representation on number line

Correct answer: A. ( [-6,-1)\cup[4,9] )

The direct answer is option A: \([-6,-1)\cup[4,9]\). Set difference means keep the first set and remove every number belonging to the second set. Start with the closed interval \([-6,9]\), which includes both -6 and 9. Remove \([-1,4)\): this removed interval includes -1, includes all numbers up to 4, but does not include 4 itself. The portion left on the left is from -6 through numbers just before -1, so it is \([-6,-1)\): -6 stays, while -1 is removed. The portion left on the right begins at 4 and goes through 9, so it is \([4,9]\): 4 stays because it was excluded from the removed interval, and 9 stays because it belongs to the original interval. Option A is therefore correct. Option B, \([-6,-1]\cup(4,9]\), wrongly keeps -1 and wrongly removes 4. Option C excludes -6 and includes 4 only up to but not including 9, so both endpoint choices are wrong. Option D is a union that overlaps the removed middle part and does not represent subtraction at all. On a number line, shade the original interval, erase the section from -1 including -1 up to but not including 4, and observe the two remaining pieces. Memory cue: in set difference, inspect the brackets at both ends of the part being removed; an included removed endpoint disappears, while an excluded one remains.

Correct answer: D. No solution

No number can be both greater than or equal to ( 3 ) and less than ( 3 ). For AND, the intersection is empty.

Correct answer: C. Closed at ( -2 ) shaded left and open at ( 10 ) shaded right

In ( x\le -2 ), ( -2 ) is closed, and in ( x>10 ), ( 10 ) is open. OR forms the union of separate parts.

Correct answer: A. ( \left[\frac{4}{5},\frac{19}{5}\right) )

Solving gives ( -19<-5x\le -4 ), then signs reverse to ( \frac{19}{5}>x\ge \frac{4}{5} ). In order, write ( \frac{4}{5}\le x<\frac{19}{5} ).

Correct answer: C. ( \mathbb{R} )

The first part gives values less than ( -9 ), and the second gives ( -9 ) and larger values. Together they form the whole number line.

Correct answer: D. ( (1,12] )

The first interval includes ( 4 ), and the second begins after ( 4 ). Together they form ( 1<x\le 12 ).

Correct answer: C. ( x>7 ), open at ( 7 ), shaded right

The direct answer is option C: \(x>7\), shown by an open point at 7 and shading to the right. Start with \((13-3x)/(-4)>2\). Multiply both sides by \(-4\); because it is negative, reverse the inequality: \(13-3x<-8\). Subtract 13: \(-3x<-21\). Divide by \(-3\), reversing the sign again: \(x>7\). The endpoint is open because 7 does not satisfy the original strict inequality: it gives \(2>2\), which is false. Option A wrongly includes 7 and points right. Option B has the wrong direction and excludes the wrong side. Option C has both the correct open endpoint and correct right-hand direction. Option D includes 7 and points left. The important warning is that dividing or multiplying by a negative number reverses the inequality sign.

Correct answer: A. \(x\in(-\infty,-1)\cup[4,\infty)\)

An open circle at \(-1\) means that \(-1\) is excluded, so the left ray is \(( -\infty,-1)\). A filled circle at \(4\) means that \(4\) is included, so the right ray is \([4,\infty)\). Since the shaded parts are two separate rays, they are joined using union, \(\cup\). Option B reverses the inclusion of both endpoints. Exam tip: use a round bracket for an open circle and a square bracket for a filled circle.

Correct answer: A. ( [1,6) )

The direct answer is option A: \([1,6)\). Solve the compound inequality as one chain: \(1\le3x-2<16\). Add 2 to all three parts: \(3\le3x<18\). Divide all parts by positive 3, so the signs do not change: \(1\le x<6\). Since 1 is included, use a closed endpoint at 1. Since 6 is excluded by the strict sign \(<\), use an open endpoint at 6. Option A, \([1,6)\), matches exactly. Option B reverses which endpoint is included. Option C is the interval before dividing by 3 and therefore gives values for \(3x\), not for \(x\). Option D has both the wrong interval and wrong endpoint inclusion. A useful check is to test \(x=2\): it satisfies the original inequality, while \(x=6\) gives \(16<16\), false. Memory cue: adding or dividing by a positive number keeps signs; equality makes a closed endpoint.

