\gt3x-15\), so \((x+5)(x-3)\lt0\). Thus \(-5\lt x\lt3\). The integer solutions sum to \(-7\).

14

Task Matching

Solve inequalities 1–4. Match each inequality with its solution set A–E.

А \((-\infty;2)\) Б \((-2;2)\) В \((0;2)\) Г \((-\infty;-2)\) Д \((2;+\infty)\)

1. \(5^{x-2}\gt1\) 2. \(-\dfrac2{x+2}\gt0\) 3. \(\log_2 x\lt1\) 4. \(x^2\lt4\)

Solution

1) \(5^{x-2}\gt1\Rightarrow x\gt2\) — E. 2) \(-2/(x+2)\gt0\Rightarrow x\lt-2\) — D. 3) \(\log_2x\lt1\Rightarrow0\lt x\lt2\) — C. 4) \(x^2\lt4\Rightarrow-2\lt x\lt2\) — B. Thus: **1–E, 2–D, 3–C, 4–B**.

15

Task One correct answer

Solve \(2^x\le3\).

А \((-\infty;\log_2 3]\) Б \((0;\log_2 3]\) В \((-\infty;\dfrac32]\) Г \((-\infty;\log_3 2]\) Д \([\log_2 3;+\infty)\)

Solution

Since \(2^x\) is increasing, \(2^x\le3\) is equivalent to \(x\le\log_2 3\). Answer: A.

16

Task One correct answer

Solve \(3^x\lt27\cdot3^{-x}\).

А \((-\infty;\dfrac23)\) Б \((\dfrac32;+\infty)\) В \((-\infty;3)\) Г \((\dfrac23;+\infty)\) Д \((-\infty;\dfrac32)\)

Solution

\(27\cdot3^{-x}=3^{3-x}\). Since the base is greater than \(1\), \(x\lt3-x\), hence \(x\lt\dfrac32\). Answer: E.

17

Task One correct answer

Solve \(10^{x+1}\gt0.01\).

А \((-\infty;-3)\) Б \((-\infty;-2)\) В \((-3;+\infty)\) Г \((-2;+\infty)\) Д \((1;+\infty)\)

Solution

\(0.01=10^{-2}\). Since \(10^x\) is increasing, \(x+1\gt-2\), so \(x\gt-3\). Answer: C.

18

Task One correct answer

Solve \(4\cdot3^x\lt3^x+6\).

А \((-\infty;\log_9 6)\) Б \((-\infty;\log_2 3)\) В \((-\infty;2)\) Г \((-\infty;1)\) Д \((-\infty;\log_3 2)\)

Solution

\(4\cdot3^x\lt3^x+6\Rightarrow3\cdot3^x\lt6\Rightarrow3^x\lt2\). Hence \(x\lt\log_3 2\). Answer: E.

19

Task One correct answer

Solve \(0.5^{2x+7}\le4\).

А \((-\infty;-4{,}5]\) Б \([-2{,}5;+\infty)\) В \([-4{,}5;+\infty)\) Г \((-\infty;-2{,}5]\) Д \([2{,}5;+\infty)\)

Solution

\(4=(0.5)^{-2}\). Since the base is less than \(1\), \(2x+7\ge-2\), so \(x\ge-4.5\). Answer: C.

20

Task One correct answer

Solve \(5^{8x+1}+5^{8x-1}\lt130\).

А \((-\infty;-0{,}25)\) Б \((-\infty;0{,}25)\) В \((-0{,}25;+\infty)\) Г \((0{,}25;+\infty)\) Д \((-\infty;4)\)

Solution

Factor out \(5^{8x-1}\): \(26\cdot5^{8x-1}\lt130\), so \(5^{8x-1}\lt5\). Hence \(8x-1\lt1\), giving \(x\lt\dfrac14=0.25\). Answer: B.

21

Task One correct answer

Solve \(4^{\frac{x+1}{x}}-17\cdot2^{\frac1x}+4\gt0\).

А \((0;\dfrac12)\) Б \((-\dfrac12;0)\) В \((-\dfrac12;\dfrac12)\) Г \((-\dfrac12;0)\cup(0;\dfrac12)\) Д \((\dfrac12;+\infty)\)

Solution

Domain: \(x\ne0\). Let \(a=2^{1/x}\gt0\). Then the inequality becomes \(4a^2-17a+4\gt0\), whose roots are \(\frac14\) and \(4\). Thus \(a\lt\frac14\) or \(a\gt4\), giving \(x\in(-\frac12,0)\cup(0,\frac12)\). Answer: D.

