|x - 9| < 2 implies -2 < x - 9 < 2 Thus, 7 < x < 11

Given \(\frac { 3+\sqrt { 7 } }{ 3-\sqrt { 7 } } =a+b\sqrt { 7 } \)
Multiplying the numerator and denominator by the conjugate of the denominator we get
\(\frac { (3+\sqrt { 7 } )(3+\sqrt { 7 } ) }{ (3-\sqrt { 7 } )(3+\sqrt { 7 } ) } =a+b\sqrt { 7 } \)
⇒ \(\frac { 9+7+6\sqrt { 7 } }{ { 3 }^{ 2 }-(\sqrt { 7 } )^{ 2 } } =a+b\sqrt { 7 } \)
⇒ \(\frac { 16+6\sqrt { 7 } }{ 9-7 } =a+b\sqrt { 7 } \Rightarrow \frac { 2(8+3\sqrt { 9 } ) }{ 2 } =a+b\sqrt { 7 } \)
 \(8+3\sqrt { 7 } =a+b\sqrt { 7 } \)
Comparing the like co-efficients both sides we get a = 8 and b = 3

Given |4x - 5| > -2.
This means -2 < 4x - 5 < 2
⇒ -2 + 5 < 4x < 2 + 5
⇒ 3 < 4x < 7
\(⇒{3\over 4}\le x\le{7\over 4}\)
\(\therefore\) The Solution set is \(\left[ \frac { 3 }{ 4 } ,\frac { 7 }{ 4 } \right] \)

Let r be the radius of the spherical tank
 Then, Volume of the spherical tank = \(\frac { 32\pi }{ 3 } \)
⇒ \(\frac { 4 }{ 3 } { \pi r }^{ 3 }=\frac { 32\pi }{ 3 } \)
⇒ 4r³ = 32
⇒  r³ = \(\frac { 32 }{ 4 } \) = 8
⇒ r³ = 2³
⇒  r = 2
∴ Radius of the spherical tank is 2 unit

We know that |u| = |v| if and only if u = v or u = -v.
Therefore, |2x - 3| = |x - 5| implies 2x - 3 = x - 5 or 2x - 3 = 5 - x.
Solving these two equations we get x = -2 and x = \(\frac{8}{3}\). 
Hence both x = -2 and x = \(\frac{8}{3}\) are solutions.

y = x² + 6x + 9
Here a = 1, b = 6, c = 9
\(\therefore\) D = b² - 4ac = (6)² - 4 (1) (9)
= 36 - 36 = 0
Since D = 0, the parabola touches the X -axis at only one point.

Given \(log_2\left({{\sqrt [ 3 ]{4 } }\over{4^2\sqrt{8}}} \right)\)
= \({log}_{2}\sqrt [ 3 ]{4 }-{log}_{2}4^2(\sqrt{8})\)
= \(log_24^{1/3}-[log_24^2+log_2\sqrt{8}]\)
= log²(2²)^1/3- log²(2²)²- log²(2³)^1/2
= log²2^1/3- log²2⁴- log²2^3/2
\(={{2}\over{3}}(1)-4(1)-{{3}\over{2}}(1)\)   \([\because {log}^{2}_{2}=1]\)
\(={{4-24-9}\over{6}}={{-29}\over{6}}\)

Let the output two years ago be x units and annual increase = r%
\(\therefore\)  Output in the last year \(=x\left( {{100+r}\over{100}}\right) \) units.   \(\left[ I={{PNR}\over{100}} \right]\)
and output in the present year.
\(=x\left({{100+r}\over{100}} \right)^2\)
\(\therefore\) By the given conduction, \(x\left({{100+r}\over{1000}} \right)=2x\)
\(\Rightarrow\)    \(\left({{100+9r}\over{100}} \right)^2=2\)
\(\Rightarrow\)  (100 + r)² = 20,000
\(\Rightarrow\)  100² + r² + 200r - 20,000 = 0
\(\Rightarrow\) r²+ 200r - 10,000 = 0
\(\Rightarrow\)   \(r={{-200\pm\sqrt{40,000+40,000}}\over{2}}=r={{-200\pm\sqrt{80,000}}\over{2}}\)
\(\Rightarrow\)   \(r={{-200\pm200\sqrt{2}}\over{2}}\)
\(\Rightarrow\)   \(r={{2(-100\pm100\sqrt{2})}\over{2}}=-100\pm100\sqrt{2}\)
\(\Rightarrow\)   \(r=100(\sqrt{2}-1)\) or \(100(-1-\sqrt{2})\)
\(\Rightarrow\)   \(r=100(\sqrt{2}-1)\)      \([\because r > 0]\)
\(\Rightarrow\)   \(r=100(1.141-1)=41.42\%\)

Given \(\frac { 1 }{ \left| 2x-1 \right| } <6\)

Multiplying the numerator and denominator by |2x-1| we get, \({|2-1|\over |2x-1|^2}<6\)
⇒ |2x-1| < 6|2x-1|²
⇒ 1< 6 |2x-1|
\(⇒\ {1\over 6}<|2x-1|\)
\(⇒\ |2x-1|> {1\over 6}\)
\(⇒\ {-1\over 6}\ge2x-1\ge{1\over 6}\)
\(⇒\ {-1\over 6}+1\ge2x\ge{1\over 6}+1\)
\(⇒\ {5\over 6}\ge2x\ge{7\over6}\)
\(⇒\ {5\over12}\ge x\ge {7\over 12}\)
∴ The solution set is \(\left( -\infty,{5\over 12}\cup [ {7\over 12},\infty\right)\)

