(a) 6x² - 9x + 2 = 0
(b) x² + 2x + 9 = 0
(c) 9x² + 2x + 1 = 0

Given roots are 7 and -3
Sum of the roots = 7+(-3) = 4
Product of the roots = 7(-3) = -21
The quadratic equation is x²- x (Sum of the roots) + Product of the roots = 0
Hence, the required quadratic equation is x²- 4x - 21 = 0

 Given y = b^x
Converting this into logarithmic form, we get

\({ log }_{ b }^{ y }\)   = x

The domain logarithmic function is the set of positive real numbers and the range is the set of real numbers.
Domain (0,\(\infty\)), and Range (0,\(\infty\))

x > - 1 and x < 4.
⇒ x ∈ (-1, 4)

[a < b ⇒ ay > by for all y< 0]
Here y = -1
Given |3-x| < 7
This means -7 < 3 -x < 7
⇒-7-3 < -x < 7-3
⇒-10 < -x < 4
⇒10 > x > -4
\(\therefore\) The Solution set is \(x\in \left( -\infty ,-4 \right) \cup \left( -4,10 \right) \)

Since \(\sqrt { 7 } \) is an irrational number,  \(\sqrt { 7 } \) \(\in\) R.
Since \(\frac { -1 }{ 4 } \) is a negative rational number \(\frac { -1 }{ 4 } \) \(\in\) Q
0 is an integer and 0 \(\in\)Z.
3.14 = \(\pi \) is a non-recurring and non-terminating decimal.
\(\therefore\) 3.14 is an irrational number \(\Rightarrow\) 3.14 \(\in\) R-Q
4 is a positive integer \(\Rightarrow\) 4 \(\in\)R-Q.
\(\frac { 22 }{ 7 } \) = 3.14 \(\in\) R. Which is an irrational number.

= \({ 3 }^{ -6\times \frac { 1 }{ 3 } }={ 3 }^{ 2 }=\frac { 1 }{ { 3 }^{ 2 } } =\frac { 1 }{ 9 } \)

\(\left( 5^{ 3 } \right) ^{ \frac { 2 }{ 3 } }={ 5 }^{ 3\times \frac { 2 }{ 3 } }\) 
= 5² = 25  \(\left[ \because ({ a }^{ m })^{ n }={ a }^{ mn } \right] \)

Note that the degree of f(x) is 3 and the leading coefficient is 1. Since g(x) divides f(x), we have f(x) = (x + b) g(x), for some b ∈ R. Thus, x³ - 3px + 2q = (x + b) (x² + 2ax + a² ).
Equating like coefficients on both sides, we have 2a + b = 0, a² + 2ab = -3p and 2q = ba², Thus b = -2a, p = a² and q = -a³
Now, q = -a³ = -a(a²) = -ap, which gives ap + q = 0.

Let x denote the number of units used. Note that x ≥ 0. Then, his electricity bill will be Rs. 110 + 4x.
The person wants his bill to be below Rs. 250. Let us solve the inequality 110 + 4x < 250. Thus, 4x < 140; which gives 0 ≤ x < 35. 
The person should keep his usage below 35 units in order to keep his bill below Rs. 250.

3|x - 2| + 7 = 19, So that we have, |x - 2| = \(\frac { 19-7 }{ 3 } =4\). Thus, we have either, x - 2 = 4 (or) x - 2 = -4
Therefore the solutions are x = -2 (or) x = 6

\({{1}\over{x^2-a^2}}={{1}\over{(x+a)(x-a)}}-={{A}\over{x+a}}+{{B}\over{x+a}}\)
\({{1}\over{x^2-a^2}}={{A(x-a)B(x+a)}\over{(x+a)(x-a)}}\)
\(\Rightarrow\) 1 = A(x - a) + B(x + a)
Putting x = a in (1) we get,
\(1=B(2a)\Rightarrow A=-{{1}\over{2a}}\)
Putting x = -a in (1) we get,
\(1=A(-2a)\Rightarrow A=-{{1}\over{2a}}\)
\(\therefore{{1}\over{x^2-a^2}}={{-{{1}\over{2}}a}\over{x+a}}+{{{{1}\over{2}}a}\over{x-a}}={{-1}\over{2a(x+a)}}+{{1}\over{2a(x-a)}}\)

Let f(x) = 3x³ + 8x² + 8x + a
Since (x² + x + 1) is a factor of f(x), f(x) is divisible by x² + x + 1

Since f(x) is divisible by x² + x + 1, the remainder is zero.
∴ a - 5 = 0
⇒ a = 5

Given \(\frac { { 3 }^{ 2n }{ 9 }^{ 2 }{ 3 }^{ -n } }{ { 3 }^{ 3n } } \) = 27
⇒  \(\frac { { 3 }^{ 2n-n }.{ 9 }^{ 2 } }{ { 3 }^{ 3n } }\)   = 27  [∵ a^m.aⁿ = a^m+n]
⇒  \(\frac { { 3 }^{ n }.({ 3 }^{ 2 })^{ 2 } }{ { 3 }^{ 3n } } \) = 27
⇒  3^n-3n (3⁴) = 27 \(\left[ \because \frac { a^{ m } }{ a^{ n } } ={ a }^{ m-n }\& ({ a }^{ m })^{ n }={ a }^{ mn } \right] \)
⇒  3^-2n.3⁴ = 27
⇒  3^-2n+4 = 3³
Equating the powers both sides we get
 -2n+4 = 3
⇒ -2n = 3-4 = -1
⇒ 2n = 1
⇒ n = \(\frac { 1 }{ 2 } \)

2x + y = 8

x + 2y + 8

x + y = 6

 
All points bounded between 2x + 3y = 8, x + 2y 8 and x + y = 6 is required region. Darkly shaded area will represents the solution set of the given linear inequalities.

