\(\text { Let } f(x)=\frac{1}{a x^{2}+b x+c}\)
\(\text { By quotient rule, }\)
\(f^{\prime}(x) =\frac{\left(a x^{2}+b x+c\right) \frac{d}{d x}(1)-\frac{d}{d x}\left(a x^{2}+b x+c\right)}{\left(a x^{2}+b x+c\right)^{2}} \)
\(=\frac{\left(a x^{2}+b x+c\right)(0)-(2 a x+b)}{\left(a x^{2}+b x+c\right)^{2}} \)
\(=\frac{-(2 a x+b)}{\left(a x^{2}+b x+c\right)^{2}}\)

Total number of outcomes = 36 \(\Rightarrow \) n(s) = 36
Possible outcomes are (1,1)(1,3)(3,1),(2,2),(1,5),,(2,4),(4,2),(3,3),(3,5),(5,3),(4,4),(4,6),(2,6),(6,2),(6,4),(5,5),(6,6) n(E)=18
\(\frac { 1 }{ 2 } \)

Let us make the following table from the given data.

Here, \(\frac { N }{ 2 } =\frac { 40 }{ 2 } =20\). The cumulative frequency just greater than 20 is 23, so the median class is 40-50.
So, we have l=40, f=12, cf=11, h=10 and N=40
Now, Median(M)=\(l+\frac { \frac { N }{ 2 } -cf }{ f } \times h=40+\frac { 20-11 }{ 12 } \times 10\)
\(=40+\frac { 90 }{ 12 } =40+7.5=47.5\)
and Mean deviation from the median = \(\frac { \sum { f_{ i }|{ x }_{ i }-M| } }{ N } \)
\(=\frac { 405 }{ 40 } =10.125\)

Let D be the mid-point of BC and E be the centroid of \(\Delta ABC\), which divides AD in the ratio of 2:1
\(\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 2 } \frac { { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } }{ 2 } \right) \)

\(\left| x \right| +\left| y \right| =1\),
\(\Rightarrow \pm x\pm y=1\)  which forms a square
The locus of the point is a square.

Here, r>1, n>2
\(\therefore \)  T_3r = ²ⁿC_3r-1 x^3r-1; T_{r+2 =} ²ⁿC_r+1 x^r+1
Then, ²ⁿC_3r-1= ²ⁿC_r+1 
\(\Rightarrow \) 3r - 1 + r + 1 = 2n \(\Rightarrow \) n = 2r

Let y = \(|x|\) 
Ans. (-3, 3)

Given, diameter = 44 cm
\(\therefore \)   Radius, r = \(\frac { 44 }{ 2 } \) = 22 cm
Let chord AB = 22 cm

Since OA = OB = AB = 22 cm
\(\therefore \) \(\Delta \)OAB is equilateral
\(\Rightarrow \theta =\angle AOB=60^{ 0 }\)
Now, we convert 60^o into radian measure.
\(\therefore 60^{ 0 }=\left( 60\times \frac { \pi }{ 180 } \right) rad\)
\(\left[ \because radian\quad measure=\frac { \pi }{ 180 } \times degree\quad measure \right] \)
\(\Rightarrow 60^{ 0 }=\frac { \pi }{ 3 } radian\)
\(\therefore Length\quad of\quad minor\quad arc\quad AB=r\theta \left[ \because \theta =\frac { arc\quad length(l) }{ radius\quad (r) } \therefore arc\quad length=r\theta \right] \)
\(=22\times \frac { \pi }{ 3 } =\frac { 22 }{ 3 } \times \frac { 22 }{ 7 } \quad \quad \left[ \because \pi =\frac { 22 }{ 7 } \right] \)
\(=\frac { 484 }{ 21 } cm\)

1+tan²2x=1-tan2x=>tan²2x+tan2x=0
\(\Rightarrow\)tan2x=0 or tan2x=-1
\(\Rightarrow\)2x=\(n\pi \)  or  2x=\(n\pi +\frac { \pi }{ 4 } \)
Ans
\(x = \frac { n\pi }{ 2 } \quad or\quad\frac { n\pi }{ 2 } +\frac { 3\pi }{ 8 } ,n\in Z\)

