Here \(3{ x }^{ 2 }-4x+\frac { 20 }{ 3 } =0\)
Comparing the given quadratic equation with ax² + bx + C = 0, we have
a=3,b=-4 and c=\(\frac { 20 }{ 3 } \)
∴  \(x=\frac { -(-4)\pm \sqrt { (-4)^{ 2 }-4\times 3\times \frac { 20 }{ 3 } } }{ 2\times 3 } \)
= \(\frac { 4\pm \sqrt { 16-80 } }{ 6 } =\frac { 4\pm \sqrt { -64 } }{ 6 } \)
= \(\frac { 4\pm 8\sqrt { -1 } }{ 6 } =\frac { 4\pm 8i }{ 6 } =\frac { 2\pm 4i }{ 3 } \)
Thus x = \(\frac { 2+4i }{ 3 } \)and x=\(\frac { 2-4i }{ 3 } \)

\(\frac { { x }^{ 2 } }{ 4 } -\frac { { y }^{ 2 } }{ 12 } \)=1

Number of revolutions in one minute = 360 revolutions
Number of revolutions in 60 seconds = 360 revolutions
Number of revolutions in 1 second =\(360\over 60\)= 6 revoIutions
Angle made by wheel in 6 revolutions
= 360 x 6 = 2160°
Now 2160° =\((2160\times {\pi\over 180})^C=(12\pi)^C\)

Use P(A∪B) = P(A) + P(B),for mutually exclusive events.
Ans.0.3

Given,\(f(x)={ x }^{ 2 }-6x+8\)    
On differentiating both sides w.r.t x we get
\(f^{ ' }(x)=2x-6\)  ..(i)
Now,  \(f^{ ' }(5)-3{ f }^{ ' }(2)=2\times 5-6-3(2\times 2-6)\)  [From Eq.(i)]
\(=10-6-12+18=10\)   ...(ii)
and\({ f }^{ ' }(8)=2\times 8-6\)    [From Eq.(ii)]
=10 ..(iii)
From Eqs.(ii) and (iii),we get
\({ f }^{ ' }(5)-3{ f }^{ ' }(2)={ f }^{ ' }(8)\)

On putting x =\( \frac{1} {2}\), we get the form\( \frac{0}{0}\)
So,let us first factorise it
Consider, \(\lim _{x \rightarrow \frac{1}{2}} \frac{4 x^{2}-1}{2 x-1}=\lim _{x \rightarrow \frac{1}{2}} \frac{(2 x+1)(2 x-1)}{(2 x-1)}\) [using factorisation method]
\(=\lim _{x \rightarrow \frac{1}{2}}(2 x+1) \)
\(=2\left(\frac{1}{2}\right)+1=2\)

Transverse axis = 2a = 10 \(\Rightarrow \) a = 5
and conjugate axis = 2b = 8 \(\Rightarrow \) b = 4
Equation of hyperbola is
\(\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\quad i.e\quad \frac { { x }^{ 2 } }{ 25 } -\frac { { y }^{ 2 } }{ 16 } \) = 1.

Given, first AP is 2, 5, 8,.... upto 50 terms, we have a₁ = 2, d₁ = 5 - 2 = 3 and second AP is 3, 5, 7, 9.... upto 60 terms,.
Here a₂ = 3, d₂ = 5 - 3 = 2
Let the mth term of the first AP be equal to the nth term of the second AP.
Then, 2 + (m - 1)\(\times \) 3 = 3 + (n - 1) \(\times \) 2
\(\Longrightarrow \)   3m - 1 = 2n + 1
\(\Longrightarrow \)   3m = 2n + 2
\(\Longrightarrow \)   \(\frac { m }{ 2 } \) = \(\frac { n+1 }{ 3 } \) = k (say)
\(\Longrightarrow \)   m = 2k and n = 3k - 1
\(\Longrightarrow \)   2k\(\le \) 50 and 3k - 1 \(\le \) 60   [ \(\because \) m\(\le \) 50 and n\(\le \) 60]
\(\Longrightarrow \)   k\(\le \) 25  and k \(\le \) 20 \(\frac { 1 }{ 3 } \) 
\(\Longrightarrow \)   k\(\le \) 20    [\(\because \) k is a natural number]
k=1, 2, 3, ...., 20
Corresponding to each value of k, we get a pair of identical terms. Hence, there are 20 identical terms in the two AP's.

We have, \(\frac { 8! }{ 4! } \) =\(\frac { 8\times 7\times 6\times 5\times 4! }{ 4! } \) \([\because n!=n(n-1)(n-2)....3.2.1]\)
\(=8\times 7\times 6\times 5=1680\)
Also, \(2!=2\times 1=2\neq 1680\)
\(\therefore \)  \(\frac { 8! }{ 4! } \neq 2!\) 

D = {-9, -8, -7,........}, so D is an infinite set.