If 2x + 3y = 6

x + 4y = 4

x > y > 0 represents the area in the 1 quadrant.
 
All points bounded between x = 0, y = 0, x + 4y = 4 and 2x + 3y = 6 is required region. Darkly shaded area will represents the solution set of the given linear inequalities.

\({{{(x-1)}^{2}}\over{x^3+x}}={{{(x-1)}^{2}}\over{x(x^2+1)}}={{A}\over{x}}+{{Bx+C}\over{x^2+1}}\)
\(\Rightarrow\)  (x-1)² = A(x²+ 1) + (Bx + C)(x)
Putting x = 0 in (1) we get,
\(\boxed{1=A}\)
Putting x = 1 in (1) we get. 
0 = A(2) + (B + C)(1)
\(\Rightarrow\)   0 = 2A + B + C
Equating the Coefficient of x² in (1) we get,
1 = A + B \(\Rightarrow\) 1 = 1  B \(\Rightarrow\) \(\boxed{B=0}\)
Substituting A = 1, B = 0 in (2) we get,
0 = 2 + 0 + C \(\Rightarrow\)  \(\boxed{C=-2}\)
\(\therefore\)  \({{{(x-1)}^{2}}\over{x^3+x}}={{1}\over{x}}+{{0x-2}\over{x^2+1}}\)
\(\Rightarrow\)  \({{{(x-1)}^{2}}\over{x^3+x}}={{1}\over{x}}-{{2}\over{x^3+1}}\)

Suppose y = 2x

-2x + 3y = 6
-2x = 6 - 3y
\(x={6-3y\over -2}\)

All the points above the y ≥ 2x and all the points below -2x + 3y ≤ 6 is required region. Darkly shaded area will represents the solution set of the given linear inequalities.

Given equation is 2x²-(a+1)x + a - 1 = 0
Let α and β be the roots of the equation
Given α - β = αβ
\(α - β={a-1\over 2}[a=2, b=-(a+1),c=a-1]\ \ \ [∵\ αβ={c\over a}={a-1\over 2}]\)....(1)
and \(α + β={-b\over a}={a+1\over 2}\)  ....(2)

Substituting \(α={a\over 2}\)in (2) we get
\({a\over 2}+\beta={a+1\over 2}\)
\(⇒\ β={a+1\over 2}-{a\over2}={1\over 2}\)
\(∴\ β={a\over 2}\)
Substituting the value of α and β in α - β = αβ
we get \({a\over 2}-{1\over 2}=\left(a\over 2\right)\left(1\over 2\right)\)
\(⇒\ {a-1\over 2}={a\over 4}\)
⇒ a-1 = \(a\over 2\)
⇒ 2a - 2 = a
⇒ 2a - a = 2
⇒ a = 2
Hence proved

The value of
\(\log _{3} 11 \log _{11} 13 \log _{13} 15 \log _{15} 27 \log _{27} 81=\log _{3} 81 \)
\(=\log _{3} 3^{4}=4 \log _{3} 3=4 \)

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

\(3 x^{2}-5 x-7=0\)
\(\text { Let the roots be } \alpha, \beta\)
\(\Rightarrow \operatorname{Sum}: \alpha+\beta=\frac{5}{3}\)
\(\text { Product : } \alpha \beta=\frac{-7}{3}\)
\(\text { Now take the roots are }-\alpha,-\beta\)
\(\Rightarrow \text { Sum : }-\alpha-\beta \quad-(\alpha+\beta)=-\frac{5}{3}\)
\(\text { Product : }(-\alpha)(-\beta)=\alpha \beta=\frac{-7}{3}\)
\(\text { The reouired equation }\)
\(x^{2}+\frac{5}{3} x-\frac{7}{3}=0 \)
\(3 x^{2}+5 x-7=0 \)

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

\(\text { Let } \log _{3} \frac{1}{81}=x \Rightarrow 3^{x}=\frac{1}{81}\)
\(\Rightarrow 3^{x}=3^{-4}\)
\(\Rightarrow x=-4 \)

\(\text { Let } \log _{\sqrt{2}} 512=x\)
\(\text { Then }(\sqrt{2})^{x}=2^{9}\)
\(\Rightarrow 2^{\frac{x}{2}}=2^{9} \Rightarrow x / 2=9 \Rightarrow x=18\)

\(5 x-1 <24 \ \ \ 5 x+1 >-24 \)
\(5 x <25 \ \ 5 x >-25\)
\(x <5 \ x >-5 \)
\(x \in(-5,5)\)