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

Question 1.

Solve for x.

(i) |3 – x| < 7

Solution:

⇒ -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

Solution:

(iii)

Solution:

(iv) |x| – 10 < -3

Solution:

|x| < -3 + 10 (= 7)

|x| < 7 ⇒ -7 < x < 7

Question 2.

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

Solution:

Question 3.

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

Solution:

-3|x| + 5 ≤ – 2

⇒ -3 |x| ≤ – 2 – 5 (= -7)

-3|x| ≤ – 7 ⇒ 3 |x| ≥ 7

Question 4.

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

Solution:

Question 5.

Solution:

Question 6.

Solve |5x – 12| < -2

Solution:

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

Question 1.

Solution:

Question 2.

Solution:

⇒ x – 2< – 1 (or) x – 2 > 1 and – 2 < x – 2 < 2

⇒ x < 1 (or) x > 3 and -2 + 2 < x < 2 + 2

⇒ x < 1 (or) x > 3 and 0 < x < 4 Hence, the required solution is (0, 1) ∪ (3, 4)

Question 3.

Solution:

Question 4.

Solve: |x – 1| ≤ 5, |x| ≥ 2

Solution:

|x – 1| ≤ 5 and |x| ≥ 2

⇒ -5 ≤ x – 1 ≤ 5 and x ≤ -2 (or) x > 2

⇒ – 5 + 1 ≤ x ≤ 5 + 1

⇒ -4 ≤ x ≤ 6 and x ≤ -2 (or) x ≥ 2

Hence x < [-4, -2] ∪ [2, 6]

Question 5.

Solution: