Class 11th

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].

73. Given, f (x) = $\frac{x}{si{n}^{n}x}$

So, f?(x) = $\frac{\left(si{n}^{x}x\frac{dx}{dx}\right)-\left(x\frac{d}{dx}si{n}^{x}x\right)}{{\left(si{n}^{x}x\right)}^2}$

= $\frac{si{n}^{x}x-x·n{(sinx)}^{x-1}\frac{d}{dx}sinx}{{(sinx)}^{2x}}$

$=\frac{{(sinx)}^n-x·x{(sinx)}^{n-1}·cosx}{{(sinx)}^{2n}}$

$=\frac{{(sinx)}^{n-1}[sinx-nx·cosx]}{{(sinx)}^{2n}}$

$=\frac{sinx-n·x·cosx}{{(sinx)}^{2n-(n-1)}}$

$=\frac{sinx-nxcosx}{{(sinx)}^{n+1}}$

11. Given, $\frac{3\left(x-2\right)5}{5}\le \frac{5\left(2-x\right)}{3}$

=> 3 x 3 (x – 2) ≤ 5 x 5 (2 – x)

=> 9x – 18 ≤ 50 – 25x

=> 9x + 25x ≤ 50 + 18

=> 34x ≤ 68

=> $x=\frac{68}{34}$

=> x ≤ 2

So,  $x\in (-\infty ,2]$

72. Given, f (x) = (x + sec x) (x tan x)

So, f?(x) = (x + see x) $\frac{d}{dx}(x-tanx)+(x-tanx)\frac{d}{dx}(x+secx).$

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

Let g(x) = tan x.

So, g?(x) =

$=\underset{h\to 0}{lim}\frac{1}{h}[tan(x+h)-tanx]$

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

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

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

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

$=\frac{1}{co{s}^{2}x}=se{c}^{2}x$ . _____ (2)

And P(x) = see x.

$=\underset{h\to 0}{lim}\frac{sec(x+h)-secx}{h}$

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

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

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{-2sin\left(\frac{x+x+h}{2}\right)sin\left(\frac{x-(x+h)}{2}\right)}{cos(x+h)cosx}\right]$

$=\frac{sinx}{co{s}^{2}x}$ = tan x sec x. ____ (3)

   Putting (2) and (3) in (1) we get,

   f?(x) = (x + sec x) (1 - sec²x) + (x- tan x) (1 + tan x sec x)

71. Given, f (x) = $\frac{x}{1+tanx}$

$=\frac{x}{1+\frac{sinx}{cosx}}$

$=\frac{cosx·x}{cosx+sinx}$

$So,f^{\prime }(x)=\frac{(cosx+sinx)\frac{d}{dx}(x\cdot cosx)-\left(x\cdot cosx\right)\frac{d}{dx}\left(cosx+sinx\right)}{{(cosx+sinx)}^2}$

$=\frac{(cosx+sinx)(-xsinx+cosx)-\left(-x·sinxcosx+xco{s}^{2}x\right)}{{(cosx+sinx)}^2}$

$=\frac{-xcosxsinx+co{s}^{2}x-xsi{n}^{2}x+sinxcosx+xsinxcosx-xco{s}^{2}x}{{(cosx+sinx)}^2}$

$=\frac{co{s}^{2}x+sinxcosx-x\left(si{n}^{2}x+co{s}^{2}x\right)}{{(cosx+csinx)}^2}$

Dividing numerator and denominator by cos2x we get,

$=\frac{\frac{co{s}^{2}x}{co{s}^{2}x}+\frac{sinx}{cosx}·\frac{cosx}{cosx}-x\frac{1}{co{s}^{2}x}}{{\left(\frac{cosx}{cosx}+\frac{sinx}{cosx}\right)}^2}$

$=\frac{1+tanx-xse{c}^{2}x}{{(1+tanx)}^2}$

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]