Required number of individuals=\(n({ C }_{ 1 }\cap { C }_{ 2 }^{ ' })\)
=\(n({ C }_{ 1 }) -n({ C }_{ 1 }\cap { C }_{ 2 })\) = 120-30= 90

We know that 0.3 = 0.3333 ...
\(\Rightarrow\)  0.3 + 0.03 + 0.003 +.......\(\infty\)
Clearly, it is Geometric series
Here a = 0.3 and r = \(\frac { 1 }{ 10 } \)
\(\therefore\) S\(\infty\) = \(\frac { a }{ 1-r } =\frac { 03 }{ 1-\frac { 1 }{ 10 } } \)
\(\therefore\) 0.3 = \(\frac { 1 }{ 3 } \)(in rational form)

Here \(f^{'}(x)=(ax^{2}+b)^{2}=a^{2}x^{4}+b^{2}+2abx^{2}\)
∴ \(f^{'}(x)=\frac{d}{dx}[a^{2}x^{4}+b^{2}+2abx^{2}]\)
=\(a^{2}\frac{d}{dx}(x^{4})+\frac{d}{dx}(b^{2})+2ab \frac{d}{dx}(x^{2})\)
=\(a^{2} \times 4 x^{3}+0+2ab\times 2x\)
=4\(a^{2}x^{3}+4abx\)
=4ax(ax² + b).

\(\frac { 5 }{ 9 } \)

\(\left( 0,\frac { -9 }{ 2 } \right) \); x=0; 2y-9=0; 18

Here \(\overset{lim}{x\rightarrow -1}\)\(\frac{x^{10}+x^{5}+1}{x-1}\)
putting x = -1
\(=\frac{(-1)^{10}+(-1)^{5}+1}{-1-1}=\frac{1-1+1}{-2}=\frac{0+1}{2}=\frac{-1}{2}\)

{x:x∈R,2<x<10}

Given equation of parabola is  y²=8x,
Here,  a = 2
 \(\therefore\)   Focus=(2, 0)
Now, focal distance = 4
We know that focal distance is a distance of any point P(x, y) on the parabola from the focus F.
\(\therefore\)   PF= Focal distance
\(\Rightarrow\) PF=   \(\sqrt { { \left( x-a \right) }^{ 2 }+{ y }^{ 2 } } \)  [by distance formula]
\(\Rightarrow\)  4=\(\sqrt { { \left( x-2 \right) }^{ 2 }+{ 8x } } \)  [\(\because\) y²=8x]
\(\Rightarrow\) 16=x²-4x+4+8x [on squaring both sides]
\(\Rightarrow\) 16=(x+2)²
\(\Rightarrow\) x+2=4    [taking positive square root on both sides]
\(\Rightarrow\)  x = 2
Now,  y² = 8\(\times\)2\(\Rightarrow\)y² = 16\(\Rightarrow\) y =\(\pm\)4   
Hence coordinates of a point are (2, 4) and (2, -4)
Note: The distance of a point P on the parabola from the focus is called focal distance of the point