We have
\(L.H.S =tan \ 4x={2tan\ 2x\over 1-tan^2 \ 2x }\)
\([\because tan 2A={2tan A\over 1-tan^2A} ]\)
\(={2.{2tan \ x\over 1-tan ^2x}\over 1-({2tan \ x\over 1-tan^2 \ x})^2}\)
\(={{4 tan x\over 1-tan^2 x}\over {(1-tan^2 \ x)^2-4tan^2 \ x \over (1-tan^2 \ x)^2}}\)
\(={4tan \ x \over 1-tan^2 x} \times {(1-tan^2 x)^2\over 1+tan^4x-2tan ^2-4tan^2 x}={4tan \ x(1-tan^2 \ x)\over 1-6tan^2 \ x+tan ^4 \ x}\)
=R.H.S

(i) Here f(x) = 2 - 3x
Since x \(\in\) R and x > 0
\(\therefore\) 3x> 0 \(\Rightarrow\)-3x<0 \(\Rightarrow\)2-3x<2
\(\therefore\) Range of function
= {a \(\in\) R: a < 2}
=(-\(\infty\),2)
(ii) Here (x) = x² + 2
Since x \(\in\) R
\(\therefore\) x² \(\ge\) 0 for all x \(\in\) R
\(\Rightarrow\) x²+2\(\ge\)2
\(\therefore\)Range of function
= {a\(\in\) R, a \(\ge\)2 \(\forall\) a \(\in\)R}
= [2,\(\infty\))
(iii) Here f(x) = x
Since X\(\in\) R
\(\therefore\) Range of function = R.

0.98

Let us make the following table from the given data.

Hence,  \(N=\Sigma { f }_{ i }=7\) and \(\Sigma { f }_{ i }{ x }_{ i }=22A\)
\(\therefore \bar { x } =\frac { \sum _{ i=1 }^{ 6 }{ { f }_{ i }{ x }_{ i } } }{ N } =\frac { 22A }{ 7 } \)
Now, variance \(=\frac { 1 }{ N } \sum _{ i=1 }^{ 6 }{ { f }_{ i }({ x }_{ 1 }-\bar { x } )^{ 2 } } \)
\(\Rightarrow \ 160=\frac { 1 }{ 7 } \left( \frac { 1120 }{ 49 } \right) { A }^{ 2 }\)
[ \(\because \) variance = 60, given]
Ans. A=7 

\(PQ=QR\quad and\quad { PQ }^{ 2 }+{ QR }^{ 2 }={ PR }^{ 2 }\)

Slope of the required perpendicular line=1/3
\(\therefore \) Equation of the required perpendicular line is
y-2=(1/3)(x-2)
For y-intercept put x=0 ,y = 4/3

(1.025)^-1/3 = (1+0.025)^-1/3
\(=1-\left( \frac { 1 }{ 3 } \right) \left( 0.025 \right) +\frac { \left( \frac { 1 }{ 3 } \right) \left( \frac { 1 }{ 3 } +1 \right) }{ 2 } \left( 0.025 \right) ^{ 2 }\)
= 1 - 0.0083 + 0.000139 - ...
= 0.991

(2,4) \(\cup \) (4, 6)

\(\theta =2n\pi +\frac { 5\pi }{ 12 } or\quad 2n\pi -\frac { \pi }{ 12 } ,n\in Z\)

The sets {a, e, i, o, u} and {c, d, e, f } have one element 'e' common.

x = 2 and x² \(\neq \) 4

Let P(n) = (xⁿ-yⁿ) is divisible by (x - y) for all n ∊ N.
For n =1
P(1) = (x¹-y¹) is divisible by (x-y)
∴ P(1) is true
Let P(n) be true for n = k
∴ P(k)=(x^k-y^k) is divisible by (x-y)
⇒ (x^k-y^k) =m(x-y) for some m ∊ Z     ..(i)
For n = k+1
∴  P(k+1)=x^k+1-y^k+1 is divisible by (x-y)
x^k+1-y^k+1 = x^k+1-x^ky+x^ky-y^k+1
= x^k(x-y)+y(x^k-y^k)
= x^k(x -y)+y.m(x-y)
= (x-y)[x^k+my]
which is divisible by (x-y)
∴  P(k+1) is true
Thus P(k) is true ⇒ P(k + 1) is true
Hence by principle of mathematical induction, pen) is true for all n∊N.

