Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.2 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.2

Question 1.

Solve for x

(i) |3 – x| < 7

Answer:

-7 < 3 – x < 7 3 – x > -7

-x > -7 -3 (= -10)

-x > -10 ⇒ x < 10

3 – x < 7

– x < 7 – 3 (= 4)

– x < 4x > -4 … .(2)

From (1) and (2)

⇒ x > -4 and x < 10

⇒ -4 < x < 10

(ii) |4x – 5| ≥ – 2

Answer:

|4x – 5| ≥ -2

(4x – 5) ≤ -(-2) or (4x – 5) ≥ -2

(4x – 5) ≤ 2 or (4x – 5) ≥ -2

4x ≤ 2 + 5 or 4x ≥ -2 + 5

∴ x ∈ (-∞, ∞) = R

(iii) |3 – \(\frac{3}{4}\)x| ≤ \(\frac{1}{4}\)

Answer:

Multiplying by 4, we have

– 13 ≤ – 3x ≤ – 11 ——– (1)

We know that a < b ⇒ \(\frac{\mathrm{a}}{\mathrm{y}}\) > \(\frac{\mathrm{b}}{\mathrm{y}}\) when y < 0

Divide equation (1) by – 3, we have

(iv) |x| – 10 < – 3

Answer:

|x| – 10 < – 3

|x| < – 3 + 10

|x| < 7

– 7 < x < 7

∴ The solution set is (-7, 7)

Question 2.

Solve \(\frac{1}{|2 x-1|}\) < 6 and express the solution using the interval notation.

Answer:

Question 3.

Solve – 3 |x| + 5 ≤ – 2 and graph the solution set in a number line.

Answer:

-3|x| + 5 ≤ – 2

⇒ -3 |x| ≤ – 2 – 5 (= -7)

-3|x| ≤ – 7 ⇒ 3 |x| ≥ 7

Question 4.

SoIve 2|x + 1| – 6 ≤ 7 and graph the solution set in a number line.

Answer:

Given 2|x + 1| – 6 ≤ 7

2|x + 1| ≤ 7 + 6

2|x + 1| ≤ 13

|x + 1| ≤ \(\frac{13}{2}\)

Question 5.

Solve \(\frac{1}{5}\) |10x – 2| < 1

Answer:

Given \(\frac{1}{5}\) |10x – 2| < 1

|10x – 2| < 5

-5 < (10x – 2) < 5

– 5 + 2 < 10x < 5 + 2

– 3 < 10x < 7

\(-\frac{3}{10}\) < x < \(\frac{7}{10}\)

∴ The solution set is x ∈ \(\left(-\frac{3}{10}, \frac{7}{10}\right)\)

Question 6.

Solve |5x – 12| < – 2

Answer:

By the definition of modulus function. |5x – 12| always positive.

∴ |5x – 12| < -2 is not possible.

∴ Solution does not exist.