NCERT Exemplar Solutions Class 11 Maths Linear Inequalities — Objective, True or False and Fill in the Blanks Q19–32

NCERT Exemplar Problems, Class 11 Mathematics — Linear Inequalities, Objective Type, True or False and Fill in the Blanks questions 19 to 32. All 14 questions solved.

Questions 19 to 25 test the two basic rules: multiplying by a negative number reverses an inequality, and $|u| < a$ versus $|u| > a$. Questions 26 to 30 ask you to read a picture, either a shaded strip or a thick ray on a number line. Question 31 has fifteen true or false statements about signs, intervals and graphs, and Question 32 has eight blanks to fill with the right inequality sign.

Two statements in Question 31 are marked wrongly in the printed key, parts (iii) and (xiv). Both are explained where they occur.

Key insight. Most of these are settled by one test: pick a number and try it. If a claim says “$xy > 0$ means both are negative”, try $x = y = 1$. If a picture claims to show $y \le 1$, test a point and see whether it sits on the shaded side. A single counter-example kills a “true or false” statement.

Question 19

Choose the correct option. If $x < 5$, then

Solution. Multiplying both sides of $x < 5$ by $-1$ reverses the sign: $-x > -5$. Check with $x = 0$: $0 > -5$ is true, while (A) would say $0 < -5$.

Question 20

Given that $x$, $y$ and $b$ are real numbers and $x < y$, $b < 0$, then

Solution. Dividing by a negative number reverses the inequality, so $x < y$ gives $\frac xb > \frac yb$. The inequality remains strict because $x \ne y$. Check with $x = 1$, $y = 2$, $b = -1$: $-1 > -2$.

Question 21

If $-3x + 17 < -13$, then

Solution. Subtract $17$: $-3x < -30$. Divide by $-3$ and reverse: $x > 10$. The inequality is strict, so $10$ is not included.

Question 22

If $x$ is a real number and $|x| < 3$, then

Solution. $|x| < 3$ says $x$ is less than $3$ units from $0$, so $-3 < x < 3$. The endpoints are excluded because $|\pm 3| = 3$ is not less than $3$.

Question 23

$x$ and $b$ are real numbers. If $b > 0$ and $|x| > b$, then

Solution. $|x| > b$ means $x$ is farther than $b$ from $0$ in either direction: $x < -b$ or $x > b$. This is two pieces, so the answer is a union.

Question 24

If $|x – 1| > 5$, then

Solution. $|x – 1| > 5$ gives $x – 1 > 5$ or $x – 1 < -5$, that is $x > 6$ or $x < -4$. Both ends are strict. Option (D) wrongly includes $6$, where $|6 – 1| = 5$, which is not greater than $5$. (The book prints a square bracket beside $-\infty$ in (C) and (D); that is a typographical slip, and infinity is never included.)

Question 25

If $|x + 2| \le 9$, then

Solution. $-9 \le x + 2 \le 9$. Subtract $2$ from every part: $-11 \le x \le 7$. The interval is centred at $-2$, not $+2$, and “$\le$” gives a single closed interval, not the two outer pieces of (C) and (D).

Question 26

The inequality representing the following graph (Fig 6.7) is:

Fig 6.7 shows a shaded rectangular strip stretching from $x = -5$ to $x = 5$ across the plane.

Solution. The shaded strip lies between $x = -5$ and $x = 5$, so the inequality is of the form $|x| < 5$ or $|x| \le 5$; options (C) and (D) describe the outside regions and are out. Between the remaining two, the figure draws no boundary lines at all, so it cannot say whether the edges are included. The printed key gives (A), and we follow it. The strip is the same region as $-5 < x < 5$, which is the meaning of $|x| < 5$.

Question 27

Solution of a linear inequality in variable $x$ is represented on the number line in Questions 27 to 30. Choose the correct option. Fig 6.8 shows an open circle at $5$ with the thick ray going to the right.

Solution. The thick ray goes right from $5$, so the solution is $x > 5$. The circle at $5$ is open, so $5$ is excluded.

Question 28

Fig 6.9 shows a filled dot at $\frac92$ with the thick ray going to the right.

Solution. A filled dot means the endpoint is included, and the ray goes right: $x \ge \frac92$. (The book labels the third and fourth options both as (D); the third is really (C).)

Question 29

Fig 6.10 shows an open circle at $\frac72$ with the thick ray going to the left.