\(\text { Let } f(x)=x+a \text { . Accordingly, } f(x+h)=x+h+a\)
\(\text { By first principle, }\)
\(f^{\prime}(x) =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \)
\(=\lim _{h \rightarrow 0} \frac{x+h+a-x-a}{h} \)
\(=\lim _{h \rightarrow 0}\left(\frac{h}{h}\right) \)
\(=\lim _{h \rightarrow 0}(1) \)
\(=1\)

The given series is
\(\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+....\)
Let 'a_n' be nth term of the given series and 'S_n' be the sum of n^th terms of the given series.
\(\therefore a_n=\frac{1}{(n^{th}\quad term\quad of\quad 1,2,3...)(n^{th}\quad term\quad of 2,3,4...)}\)
= \(\frac{1}{[1+(n+1)\times 1][2+(n-1)\times 1]}\)
= \(\frac{1}{n(n+1)}=\frac{1}{n}+(\frac{-1}{n+1})\) [By partial fraction]
Putting n = 1, 2, 3, ....n, we have
\(a_1=\frac{1}{1}-\frac{1}{2};\)
\(a_2=\frac{1}{2}-\frac{1}{3};\)
\(a_3=\frac{1}{3}-\frac{1}{4}.....\)
\(a_n=\frac{1}{n}-\frac{1}{n+1}....\)
Adding vertically, we have:
S_n = a₁ + a₂ + a₃ + .... + a_n
= \(\frac{1}{1}-\frac{1}{n+1}=\frac{n+1-1}{n+1}=\frac{n}{n+1}\)
\(\therefore\)  S_n = \(\frac{n}{n+1}\)

(i) Let (x', y') the coordinates of the given point (x, y0 in the new system of translation of axes.
\(\therefore\) x'=x-h \(\Rightarrow\) x=x'+h=x'+1 and y=y'+k=y'+1
Substituting these values of x and y in the given equation x²+xy-3y²-y+2 = 0, we get
(x'+1)²+(x'+1)(y'+1)-3(y'+1)²-(y'+1)+2=0
\(\Rightarrow\) x'²+1+2x'+x'y'+x'+y'+1-3y'²-3-6y'-y'-1+2=0
\(\Rightarrow\)  x'²+x'y'-3y'²+3x'-6y' = 0
Hence the equation of the given pair of straight lines in a new system is
x²+xy-3y²+3x-6y=0
(ii) The orgin is shifted to a point (1, 1) and Let (x',y') be the new coordinates of the given point (x, y) in the line.
\(\therefore\) x = x' + h
\(\Rightarrow\) x = x' + 1 and y = y' + 1
Substituting these values of x and y in the given equation xy - y² - x+y=0 we get,
(x'+1) (y'+1)-(y'+1)²-(x'+1)+(y'+1) = 0
\(\Rightarrow\)  x'y' +x' +y' +1-y'²-1-2y'-x'-1+y'+1 = 0
\(\Rightarrow\)  x'y'-y'² = 0
Hence the equation of the given pair of straight lines in a new system is xy-y² = 0
(iii) Here h = 1 and k = 1
Let (x', y' ) be the coordinates of the new point
\(\therefore\)  x = x'+h, y=y' +k
\(\Rightarrow\)  x = x'+1, \(\Rightarrow\) y = y' +1
Now substituting the values of x and y is given equation xy-x-y+1 = 0, we get
(x' +1) (y'+1) - (x'+1)-(y'+1)+1=0
\(\Rightarrow\)  x'y' +x' +y' +1-x'-1-y'-1+1 =0
\(\Rightarrow\) x' y' = 0
Hence the equation of the given pair of straight lines in a new system of translation of axes is xy = 0.

Mean \(\left( \bar { x } \right) =A+\frac { \sum { fu } }{ N } \times h=27.5+\frac { 8 }{ 100 } \times 5=27.5+0.4=27.9\)
Standard deviation \(\left( \sigma \right) \) \(=\frac { h }{ N } \sqrt { N\sum { { fu }^{ 2 }-{ (\sum { fu) } }^{ 2 } } } \)
\(\sigma =\frac { 5 }{ 100 } \sqrt { 100\times 150-{ (8) }^{ 2 } } =\frac { 1 }{ 20 } \sqrt { 15000-64 } =\frac { 1 }{ 24 } \times 122.21=6.11\)
\(\therefore \ C.V.=\frac { \sigma }{ x } \times 100=\frac { 6.11 }{ 27.9 } \times 100=21.89\)

The given inequality is 2x - y > l.
Draw the graph of the line 2x - y = l.

Table of value satisfying the equation 2x - y = 1

Putting (0, 0) in the given inequation, we have
2 x 0 - 0 > 1  \(\Rightarrow\) 0 > 1 , which is false
\(\therefore\) Half plane of 2x - y > 1 is away from origin
Also the given inequality is x - 2y < -l.
Draw the graph of the line x - 2y =-l.
Table of values satisfying the equation
x - 2y = -1

Putting (0, 0) in the given inequation, we have
0-2 x 0 < -1 \(\Rightarrow\) 0 < -1, which is false
\(\therefore\) Half plane of x - 2y < -1 is away from origin