Now, \(\frac{x}{2 x+1} \geq \frac{1}{4} \Rightarrow \frac{x}{2 x+1}-\frac{1}{4} \geq 0\)
\(\Rightarrow \frac{2 x-1}{4(2 x+1)} \geq 0, \text { here } 2 x+1 \neq 0, \text { i.e. } x \neq-\frac{1}{2} \)
\(\Rightarrow (2 x-1)(2 x+1) \geq 0\)
Case I When both are non-negative
\(2 x-1 \geq 0 \text { and } 2 x+1>0 \)
\(\Rightarrow x \geq \frac{1}{2} \text { and } x>-\frac{1}{2} \)
\(\therefore x \in\left[\frac{1}{2} \infty\right)\) ..(i)
Case II When both are non-positive
\((2 x-1) \leq 0 \text { and } 2 x+1<0 \)
\(\Rightarrow x \leq \frac{1}{2} \text { and } x<\frac{-1}{2} \)
\(\therefore x \in\left(-\infty,-\frac{1}{2}\right)\) .(ii)
From Eqs. (i) and (ii), we get
\(x \in\left(-\infty,-\frac{1}{2}\right) \cup\left[\frac{1}{2}, \infty\right) \) .(iii)
now,\(\frac{6 x}{4 x-1}<\frac{1}{2} \Rightarrow \frac{6 x}{4 x-1}-\frac{1}{2}<0\)
\(\Rightarrow \frac{8 x+1}{2(4 x-1)}<0 \)
\(\Rightarrow (8 x+1)(4 x-1)<0\)
We have, \((8 x+1)<0 \text { and }(4 x-1)>0 \)
or \((8 x+1)>0 \text { and }(4 x-1)<0\)
\(\Rightarrow \frac{-1}{8}<x<\frac{1}{4}\)  ..(iv)
Now, for the solution set of given system of inequalities take common value of x from Eqs. (ill) and (iv), but there is no common value of x.
Ans. Solution set is a empty set.

Given, A = {4, 5, 7, 8, 10}, B = {4, 5, 9} and C = {1, 4, 6, 9}
(i) Now, A \(\cap\) B = {4, 5, 7, 8, 10} \(\cap\) {4, 5, 9} = {4, 5}
\(\therefore\) LHS = (A \(\cap\) B) \(\cap\) C
= {4, 5} \(\cap\) {I, 4, 6, 9} = {4} ... (i)
Now, B \(\cap\) C = {4, 5, 9} \(\cap\) {I, 4, 6, 9} = {4, 9}
\(\therefore\) RHS = A \(\cap\) (B \(\cap\) C)
= {4, 5, 7, 8, 10} \(\cap\) {4, 9}= {4} ...(ii)
From Eqs. (i) and (ii), we get
LHS = RHS = {4}
Hence, (A \(\cap\) B) \(\cap\) C = A \(\cap\) (B \(\cap\) C)
(ii) Here, B \(\cap\) C = {4, 9}
\(\therefore\) LHS = A \(\cup\) (B \(\cap\) C)
= {4, 5, 7, 8, 10} \(\cup\) {4, 9}
\(\Rightarrow\) LHS = {4, 5, 7, 8, 9,10} ...(iii)
Now, A \(\cup\) B = {4, 5, 7, 8, 10} \(\cup\) {4, 5, 9}
= {4, 5, 7, 8, 9,10}
and A \(\cup\) C = {4, 5, 7, 8, 10} \(\cup\) {1, 4, 6, 9}
= {1,4, 5, 6, 7, 8, 9,10}
\(\therefore\) RHS = (A \(\cup\) B) \(\cap\) (A \(\cup\) C)
= {4, 5, 7, 8, 9,10} \(\cap\) {1, 4, 5, 6, 7, 8, 9,10}
= {4, 5, 7, 8, 9, 10} ...(iv)
From Eqs. (iii) and (iv), we get
LHS = RHS = {4, 5, 7, 8, 9,10}
Hence, A \(\cup\) (B \(\cap\) C) = (A \(\cup\) B) \(\cap\) (A \(\cup\) C)
(iii) Now, B \(\cup\) C = {4, 5, 9} \(\cup\) {I, 4, 6, 9}= {1, 4, 5, 6, 9}
\(\therefore\) LHS = A \(\cap\) (B \(\cup\) C)
= {4, 5, 7, 8, 10} \(\cap\) {1, 4, 5, 6, 9} = {4, 5} ... (v)
Now, A \(\cap\) B = {4, 5, 7, 8, 10} \(\cap\) {4, 5, 9}= {4, 5}
and A \(\cap\) C = {4, 5, 7, 8, 10} \(\cap\) {1, 4, 6, 9}= {4}
\(\therefore\) RHS = (A \(\cap\) B) \(\cup\) (A \(\cap\) C) = {4, 5} \(\cup\) {4}
= {4, 5} .....(vi)
From Eqs. (v) and (vi), we get
LHS = RHS = {4, 5}
Hence, A \(\cap\) (B \(\cup\) C) = (A \(\cap\) B) \(\cup\) (A \(\cap\) C)