Solution. The thick part lies to the left of an open circle: $x < \frac72$, endpoint excluded. (Option (C) is printed in the book as $\left[\frac72, -\infty\right)$, a typographical slip.)

Question 30

Fig 6.11 shows a filled dot at $-2$ with the thick part of the line to the left.

Solution. In this figure the heavy line runs left from the filled dot, so $x \le -2$ and the endpoint is included. Option (D) is the mirror image and would need the thick part on the right.

Question 31

State which of the following statements is True or False.

The first nine statements are about signs and intervals, and the last six ask whether a picture matches the stated inequality. For the pictures, “the shaded side” means the side containing the shading in the book’s figure.

Question 31 (i)

If $x < y$ and $b < 0$, then $\frac xb < \frac yb$.

Solution. Dividing by a negative number reverses the inequality, so $\frac xb > \frac yb$. (Try $x = 1$, $y = 2$, $b = -1$: $-1 < -2$ is false.)

Question 31 (ii)

If $xy > 0$, then $x > 0$ and $y < 0$.

Solution. A positive product means $x$ and $y$ have the same sign. Here the signs are opposite, so this is impossible. Take $x = y = 1$: $xy > 0$ but $y < 0$ fails.

Question 31 (iii)

If $xy > 0$, then $x < 0$ and $y < 0$.

Solution. A positive product happens when both are negative or both are positive. The statement forces only the first case. Take $x = y = 1$: $xy = 1 > 0$, yet $x < 0$ fails. The printed key marks this True, which is wrong; one counter-example is enough.

Question 31 (iv)

If $xy < 0$, then $x < 0$ and $y < 0$.

Solution. A negative product needs opposite signs, but the statement makes both negative, giving a positive product. Take $x = 1$, $y = -1$: $xy < 0$ yet $x < 0$ fails.

Question 31 (v)

If $x < -5$ and $x < -2$, then $x \in (-\infty, -5)$.

Solution. Both must hold. A number below $-5$ is automatically below $-2$, so the stricter condition decides: $x < -5$.

Question 31 (vi)

If $x < -5$ and $x > 2$, then $x \in (-5, 2)$.

Solution. No number is below $-5$ and above $2$ at once, so the solution set is empty. The interval $(-5, 2)$ is not even related to these conditions.

Question 31 (vii)

If $x > -2$ and $x < 9$, then $x \in (-2, 9)$.

Solution. This is the definition of the open interval.

Question 31 (viii)

If $|x| > 5$, then $x \in (-\infty, -5) \cup [5, \infty)$.

Solution. $|x| > 5$ means $x < -5$ or $x > 5$, with both ends open. The statement closes $5$ on the right, but $|5| = 5$ is not greater than $5$.

Question 31 (ix)

If $|x| \le 4$, then $x \in [-4, 4]$.

Solution. $|x| \le 4$ means $-4 \le x \le 4$, with both ends included.

Question 31 (x)

Graph of $x < 3$ is (Fig 6.12).

Fig 6.12 shows a shaded vertical strip to the left of the line $x = 3$, which is drawn as a solid line.

Solution. The shading is on the correct side: it lies to the left of $x = 3$. What the figure gets wrong is the boundary. For a strict inequality such as $x < 3$, the points on $x = 3$ are not solutions, so the line must be drawn dashed; a solid line is the picture of $x \le 3$. The printed key marks False, and the solid boundary is the only fault in the picture that supports it, so we follow it. This is a judgement about drawing convention, not arithmetic.

Question 31 (xi)

Graph of $x \ge 0$ is (Fig 6.13).

Fig 6.13 shows a shaded region lying to the right of the $y$-axis.

Solution. $x \ge 0$ is every point on or to the right of the $y$-axis, which is what is shaded.

Question 31 (xii)

Graph of $y \le 0$ is (Fig 6.14).

Fig 6.14 shows a shaded region lying above the $x$-axis.

Solution. $y \le 0$ means on or below the $x$-axis, but the shading is above it. That region is $y \ge 0$.

Question 31 (xiii)

Solution set of $x \ge 0$ and $y \le 0$ is (Fig 6.15).

Fig 6.15 shows a shaded region in the first quadrant, above the $x$-axis and right of the $y$-axis.

Solution. $x \ge 0$ and $y \le 0$ is the fourth quadrant (right of the $y$-axis, below the $x$-axis). The figure shows the first quadrant, which is $x \ge 0$ and $y \ge 0$.