The means \(-{{1}\over{4}}-3\le{3\over4}x\le{1\over 4}.\)
\(⇒-{{1}\over{4}}-3≤-{3\over4}x≤{1\over4}-3\)
\(⇒-{13\over 4}≤-{3\over4}x≤-{11\over4}\)
Multiplying by 4 throughout we get,
\(-13≤-3x≤-11\)
\({-13\over -3}≥x≥{-11\over3}.\)
\(\therefore\) The Solution set is \(\left[ \frac { -11 }{ 3 } ,\frac { 13 }{ 3 } \right] \)

\((x+1)^{1\over 3} = (x-3)^{1\over 2}\)
L.C.M. of 2 and 3 is 6 & Raising to the power 6
\(\left\{ (x+1)^{ \frac { 1 }{ 3 } } \right\} ^{ 6 }=\left\{ (x-3)^{ \frac { 1 }{ 2 } } \right\} ^{ 6 }\)
(x+1)² = (x-3)³
x²+2x+1 =x³-9x²+27x-27
0 = x³-9x²+27x-27-x²-2x-1
x³-10x²+25x-28 =0
since constant term is - 28
we can have a factor as (x ± 2) or (x ± 4) or (x ± 7)
By trial and error method we find that (x - 7) is a factor
Using synthetic division

we get x³-10x²+25x-28=(x-7)(x²-3x+4)
solving x²-3x+4 =0
x=\(\frac { 3\pm \sqrt { 9-16 } }{ 2 } =\frac { 3\pm \sqrt { -7 } }{ 2 } \)
the roots are x =7 x=\(\frac { 3\pm \sqrt { -7 } }{ 2 } \)

LHS= \({{1}\over{3-\sqrt{8}}}-{{1}\over{\sqrt{8}-\sqrt{7}}}+{{1}\over{\sqrt{7}-\sqrt{6}}}-{{1}\over{\sqrt{6}-\sqrt{5}}}+{{1}\over{\sqrt{5}-2}}=5\)
Multiplying each term by the conjugate of the denominator we get
\(={3+\sqrt8\over (3-\sqrt8)(3+\sqrt8)}-{\sqrt8+\sqrt7\over(\sqrt8-\sqrt7)(\sqrt8+\sqrt7)}+{\sqrt7+\sqrt6\over( \sqrt7-\sqrt6)(\sqrt7+\sqrt6)}-{\sqrt6+\sqrt5\over(\sqrt6-\sqrt5)(\sqrt6+\sqrt5)}+{\sqrt5+2\over(\sqrt5-2)(\sqrt5+2)}\)
\(={3+\sqrt8\over3^2-(\sqrt8)^2}-{\sqrt8+\sqrt7\over (\sqrt8)^2-(\sqrt7)^2}+{\sqrt7+\sqrt6\over (\sqrt7)^2-( \sqrt6)^2}-{\sqrt6+\sqrt5\over(\sqrt6)^2-(\sqrt5)^2}+{\sqrt5+2\over (\sqrt5)^2-2^2}\)
\(={3+8\over 9-8}-{\sqrt8-\sqrt7\over 8-7}+{\sqrt7+\sqrt6\over 7-6}-{\sqrt6+\sqrt5\over 6-5}+{\sqrt5+2\over5-4}\)

Hence proved

Given log_5-x (x²- 6x + 65) = 2
(5 - x)² = x²- 6x + 65  [Converting into expotential form]
⇒ 25 + x²-10x = x²- 6x + 65
⇒ -6x + 65 - 25 + 10x = 0
⇒ 4x + 40 = 0
⇒ 4x = -40
⇒  \(\frac { -40 }{ 4 } \)  = -10
∴  x = -10

Given log₂ x-3log_1/6 = 6 [using quotient rule]
\(⇒\ {1\over log_x^2}-{3\over log_x^{1\over 2}}=6\)
\(⇒\ {1\over log_x^2}-{3\over log_x^1-log_x^2}=6\) [using quotient rule]
\(⇒\ {1\over log_x^2}-{3\over 0-log_x^2}=6\)
\(⇒\ {1\over log_x^2}+{3\over log_x^2}=6\)
\(⇒\ {1\over log_x^2}(1+3)=6\)
\(⇒\ {1\over log_x^2}(4)=6\)
\(⇒\ {1\over log_x^2}={6\over 4}={3\over 2}\)
\(⇒\ log_2^x={3\over2}\)
\(⇒\ 2^{3\over2}=x\)
⇒ (2³)^1/2 = x ⇒ (8)^1/2 = x
\(⇒\ x=\sqrt8=\sqrt{4\times2}\)
⇒  \(x=x\sqrt { 2 } \)

\(x^{2}+a x+c=0 \)
\(x^{2}+d x+b=0 \)
\(8 \& 2 \text { are the roots }\)\(\text { 3. } 3 \text { are the roots }\)
\(\therefore a=-10 ; c=16 \quad d=-6, \quad b=9\)
\(x^{2}+a x+b =0 \)
\(x^{2}-10 x+9 =0 \)
\(\Rightarrow(x-1)(x-9) =0 \)
\(\therefore x =1 \text { (or) } 9 \)

\(\log _{a} b \log _{b} c \log _{c} a=\log _{a} c \log _{c} a=\log _{a} a=1\)

\(\log _{\sqrt{x}} 0.25 =4 \)
\((\sqrt{x})^{4} =0.25 \)
\((\sqrt{x})^{4} =\frac{1}{4} \)
\(x^{2} =\frac{1}{4} \)
\(\Rightarrow x=\frac{1}{2}=0.5\)