Interactive ISC Mathematics Class 12 Solutions Chapter 7 Applications of Derivatives Ex 7.4 engage students in active learning and exploration.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 7 Applications of Derivatives Ex 7.4

Question 1.

Solve for x :

(i) x (x – 2) (x – 5) (x + 3) > 0

(ii) x^{4} – 5x^{2} + 4 ≥ 0.

Solution:

(i) Given, x (x – 2) (x – 5) (x + 3) = 0 ………….(1)

Here, mark the points – 3, 0, 2, 5 on real line.

By method of intervals, the inequality (1) is satisfied when

x < – 3 or 0 < x < 2 or x > 5

∴ The required solution set is (- ∞, – 3) ∪ (0, 2) ∪ (5, ∞).

(ii) Given inequality be x^{4} – 5x^{2} + 4 ≥ 0

(x^{2} – 1) (x^{2} – 4) ≥ 0

⇒ (x- 1) (x – 2) (x + 1) (x + 2) ≥ 0 ………….(1)

Here mark the real numbers – 1, – 2, 1, 2 on real line and by using method of intervals, eqn. (1) is satisfied

When x ≤ – 2 or – 1 ≤ x ≤ 1 or x ≥ 2

∴ required solution set be (- ∞, – 2] ∪ [- 1, 1] ∪ [2, ∞)

Question 2.

Find all real values of x which satisfy

(i) x^{3} (x – 1) (x – 2) > 0

(ii) x^{2} (x – 1) (x – 2) ≤ 0

Solution:

(i) Given inequality be x^{3} (x – 1) (x – 2) > 0

⇒ x (x – 1) (x – 2) > 0 ……….(1)

[∵ x^{2} ≥ 0]

Mark the real numbers 0, 1, 2 on real line.

Using method of intervals, eqn. (1) is satisfied when 0 < x < 1 or x > 2

∴ required solution set be given by (0, 1) ∪ (2, ∞).

(ii) Given inequality be

x^{2} (x – 1) (x – 2) ≤ 0

(x – 1) (x – 2) ≤ 0 ……….(1)

[∵ x^{2} ≥ 0]

Mark the numbers 1, 2 on real line and by using method of intervals, eqn. (1) is satisfied when 1 ≤ x ≤ 2

also given eqn. is satisfied when x = 0

∴ required solution set = [1, 2] ∪ {0}

Question 3.

(i) \(\frac{1}{x-2}\) ≤ 1

(ii) \(\frac{(x+1)(x-3)}{x+2}\) ≥ 0

Solution:

(i) Given inequality be \(\frac{1}{x-2}\) ≤ 1 ……………(1); x ≠ 2

Since (x – 2)^{2} > 0 ∀ x ∈ R, x ≠ 2

⇒ (x – 2) ≤ (x – 2)^{2}

[multiplying by (x – 2)^{2}]

⇒ (x – 2)^{2} – (x – 2) ≥ 0

⇒ (x – 2) (x – 3) ≥ 0

Mark the numbers 2, 3 on real line.

Using method of intervals, eqn. (2) is satisfied when x ≤ 2 or x ≥ 3

But x ≠ 2.

Thus, required solution set be (- ∞, 2) ∪ [3, ∞)

(ii) Given inequality \(\frac{(x+1)(x-3)}{x+2}\) ≥ 0 ; x ≠ – 2

∴ (x + 2)^{2} > 0 ∀ x ∈ R, x ≠ – 2

⇒ (x + 1) (x – 3) (x + 2) ≥ 0 ………..(1)

[Multiplying both sides by (x + 2)^{2} > 0]

Mark the numbers – 1, – 2, 3 on real line and by using method of intervals, eqn. (1) is satisfied

When – 2 ≤ x ≤ – 1 or x ≥ 3

But x ≠ – 2

∴ required solution is given by (- 2, -1] ∪ [3, ∞).

Question 4.

Find the values of x for which:

(i) \(\frac{x^2+6 x-11}{x+3}\)

(ii) \(\frac{x^2-3 x+24}{x^2-3 x+3}\) < 4

Solution:

(i) Given inequality be \(\frac{x^2+6 x-11}{x+3}\) < – 1

⇒ \(\frac{x^2+6 x-11}{x+3}\) + 1 < 0

⇒ \(\frac{x^2+7 x-8}{x+3}\) < 0

⇒ \(\frac{(x-1)(x+8)}{x+3}\) < 0 ;

Since (x + 3)^{2} > 0 ∀ x ∈ R, x ≠ – 3

(x – 1) (x + 8) (x + 3) < 0

[multiplying by (x + 3)^{2}]

Mark the numbers – 3, – 8 and 1 on real line.

Using methods of intervals, given inequality satisfied when

– 3 < x < 1 or x < – 8

∴ required solution set be (- ∞, – 8) ∪ (- 3, + 1).

(ii) Given \(\frac{x^2-3 x+24}{x^2-3 x+3}\) < 4

∴ from (1) ;

– 3 (x^{2} – 3x – 4) < 0

⇒ (x – 3x – 4) > 0

⇒ (x + 1) (x – 4) > 0 ………….(2)

Mark the real numbers – 1 and 4 on real line.

Using method of intervals, eqn. (2) is satisfied when x < – 1 or x > 4

∴ required solution set be (- ∞, – 1) ∪ (4, ∞).