Ncert Solutions Maths class 11th

19. Given, 3 (1–x) < 2 (x + 4)

3 –3x < 2x + 8

3–8 < 2x+ 3x

–5 < 5x

$\frac{-5}{5}<x$

-1< x

So, x ∈ (–1, ∞ )

18. Given, 5x – 3 > 3x 5

5x –3x≥ –5 + 3.

2x ≥ –2.

x $\frac{-2}{2}$

x ≥ –1.

So, x [1, - ∞ ).

17. Given, 3x – 2 < 2x + 1

3x – 2x< 2 + 1

x< 3.

So, x (- ∞, 3).

18. $\underset{}{}$$\underset{x?0}{lim}\frac{ax+xcosx}{bsinx}=\underset{x?0}{lim}\frac{x(a+cosx)}{bsinx}$

$=\frac{1}{b}*\frac{(a+cos0)}{1}$

$=\frac{a+1}{b}$

16. Given,

$\frac{2x-1}{3}\ge \frac{3x-2}{4}-\frac{2-x}{5}$

$\frac{(2x-1)}{3}\ge \frac{5(3x-2)-4(2-x)}{4*5}$

20 (2x- 1) ≥3 [15x- 10 - 8 + 4x].

40x-20 ≥3 [19x-18]

40x-20 ≥ 57x – 54.

54 - 20 ≥ 57x- 40x

34 ≥17x

$\Rightarrow \frac{34}{17}\ge x.$

2 ≥x

So, x (- ∞, 2].

14. Given, 37 – (3x + 5) ≥9x –8 (x –3)

37– 3x–5 ≥9x –8x+ 24.

37 – 24–5 ≥9x –8x+ 3x

8 ≥4x.

$\Rightarrow \frac{8}{4}\ge x$

2 ≥x

So, x∈ (–∞, 2]

13. Given 2 (2x + 3) –10 < 6 (x 2)

4x + 6 –10 < 6x 12.

4x –4 < 6x 12

12 –4 < 6x 4x

8 < 2x

$\Rightarrow \frac{8}{2}<\frac{2x}{2}\Rightarrow 4<x$

So, x∈ (4, –∞)

12. Given, $\frac{1}{2}$ $\left(\frac{3x}{5}+4\right)\ge \frac{1}{3}(x-6)$

$\frac{6}{2}\left(\frac{3x}{5}+4\right)\ge \frac{6}{3}(x-6)$

$\Rightarrow 3\left(\frac{3x}{5}+4\right)\ge 2(x-6)$

$\Rightarrow \frac{9}{5}x+12\ge 2x-12$

$\Rightarrow \frac{9}{5}x-2x\ge -12-12.$

$\Rightarrow \frac{9x-10x}{5}\ge -24.$

$\Rightarrow \frac{-x*5\ge }{5}-24*5.$

x ≥–120.

(–1) x (–x)≤ ( –120)* (–1)

x ≤ 120.

So, x ∈ (–∞, 120].