Question 31 (xiv)

Solution set of $x \ge 0$ and $y \le 1$ is (Fig 6.16).

Fig 6.16 shows a line through $y = 1$ and a shaded region to the right of the $y$-axis lying above that line.

Solution. $y \le 1$ is the region on or below the line $y = 1$, and with $x \ge 0$ that is the area to the right of the $y$-axis below the line. The figure shades the area above the line, which is $x \ge 0$ and $y \ge 1$. The printed key marks this True, which is wrong. Test the point $(1, 2)$, which is shaded: it fails $y \le 1$.

Question 31 (xv)

Solution set of $x + y \ge 0$ is (Fig 6.17).

Fig 6.17 shows the line $x + y = 0$ through the origin, sloping down to the right, with the region on its upper-right side shaded.

Solution. Test the point $(1, 1)$, which lies on the shaded side: $1 + 1 = 2 \ge 0$ holds. Test $(-1, -1)$, on the unshaded side: $-2 \ge 0$ fails. So the shaded side is $x + y \ge 0$, and the line is solid because the inequality includes equality.

Question 32

Fill in the blanks of the following.

The idea in every part is the same: transform both sides, and flip the sign each time you multiply or divide by a negative number.

Question 32 (i)

If $-4x \ge 12$, then $x \ \_\_\_\_\ -3$.

Solution. Divide by $-4$ and reverse: $x \le -3$.

Question 32 (ii)

If $\frac{-3}{4}x \le -3$, then $x \ \_\_\_\_\ 4$.

Solution. Multiply by $-\frac43$ and reverse: $x \ge 4$.

Question 32 (iii)

If $\frac{2}{x+2} > 0$, then $x \ \_\_\_\_\ -2$.

Solution. The numerator $2$ is positive, so the fraction is positive only if $x + 2 > 0$, that is $x > -2$.

Question 32 (iv)

If $x > -5$, then $4x \ \_\_\_\_\ -20$.

Solution. Multiply by the positive number $4$: $4x > -20$, and the direction stays.

Question 32 (v)

If $x > y$ and $z < 0$, then $-xz \ \_\_\_\_\ -yz$.

Solution. Since $z < 0$, we have $-z > 0$. Multiplying $x > y$ by the positive number $-z$ keeps the direction: $-xz > -yz$.

Question 32 (vi)

If $p > 0$ and $q < 0$, then $p – q \ \_\_\_\_\ p$.

Solution. $-q$ is positive, so $p – q = p + (-q)$ adds something positive to $p$, making it larger.

Question 32 (vii)

If $|x + 2| > 5$, then $x \ \_\_\_\_\ -7$ or $x \ \_\_\_\_\ 3$.

Solution. $x + 2 > 5$ gives $x > 3$, and $x + 2 < -5$ gives $x < -7$. So in the order given: $x < -7$ or $x > 3$.

Question 32 (viii)

If $-2x + 1 \ge 9$, then $x \ \_\_\_\_\ -4$.

Solution. $-2x \ge 8$; divide by $-2$ and reverse: $x \le -4$.

Common mistakes

• Questions 19 and 20: Multiplying or dividing by a negative number and keeping the sign. This is the single most common error in the chapter, and Question 31(i) tests exactly the same slip.
• Question 24: Closing the bracket at $6$ in option (D). $|x – 1| > 5$ is a strict inequality, so $6$ (where the value is exactly $5$) is not a solution.
• Question 25: Centring the interval at $2$ instead of $-2$. Subtract $2$ from every part of $-9 \le x + 2 \le 9$ and you get $[-11, 7]$.
• Questions 28 and 30: Mixing up filled and open circles with square and round brackets. Filled dot means the endpoint is included, so the bracket is square.
• Question 31(iii), (iv): Treating “$xy > 0$” as “both negative”. The product of two negatives is positive, but so is the product of two positives; one counter-example such as $x = y = 1$ settles it.
• Question 31(xii) to (xiv): Accepting a picture because the line is in the right place. Always test a point inside the shading, such as $(1, 2)$ in Fig 6.16, against the stated inequality.

Practise next

• The other page of this unit: Short and Long Answer, Q1–18.
• NCERT textbook: Exercise 5.1, Linear Inequalities for sign-reversal drills like Questions 19 to 21 and 32.
• NCERT textbook: Miscellaneous Exercise, Linear Inequalities for modulus inequalities like Questions 22 to 25.