There are 9 ladies and 8 gentelmen. We have to form a commitee of 1 , consiting of atleat 5 ladoies
this can be formed by selecting
(a)5 ladies and 5 gentelmen 
(b)6 ladies and 4 gentelmen
(c)7 ladies and 3 gentelmen 
(d)8 ladies and 2 gentelmen 
(e)9 ladies and 1 gentelmen 
Noe , the number of ways of forming a committee
=⁹C₅ X ⁸C₅ + ⁹C6 X ⁸C₄ + ⁹C₇X ⁸C₃ + ⁹C₈ X ⁸C₂+ ⁹C₉ X ⁸C₁
=⁹C₄ X ⁸C₃ + ⁹C₃ X ⁸C₄ + ⁹C₂ X ⁸C₃ + ⁹C₁ X ⁸C₂ + ⁹C₀ X⁸C¹ [ⁿC_r=ⁿC_n-r]
=\(\frac { 9\times 8\times 7\times 6 }{ 4\times 3\times 2\times 1 } \times \frac { 8\times 7\times 6 }{ 3\times 2\times 1 } +\frac { 9\times 8\times 7 }{ 3\times 2\times 1 } \times \frac { 8\times 7\times 6\times 5 }{ 4\times 3\times 2\times 1 } +\frac { 9\times 8 }{ 2\times 1 } \times \frac { 8\times 7\times 6 }{ 3\times 2\times 1 } +\)
\(9\times \frac { 8\times 7 }{ 2\times 1 } \)
=126 X 56 + 84 X 70 + 36 X 56 + 9 X 28 + 8
=7056+5880+2016+252+8=15212
(i) Clearly, ladies are in majority in (b),(c),(d) and (e) case as discussed above.
 nUmber of committees in which ladies are in majority = 15212-⁹C₅ x ⁸C₅
(ii) Clearly, gentlemen are not in majority in the cases discussed above.
Thus, there is no committee in which gentlemen are in majority.Hence, number of such committee = 0.

\(Given,\ { 25x }^{ 2 }-30x+11=0\)
On comparing Eq.(i) with ax²+bx+c=0, we get
\(a=25,\quad b=-30\quad and\quad c=11\)
\( \because x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \)
\(\therefore x=\frac { 30\pm \sqrt { { (-30) }^{ 2 }-4\times 25\times 11 } }{ 2\times 25 } \)
\(\Rightarrow x=\frac { 30\pm \sqrt { 900-1100 } }{ 50 } \)
\( \Rightarrow x=\frac { 30\pm \sqrt { -200 } }{ 50 } \)
\( \Rightarrow x=\frac { 30\pm 10i\sqrt { 2 } }{ 50 } \quad [\because \sqrt { -1 } =i]\)
\( \Rightarrow x=\frac { 3 }{ 5 } \pm \frac { \sqrt { 2 } }{ 5 } i\)
\(\text{Hence, the roots are}\frac { 3 }{ 5 } +\frac { \sqrt { 2 } }{ 5 } i\ and\ \frac { 3 }{ 5 } -\frac { \sqrt { 2 } }{ 5 } i\)
\(Now,\ let\ \alpha =\frac { 3 }{ 5 } +\frac { \sqrt { 2 } }{ 5 } i \ and\ \beta =\frac { 3 }{ 5 } -\frac { \sqrt { 2 } }{ 5 } i\)
\(\because \ \alpha =\bar { \beta } and\ \beta =\bar { \alpha } \)
So,we can say that,the roots are complex conjugate of each other.

We have,A={1,4}, B={2,3,6} and C={2,3,7}
To find \(A\times (B\cap C)\)
\(B\cap C\) = {2,3}
\(\therefore\) \(A\times (B\cap C)\)={1,4}\(\times \){2,3}
= {(1,2),(1,3),(4,2),(4,3}....(iii)
To find \((A\times B)\cap (A\times C)\)
\(A\times B\) ={(1,2),(1,3),(1,6),(4,2),(4,3),(4,6)}
\(A\times C\)= {(1,2),(1,3),(1,7),(4,2),(4,3),(4,7)}
\(\therefore \) \((A\times B)\cap (A\times C)\)= {(1,2),(1,3),(4,2),(4,3)}......(iv)
From Eqs.(iii) and (iv), we get 
\(A\times (B\cap C)=(A\times B)\cap (A\times C)\)

Compound statement is true and its component statements are
p: The Sun is a star
q: The Sun is a plant

Consider P(k): (2k+7)<(k+3)²
\(\Rightarrow \) (2k+7)+2 < (k+3)² +2
\(\Rightarrow \) 2(k+1)+7 < (k+4)²  [(k+3)³+2<(k+4)²]

(2n + 1)\({\pi\over 4},2m\pi;m,n \in Z\)

