x=15, y=20

\(\frac { \alpha }{ \beta } +\frac { \alpha }{ \alpha } +2\left( \frac { 1 }{ \alpha } +\frac { 1 }{ \beta } \right) +3\alpha \beta \)
\(=\frac { \alpha ^{ 2 }+\beta ^{ 2 } }{ \alpha \beta } +2\left( \frac { \alpha +\beta }{ \alpha \beta } \right) +3\alpha \beta =-\frac { 2 }{ 3 } \)

\((i){9\over19}(ii){1\over19}(iii){6\over19}(iv){3\over19}\)

Given XY and X'Y' are two parallel tangents. Another tangent AB touches the circle at C and intersect XY at A and X'Y'at B.
To prove \(\angle\)AOB = 90°
Proof We know that, tangents drawn from an external point to a circle are equal in length.
\(\therefore\)   AP = AC   [\(\because\) A is an external point] ...(i)
Thus, in \(\Delta\)APO and \(\Delta\)ACO, AP = AC   [from Eq. (i)]
AO = AO  [common sides]
OP = OC [radii of circle]
\(\Delta\)APO \(\cong\)\(\Delta\)ACO  [by SSS congruence rule]
Then, \(\angle\)OAP = \(\angle\)OAC    [by CPCT]        ...(ii)
\(\Rightarrow\) \(\angle\)PAC = 2 \(\angle\)CAO                             ....(iii)
Similarly, we can prove that \(\angle\)CBO = \(\angle\)OBQ
\(\Rightarrow\)   \(\angle\)CBQ = 2 \(\angle\)CBO                           ...(iv)
since, XY || X'Y'      [given]
\(\therefore\) \(\angle\)PAC +  \(\angle\)QBC = 180°
[\(\because\) sum of interior angles on the same side of transversal is 180°]
\(\Rightarrow\) 2 \(\angle\)CAO + 2 \(\angle\)CBO = 180°    [from Eqs. (iii) and (iv)]
\(\Rightarrow\)   \(\angle\)CAO + \(\angle\)CBO = 90°                    ...(v)
Now, in \(\Delta\)AOB, \(\angle\)CAO + \(\angle\) CBO + \(\angle\)AOB = 180°
[by angle sum property of triangle]
\(\Rightarrow\) \(\angle\)CAO + \(\angle\)CBO = 180° - \(\angle\)AOB    ...(vi)
From Eqs. (v) and (vi), we get
180° - \(\angle\)AOB = 90° \(\Rightarrow\) \(\angle\)AOB = 90° Hence proved.

\(\left[x^{2}-\left(\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}\right) x+\left(\frac{\alpha^{2}}{\beta}\right)\left(\frac{\beta^{2}}{\alpha}\right)\right]\) 
\(=\left[x^{2}-\left(\frac{\alpha^{3}+\beta^{3}}{\alpha \beta}\right) x+\alpha \beta\right]\) 
\(=x^{2}-\left[\frac{(\alpha+\beta)\left\{(\alpha+\beta)^{2}-3 \alpha \beta\right\}}{\alpha \beta}\right] x+\alpha \beta\) 
\(=x^{2}-\left[\frac{3\left\{(3)^{2}-3(-2)\right\}}{-2}\right] x+(-2)=x^{2}+\frac{45}{2} x-2\) 
= \(\frac{1}{2}\left(2 x^{2}+45 x-4\right)\)

This can be done by trial and error. But to do it systematically, we find HCF (420, 130). Then this number will give the maximum number of barfis in each stack and the number of stacks will then be the least. The area of the tray that is used up will be the least. Now, let us use Euclid’s algorithm to find their HCF. We have :
420 = 130 x 3 + 30
130 = 30 x 4 + 10
30 = 10 x 3 + 0
So, the HCF of 420 and 130 is 10.
Therefore, the sweetseller can make stacks of 10 for both kinds of barfi.