cos 6\(\theta\) + 6 cos 4\(\theta\) + 15 cos 2\(\theta\) + 10
= (cos 6\(\theta\) + cos 4\(\theta\)) + (5 cos 4\(\theta\) + 5 cos 2\(\theta\)) + (10 cos 2\(\theta\) + 10)
= (cos 6\(\theta\) + cos 4\(\theta\)) + 5 (cos 4\(\theta\) + cos 2\(\theta\)) + 10 (cos 2\(\theta\)+ 1)
= 2cos 5\(\theta\) cos \(\theta\) + 5 x 2 cos 3\(\theta\) cos \(\theta\)+ 10 x 2 cos \(\theta\) cos \(\theta\)
= 2cos \(\theta\) [cos5\(\theta\) + 5 cos 3\(\theta\) + 10 cos \(\theta\)]
\(\therefore {cos \theta + 6 cos 4 \theta + 15 cos 2 \theta + 10 \over cos 5 \theta + 5 cos 3 \theta + 10 cos \theta}=2cos \theta\)

\(\frac { 1 }{ { z }_{ 1 }\bar { { z }_{ 1 } } } =\frac { 1 }{ (2-i)(2+i) } =\frac { 1 }{ 4-{ i }^{ 2 } } =\frac { 1 }{ 5 } \)
On comparing imaginary parts , we obtain
∴ \(Im\left( \frac { 1 }{ { z }_{ 1 }\bar { { z }_{ 1 } } } \right) =0\)

\(cos \theta ={-1\over5},tan \theta=2\sqrt{6},cot\theta ={1\over 2\sqrt{6}},sec\theta=-5,cosec\theta={-5\over2\sqrt{6}}\)

Domain of R = {3, 5, 7, 9}
Range of R = {2, 4, 6, 8}

\(z+\sqrt { 2 } \left| (z+1) \right| +i=0\)
Let \( z=x+iy\)
Then, \( (x+iy)+\sqrt { 2 } \left| (x+iy+1 \right| +i\ =0\)
\( \Rightarrow x+i(y+1)+\sqrt { 2 } \left| (x+1)+iy \right| =0\)
\(\Rightarrow x+i(y+1)+\sqrt { 2 } \sqrt { { (x+1) }^{ 2 }+{ y }^{ 2 } } =0\)
\([if\quad z=a+ib,\quad then\quad \left| z \right| =\sqrt { { a }^{ 2 }+{ b }^{ 2 } } ]\)
\(\Rightarrow x+\sqrt { 2 } \sqrt { { x }^{ 2 }+1+2x+{ y }^{ 2 } } +i(y+1)=0+0i\)
On equating real and imaginary part, we get
\( x+\sqrt { 2 } \sqrt { { x }^{ 2 }+1+2x+{ y }^{ 2 } } =0\quad .......(i)\)
\(and\quad y+1=0\quad ......(ii)\)
From Eq. (ii), we get 
y = -1
Now, on substituting y = - 1 in Eq. (i), we get
\( x+\sqrt { 2 } \sqrt { { x }^{ 2 }+1+2x+{ +1 }^{ } } =0\)
\(\Rightarrow x=-\sqrt { 2 } \sqrt { { x }^{ 2 }+2x+2 } \)
On squaring both sides, we get
\({ x }^{ 2 }=2({ x }^{ 2 }+2x+2)\)
\(\Rightarrow { x }^{ 2 }=2{ x }^{ 2 }+4x+4\Rightarrow { x }^{ 2 }+4x+4=0\)
\(\Rightarrow { (x+1) }^{ 2 }=0\Rightarrow x+2=0\Rightarrow x=-2\)
\(Hence,\ z=x+iy=-2-i\)