The given equations are 3x + 4y - 7 = 0 and x - y + 2 = 0
Equation of any line passing through the intersection of the given lines is in the form
(3x + 4y - 7) + k (x - y + 2) = 0 ... (i)
⇒ 3x + 4y - 7 + kx - ky + 2k = 0
⇒ (3 + k) x + (4 - k) y - 7 + 2k = 0
Slope= \(-(3+k)\over 4-k\)
∴ \(-{(3+k)\over 4-k}=5\)
⇒ 20 - 5k = -3 - k
⇒ 4k = 23
∴ \(k={23\over 4}\)
Substituting the value of k is equation (i)
\((3x + 4y - 7) +{23\over 4}(x - y + 2) = 0\)
⇒ 12x + 16y - 28 + 23x - 23y + 46 = 0
⇒ 35x - 7y + 18 = 0

The given inequality 5x +4y \(\le\) 20
draw the graph of the line 5x + 4y = 20

table of values satisfying the equation 5x +4y = 20

Putting (0,0) in the given inequation, we have 5 x 0 +4 x 0 \(\le\) 20 \(\Rightarrow\) 0 \(\le\) 20, which is true
\(\therefore\) Half plane of 5x +4y \(\le\) 20 is towards origin
Also the given inequality ix x \(\ge\) 1
Draw the graph of the line x = 1
Putting (0,0) in the given inequation, we have 0 \(\ge\) 1 which is false
\(\therefore\) Half plane of x \(\ge\) 1 is away from origin
the given inequality is y \(\ge\) 2
Draw the graph of the line y = 2
Putting (0,0) in the given inequation, we have 0 \(\ge\)2, which is false
\(\therefore\) half plane y \(\ge\) 2 is away from origin

Mean\(\overline { x } =\frac { 1 }{ N } \sum { { f }_{ i }{ x }_{ i } } =\frac { 1 }{ 50 } \times 17900=358\)
Mean deviation about mean\(=\frac { 1 }{ N } \sum _{ i=1 }^{ n }{ { f }_{ i }\left| { x }_{ i }-\overline { x } \right| } \)
\(=\frac{1}{50}\times7896 = 157.92\) 

Let 'a' be the first term and 'd' be the common difference of given A.P.
\(\because \quad { S }_{ m }=\frac { m }{ 2 } [2a+(m-1)d]\)
and \({ S }_{ n }=\frac { n }{ 2 } [2a+(n-1)d]\)
\(\therefore \quad \frac { { S }_{ m } }{ { S }_{ n } } =\frac { \frac { m }{ 2 } [2a+(m-1)d] }{ \frac { n }{ 2 } [2a+(n-1)d] } \)
But \(\frac { { S }_{ m } }{ { S }_{ n } } =\frac { { m }_{ 2 } }{ { n }_{ 2 } } \)   [Given]
\(\Rightarrow \frac { \frac { m }{ 2 } [2a+(m-1)d] }{ \frac { n }{ 2 } [2a+(n-1)d] } =\frac { { m }^{ 2 } }{ { n }^{ 2 } } \)
\(\Rightarrow \frac { 2a+(n-1)d }{ 2a+(n-1)d] } =\frac { { m }^{ 2 } }{ { n }^{ 2 } } \times \frac { n }{ 2 } \times \frac { 2 }{ m } =\frac { m }{ n } \)
\(\Rightarrow\) 2an + n(m-1)d = 2am + m(n-1)d
\(\Rightarrow\) 2an-2am= (mn-m)d-(mn-n)d
\(\Rightarrow\) 2a[n-m]=[mn-m-mn+n]d
\(\Rightarrow\) 2a[n-m]=[n-m]d
\(\Rightarrow d=\frac { 2a[n-m] }{ [n-m] } =2a\)
Now, \(\frac { { a }_{ m } }{ { a }_{ n } } =\frac { a+(m-1)d }{ a+(n-1)d } =\frac { a+(m-1)\times 2a }{ a+(n-1)\times 2a } \)
\(=\frac { a[1+2m-2] }{ a[1+2n-2] } =\frac { 2m-1 }{ 2n-1 } \)
Thus, the ratio of 'm^th' and 'n^th' term is (2m+1): (2m-1).