22

Task Numeric answer

Solve \(3^{x-3}\gt5^{x^2-7x+12}\). Enter the smallest integer solution.

Your answer

Solution

Taking base-5 logarithms gives \((x-3)\log_5 3\gt(x-3)(x-4)\), hence \((x-3)(x-4-\log_5 3)\lt0\). Thus \(3\lt x\lt4+\log_5 3\), so the smallest integer solution is \(4\).

23

Task Numeric answer

Solve \(\left(\sqrt{4+\sqrt{15}}\right)^x+\left(\sqrt{4-\sqrt{15}}\right)^x\lt8\). Enter the greatest integer solution.

Your answer

Solution

The product of the two bases equals \(1\). Let \(t=(\sqrt{4+\sqrt{15}})^x\gt0\). Then \(t+1/t\lt8\), so \(4-\sqrt{15}\lt t\lt4+\sqrt{15}\), which is equivalent to \(-2\lt x\lt2\). The greatest integer solution is \(1\).

24

Task Numeric answer

Solve \(3^{1+\sqrt{x+1}}+3^{2-\sqrt{x+1}}\ge28\). Enter the smallest integer solution.

Your answer

Solution

Domain: \(x\ge-1\). Let \(t=3^{\sqrt{x+1}}\ge1\). Then \(3t+9/t\ge28\), so \(3t^2-28t+9\ge0\). Hence \(t\le1/3\) or \(t\ge9\). Since \(t\ge1\), we need \(t\ge9\), giving \(x\ge3\). The smallest integer solution is \(3\).

25

Task One correct answer

Solve \(3^x\cdot5^x\le\dfrac1{15}\).

А \((-\infty;-1]\) Б \([1;+\infty)\) В \((-\infty;\dfrac1{15}]\) Г \((-\infty;1]\) Д \([-1;+\infty)\)

Solution

\(3^x\cdot5^x=15^x\) and \(1/15=15^{-1}\). Since \(15\gt1\), we get \(x\le-1\). Answer: A.

26

Task One correct answer

Which of the following numbers is a solution of \(5^{-x}\gt25\)?

А \(-3\) Б \(1\) В \(-1\) Г \(0\) Д \(3\)

Solution

\(25=5^2\). From \(5^{-x}\gt5^2\), we get \(-x\gt2\), hence \(x\lt-2\). Only \(-3\) satisfies this. Answer: A.

27

Task One correct answer

Solve the system \(\begin{cases}5^x\lt25,\\2-x\lt8.\end{cases}\)

А \((2;6)\) Б \((2;+\infty)\) В \((-6;5)\) Г \((-\infty;-6)\) Д \((-6;2)\)

Solution

From \(5^x\lt5^2\), \(x\lt2\). From \(2-x\lt8\), \(x\gt-6\). Thus \(x\in(-6,2)\). Answer: E.

28

Task One correct answer

Solve the system \(\begin{cases}\left(\dfrac13\right)^x\lt81,\\|x|\le5.\end{cases}\)

А \((-\infty;4)\) Б \([5;+\infty)\) В \([-5;-4)\) Г \((-4;5]\) Д \([-5;4)\)

Solution

\(81=(1/3)^{-4}\). Since the base is less than \(1\), we get \(x\gt-4\). Also \(-5\le x\le5\). Hence \(x\in(-4,5]\). Answer: D.

29

Task One correct answer

Solve \(\left(\dfrac15\right)^x\gt\dfrac1{25}\).

А \((2;+\infty)\) Б \((0;2)\) В \((-\infty;5)\) Г \((5;+\infty)\) Д \((-\infty;2)\)

Solution

\(1/25=(1/5)^2\). Since the base is less than \(1\), we get \(x\lt2\). Answer: E.

30

Task One correct answer

Solve \(8^x\gt\dfrac1{64}\).

А \((2;+\infty)\) Б \((-\infty;-3)\) В \((-2;+\infty)\) Г \((0;2)\) Д \((-\infty;-2)\)

Solution

\(8=2^3\) and \(1/64=2^{-6}\). Thus \(3x\gt-6\), so \(x\gt-2\). Answer: C.

Exponential inequalities