Domain = R - {-3,2}

Let the length of shortest side be x cm
Then, according to question, we have 
Length of third side = (x+2) cm
Since the perimeter of the triangle is more than 166 cm.
2x+x+(x+2)>166
\(\Rightarrow 4x+2>166\)
\( \Rightarrow 4x>164\quad [subtracting\quad 2\quad from\quad both\quad sides]\)
\(\Rightarrow x>41\quad [dividing\quad both\quad sides\quad by\quad 4]\)
Hence,the length of the shortest side should be greater than 41cm.

Let z =a+ib, such that \(\left| z \right| =\sqrt { { a }^{ 2 }+{ b }^{ 2 } } =1\)
Now, consider \( \left( \frac { z-1 }{ z+1 } \right) =\left( \frac { a+ib-1 }{ a+ib+1 } \right) \)
\(=\frac { (a+ib+1) }{ (a+ib+1) } \times \frac { (a+ib-1) }{ (a+ib-1) } \)
[by rationalising the denminator]
\(=\frac { [(a-1)+ib][(a+1)-ib] }{ { (a+1) }^{ 2 }+\quad { (ib) }^{ 2 } } \quad \quad [\because ({ z }_{ 1 }+{ z }_{ 2 })({ z }_{ 1 }-{ z }_{ 2 })={ z }_{ 1 }^{ 2 }-{ z }_{ 2 }^{ 2 }]\)
\(=\frac { { a }^{ 2 }-1-iab+ib+iab+ib-{ i }^{ 2 }{ b }^{ 2 } }{ { (a+1) }^{ 2 }-{ i }^{ 2 }{ b }^{ 2 } } \)
\(=\frac { ({ a }^{ 2 }+{ b }^{ 2 }-1)+2bi }{ { (a+1) }^{ 2 }+{ b }^{ 2 } } \quad [\because { i }^{ 2 }=-1]\)
\(=\frac { (1-1)+2bi }{ { a }^{ 2 }+1+2a+{ b }^{ 2 } } \)
\([\because { ({ z }_{ 1 }+{ z }_{ 2 }) }^{ 2 }={ z }_{ 1 }^{ 2 }-{ z }_{ 2 }^{ 2 }+2{ z }_{ 1 }{ z }_{ 2 }]\)
\(=\frac { 0+2bi }{ { (a }^{ 2 }+{ b }^{ 2 })+1+2a } =\frac { 2bi }{ 2+2a } \ [\because { a }^{ 2 }+{ b }^{ 2 }=1]\)
\( =0+\frac { bi }{ 1+a } \)
Clearly, real part of \(\left(\frac{z-1}{z+1}\right)\) is zero and imaginary part of  \(\left(\frac{z-1}{z+1}\right)\)is \(\frac{b}{1+a}\)
\(\left(\frac{z-1}{z+1}\right)=\frac{i b}{1+a}\) is purely imaginary.
Again, when Z = 1,then \(\left(\frac{z-1}{z+1}\right)=\frac{1-1}{1+1}=0\)which is purely real.

(i) (b)
(ii) (d)
(iii) (a)
(iv) (a)
(v) (b)