Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 2 Basic Algebra Ex 2.5 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.5

Question 1.

Solve 2x^{2} + x – 15 ≤ 0

Answer:

The given inequality is

2x^{2} + x – 15 ≤ 0 ——— (1)

2x^{2} + x – 15 = 2x^{2} + 6x – 5x – 15

= 2x (x + 3) – 5 (x + 3)

= (2x – 5)(x + 3)

2x^{2} + x – 15 = 2\(\left(x-\frac{5}{2}\right)\))(x + 3) ——— (2)

The critical numbers are x – \(\frac{5}{2}\) = 0 or x + 3 = 0

The critical numbers are x = \(\frac{5}{2}\) or x = – 3

Divide the number line into three intervals

(i) (- ∞, – 3)

When x < – 3 say x = – 4

The factor x – \(\frac{5}{2}\) = – 4 – \(\frac{5}{2}\) < 0 and

x + 3 = – 4 + 3 = – 1 < 0

x – \(\frac{5}{2}\) < 0 and x + 3 < 0

⇒ \(\left(x-\frac{5}{2}\right)\) (x + 3) > 0

Using equation (2) 2x^{2} + x – 15 > 0

∴ 2x^{2} + x – 15 ≤ 0 is not true in (- ∞, – 3)

(ii) \(\left(-3, \frac{5}{2}\right)\)

When – 3 < x < \(\frac{5}{2}\) say x = 0

The factor x – \(\frac{5}{2}\) = 0 – \(\frac{5}{2}\) = – \(\frac{5}{2}\) < 0 and

x + 3 = 0 + 3 = 3 > 0

x – \(\frac{5}{2}\) < 0 and x + 3 > 0

⇒ \(\left(x-\frac{5}{2}\right)\) (x + 3) < 0

using equation (2) 2x^{2} + x – 15 < 0

∴ 2x^{2} + x – 15 ≤ 0 is true in \(\left(-3, \frac{5}{2}\right)\)

(iii) \(\left(\frac{5}{2}, \infty\right)\)

When x > \(\frac{5}{2}\) say x = 3

The factor x – \(\frac{5}{2}\) = 3 – \(\frac{5}{2}\) > 0 and

x + 3 = 3 + 3 > 0

x – \(\frac{5}{2}\) > 0 and x + 3 > 0

= \(\left(x-\frac{5}{2}\right)\) (x + 3) > 0

Using equation (2) 2x^{2} + x – 15 > 0

∴ 2x^{2} + x – 15 ≤ 0 is not true in \(\left(\frac{5}{2}, \infty\right)\)

We have proved the inequality 2x^{2} + x – 15 ≤ 0 is true in the interval \(\left(-3, \frac{5}{2}\right)\)

But it is not true in the interval

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Question 2.

Solve x^{2} + 3x – 2 ≥ 0

Answer:

The given inequality is

– x^{2} + 3x – 2 ≥ 0

x^{2} – 3x + 2 < 0 ——– (1)

x^{2} – 3x + 2 = x^{2} – 2x – x + 2

= x(x – 2) – 1(x – 2)

x^{2} – 3x + 2 = (x – 1) (x – 2) ——— (2)

The critical numbers are

x – 1 = 0 or x – 2 = 0

The critical numbers are

x = 1 or x = 2

Divide the number line into three intervals

(- ∞, 1), (1, 2) and (2, ∞).

(i) (- ∞, 1)

When x < 1 say x = 0

The factor x – 1 = 0 – 1 = – 1 < 0 and

x – 2 = 0 – 2 = – 2 < 0

x – 1 < 0 and x – 2 < 0

⇒ (x – 1)(x – 2) > 0

Using equation (2) x^{2} – 3x + 2 > 0

∴ The inequality x^{2} – 3x + 2 ≤ 0 is not true in the interval (- ∞, 1 )

(ii) (1, 2)

When x lies between 1 and 2 say x = \(\frac{3}{2}\)

The factor x – 1 = \(\frac{3}{2}\) – 1 = \(\frac{1}{2}\) > 0 and

x – 2 = \(\frac{3}{2}\) – 2 = – \(\frac{1}{2}\) – < 0

x – 1 > 0 and x – 2 < 0

⇒ (x – 1)(x – 2) < 0

Using equation (2) x^{2} – 3x + 2 < 0

∴ The inequality x^{2} – 3x + 2 ≤ 0 is true in the interval (1, 2 )

(iii) (2, ∞)

When x > 2 say x = 3

The factor x – 1 = 3 – 1 = 2 > 0 and

x – 2 = 3 – 2 = 1 > 0

x – 1 > 0 and x – 2 > 0

= (x – 1)(x – 2) > 0

Using equation (2) x^{2} – 3x + 2 > 0

∴ The inequality x^{2} – 3x + 2 ≤ 0 is not true in the interval (2, ∞)

We have proved the inequality x^{2} – 3x + 2 ≤ 0 is true in the interval [ 1, 2 ].

But it is not true in the interval

(- ∞, 1) and (2, ∞)

∴ The solution set is [1, 2]