Maths

9. x + $\frac{x}{2}+\frac{x}{3}$ < 11

$\Rightarrow \frac{6x+3x+2x}{6}<11$

$\Rightarrow \frac{(6x+3x+2x)*6}{6}<11*6$

11x < 66

$\frac{11x}{11}<\frac{66}{11}$

x < 6.

So, x ∈ (–∞, 6)

8. 3 (2 – x) ≥ 2 (1 – x)

6 – 3x ≥ 2 – 2x

6– 2 ≥ 3x– 2x

4 ≥ x

So, x ∈ (–∞, 4]

 70. Given, f (x) = $\frac{x^{2}cos\left(\frac{\pi }{4}\right)}{sinx}$

$=\frac{x^2}{sinx}*cos\frac{\pi }{4}$

So, f? (x) =

$=cos\frac{\pi }{4}\left[\frac{2·xsinx-x^{2}cosx}{si{n}^{2}x}\right]$

$=xcos\frac{\pi }{4}\left[\frac{2sinx-xcosx}{si{n}^{2}x}\right].$

7. 3 (x –1) ≤ 2 (x –3)

3x– 3 ≤ 2x –6

=> 3x– 2x ≤ –6 + 3

x ≤ –3

So, x ∈ (–∞, – 3]

69. Given, f (x) = $\frac{4x+5sinx}{3x+7cosx}$

So,

$f^{\prime }(x)=\frac{(3x+7cosx)\frac{d}{dx}(4x+5sinx)-(4x+5sinx)\frac{d}{dx}(3x+7cosx)}{{(3x+7cosx)}^2}$

$=\frac{(3x+7cosx)(4x+5cosx)-(4x+5sinx)(3-7sinx)}{{(3x+7cosx)}^2}$

$\begin{array}{l}\underset{\_}{=x+xx+x+{}^{}x-x+xx-x+35si{n}^{2}x}\\ {\left(3x+7cosx\right)}^2\end{array}$

$=\frac{15xcosx+28cosx+28xsinx-15sinx+35\left(co{s}^{2}x+si{n}^{2}x\right)}{{(3x+7cosx)}^2}$

$=\frac{35+15xcosx+28cosx+28xsinx-15sinx}{{(3x+7cosx)}^2}$

6. Given, 3x – 7 > 5x – 1.

7+1 > 5x – 3x

6 > 2x

$\Rightarrow \frac{-6}{2}>\frac{2x}{2}$

–3 > x

So, x ∈ (–∞, – 3)

68. Given, f (x) = (x + cos x) (x tan x)

So, f?(x) = (x + cos x) $\frac{d}{dx}-tanx)+\frac{d}{dx}(x+cosx)·(x-tanx)$

$=(x+cosx)\left(\frac{dx}{dx}-\frac{d}{dx}tanx\right)+\left(\frac{d}{dx}+\frac{dcosx}{dx}\right)(x-tanx)$

Let g (x) = tan x.

So, g?(x) =

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{sin(x+h)}{cos(x+h)}-\frac{sinx}{cosx}\right]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{sin(x+h)cosx-sinxcos(x+h)}{cos(x+h)·cosx}\right]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{sin(x+h-x)}{cosx·cos(x+h)}\right]$

$=\underset{h\to 0}{lim}\frac{1}{cosx·co}$$\frac{}{}$$\frac{s(x+h)}{}*\underset{h\to 0}{lim}\frac{sinh}{h}$

$=\frac{1}{co{s}^{2}x}*1$

= sec²x ______ (2)

     Put (2) in (1) we get,

     f?(x) = (x + cos x) (1 - sec²x) +(1 - sin x) (x- tan x)

     We know that,

     1 + tan²x = sec²x

     Þ 1 - sec²x = - tan²x

     So, f?(x) = - tan²x(x + cos x) +(x- tan x) (1 - sin x).

5. Given,

4 x + 3 < 5 x+ 7.

3 – 7 < 5 x 4 x.

–4 < x.

So, x ( –4, ∞)

4. Given, 3x + 8 > 2

3x + 8 – 8 > 2 – 8

3x > – 6

$\Rightarrow \frac{3x}{3}>-\frac{6}{3}$

x > – 2

(i) when x is an integer, x ∈ z all integer greater than – 2 are the soln. so, x = { – 1, 0, 1, 2, 3, …}.

(ii) when x is a real number, all real number greater than – 2 are the soln. so, x ( –2, ∞  )

3. Given, 5x – 3 < 7.

5x – 3 + 3 < 7 + 3

5x< 10

$\frac{5x}{5}<\frac{10}{5}$

x< 2.

(i) when x ∈ z, i. e, x is an integer all integers less than 2 are the solⁿ so, x = {…., – 3, – 2, – 1, 0, 1}

(ii) when x ∈ R i, e x is a real number all t real number less than 2. are the solⁿ so, x∈ (– ∞, 2)