Correct answer: D. ( \emptyset )

The first interval stops before ( 5 ), and the second begins after ( 5 ). Therefore, there is no common part.

Correct answer: A. Only a closed dot at ( -4 )

Absolute value cannot be less than ( 0 ), and ( |x+4|=0 ) gives ( x=-4 ). So only one closed dot appears.

Correct answer: A. ( \left[-\frac{7}{2},\frac{5}{2}\right) )

Solving gives ( -5<-2x\le 7 ), then signs reverse to ( \frac{5}{2}>x\ge -\frac{7}{2} ). In order, write ( -\frac{7}{2}\le x<\frac{5}{2} ).

Correct answer: A. An open endpoint at 2, a closed endpoint at 8, and an open hole at 5

In the interval \((2,8]\), 2 is excluded, so it has an open endpoint, while 8 is included, so it has a closed endpoint. The notation \(\setminus\{5\}\) removes 5 from the interval, so an open hole must be marked at 5. Therefore, option A is correct. Exam tip: round brackets indicate open endpoints, whereas square brackets indicate closed endpoints.

Correct answer: B. ( x\le -3 ) or ( x>10 )

The first inequality gives ( x\le -3 ), and the second gives ( x>10 ). For OR, take the union of both parts.

Correct answer: A. ( (3,7] )

The first inequality gives ( x>3 ), and the second gives ( x\le 7 ). For both conditions, the common part is ( 3<x\le 7 ).

Correct answer: A. Whole number line shaded with an open hole at ( 0 )

The two intervals give all real numbers except ( 0 ). The open hole in the middle shows the removed point.

Correct answer: C. Both endpoints closed

Both sides have square brackets, so both boundary values are included. Included boundaries are shown by closed dots.

Correct answer: C. ( \emptyset )

The first interval does not include ( 2 ), and the second starts at ( 2 ). Therefore, there is no common point.

Correct answer: C. Only a closed dot at ( 3 )

The direct answer is option C: only the closed point 3. An intersection contains numbers that belong to both intervals. In \((-4,3]\), numbers are greater than -4 and up to and including 3. In \([3,9)\), numbers start at and include 3, then go below 9. The only common number is 3, and both intervals include it, so the result is \(\{3\}\), shown by a closed dot. Option A, \((-4,9)\), is a broad union-like interval and includes numbers belonging to only one interval. Option B, \((3,9)\), excludes 3 and contains numbers not in the first interval. Option C is correct. Option D is wrong because the intersection is not empty. Memory cue: for intersection, keep the overlap; when endpoints meet and both brackets are square, keep the endpoint.

Correct answer: A. ( [-3,3] )

( |4x+1|\le 13 ) gives ( -13\le 4x+1\le 13 ). Thus ( -\frac{7}{2}\le x\le 3 ), not the listed interval.

Correct answer: A. ( (-\infty,-1)\cup(4,9] )

The complement of the given union gives values less than ( -1 ) and values greater than ( 4 ) up to ( 9 ). The point ( 9 ) remains because the second part excludes ( 9 ).

Correct answer: C. ( x\ge 3 ), closed at ( 3 ), shaded right

Simplifying gives ( 9-6x\le -9 ), so ( -6x\le -18 ) and ( x\ge 3 ). Reverse the sign when dividing by a negative.

Correct answer: D. \( x>\frac{1}{4} \)

The round bracket excludes \( \frac{1}{4} \), and the rightward arrow gives larger values. Therefore \( x>\frac{1}{4} \).

Correct answer: A. भरा हुआ वृत्त

A filled circle shows that the boundary point belongs to the solution set, so it is used with \(\le\) or \(\ge\). An open circle is used for \(<\) or \(>\). Exam tip: check for equality before drawing the endpoint.

Correct answer: B. \( (-\infty,1] \)

The expression \( A\setminus B \) means elements in \(A\) that are not in \(B\). Here \(A=(-\infty,6] \) and \(B=(1,6]\). So we remove all numbers greater than \(1\) up to \(6\), including \(6\). Therefore all numbers \(x\le 1\) remain. Hence the result is \( (-\infty,1] \).