Class 11th

22. Let 'x' be the marks obtained by Sunita in the 5th exam.

Then, average marks ≥ 90.

$\frac{87+92+94+95+x}{5}\ge 90$ .

368 + x ≥ 90 * 5.

368 + x ≥ 450.

x ≥ 450 - 368.

x ≥ 82.

Minimum mark required= 82.

21. Let x be the marks obtained by Ravi in third test.

Then, average mark ≥ 60

$\Rightarrow \frac{70+75+x}{3}\ge 60$

145 + x ≥ 180.

x ≥180 - 145

x ≥ 35.

So minimum mark required = 35.

20. Given,

$\frac{x}{2}\ge \frac{(5x-2)}{3}-\frac{(7x-3)}{5.}$

$\Rightarrow \frac{x}{2}\ge \frac{5(5x-2)-3(7x-3)}{3*15}$

15x ≥ 2 [25x –10 –21x + 9].

15x ≥ 2 [4x –1]

15x ≥ 8x –2

15x – 8x ≥ –2

7x ≥ –2

x ≥$-\frac{2}{7}.$

So, x $\left[-\frac{2}{7},\infty \right)$ $\left\{\therefore -\frac{2}{7}=-0.28\right\}$

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].