A=Assumed mean =35
Mean = A +\(\frac { \Sigma f_{ i }u_{ i } }{ \Sigma f_{ i } } \times 10\)
\(31.4=35+\frac { -2x-2 }{ 42+x } \times 10\)
(2x + 210 = (42 + x)(3.6)
20x + 20=151.2+3.6x
16.4x = 131.2
x = 8

The cumulative frequency table of given data is

Here, n=200 ∴\(\frac { n }{ 2 } =\frac { 200 }{ 2 } =100\) \(\)
Since, the cumulative frequency just greater than 100 is 160 and the corresponding class interval is 10-15.
∴ Median class=10-15, l=10, cf=80, h=5 ad f=80
Now, median\(\) \(=l+\left\{ \frac { \frac { n }{ 2 } -cf }{ f } \right\} \times h=10+\left\{ \frac { 100-80 }{ 80 } \right\} \times 5\)
\(=10+\left( \frac { 20 }{ 80 } \right) \times 5=10+1.25=11.25\) ​​​​

We know that, tangents drawn from an exterior point to a circle are equal in length.
AD = AF = x cm
BD = BE = y cm
CE = CF = z cm
Given, AB = 8 cm
⇒ AD + BD = 8cm
⇒ x + y = 8
BC = 10 cm ⇒ BE + CE = 10 cm
⇒ y + z = 10
and CA = 12 cm ⇒ CF + AF = 12 cm
⇒ z + x = 12
On adding Eqs. (i), (ii) and (iii), we get
2(x + y + z) =30
⇒ x + y + z = 15
On subtracting Eq. (ii) from Eq. (iv), we get
x =15 - 10 = 5
On subtracting Eq. (iii) from Eq. (iv), we get
y = 15-12 = 3
On subtracting Eq. (i) from Eq. (iv), we get
z=15-8=7
AD = xcm = 5 cm
BE= ycm = 3 cm
and CF = z cm = 7 cm
Hence, the length of AD, BE and CE are 5 cm, 3 cm and 7 cm, respectively.

Let ABCD is a quadrilateral circumscribing a circle with centre O. Let circle touches the sides of a quadrilatcral at points E, F, G and H.

To prove \(\angle\)AOB + \(\angle\)COD = 180°
and \(\angle\) AOD + \(\angle\)BOC = 180°
Construction Join OE, OF, OG and OH.
Proof We know that two tangents drawn from an external point to a circle subtend equal angles at the centre.
and       ....(i)
Also, we know that the sum of all angles subtended at a point is 360°.
\(\begin{array}{rlrl} \therefore \angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8 & =360^{\circ} \\ \end{array}\)    ...(ii)
\(\begin{array}{rlrl} \Rightarrow 2(\angle 2+\angle 3+\angle 6+\angle 7) & =360^{\circ} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & (\angle 2+\angle 3)+(\angle 6+\angle 7)=180^{\circ} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & \angle A O B+\angle C O D=180^{\circ} \end{array}\)
Similarly, we have
\(\begin{aligned} 2(\angle 1+\angle 8+\angle 4+\angle 5) & =360^{\circ} \\ \end{aligned}\)  [from Eq. (i) and (ii)]
\(\begin{aligned} & \Rightarrow(\angle 1+\angle 8)+(\angle 4+\angle 5)=180^{\circ} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \angle A O D+\angle B O C=180^{\circ} \end{aligned}\) Hence proved.

Let us assume that -7- 2\(\sqrt{3}\) is rational number.
Then, it will be the form \(\frac{a}{b}\), where a,b are coprime integers and b \(\neq 0\)
Now, \(-7-2 \sqrt{3}=\frac{a}{b}\)
On rearranging, we get
\(\begin{aligned} -7-\frac{a}{b}=2 \sqrt{3} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad-\frac{7}{2}-\frac{a}{2 b}=\sqrt{3} \end{aligned}\)
Since,  \(\frac{7}{2}\) and \(\frac{a}{2b}\) are rational.
So, their difference will be rational.
\(\therefore\) \(\sqrt{3}\) is rational.
But given that \(\sqrt{3}\) is an irrational number.
So, this contradicts the fact that \(\sqrt{3}\) is rational.
Therefore, our assumption is wrong.
Hence, \(-7-2 \sqrt{3}\) is irrational.

Let p(x) = 3x² + 11x - 4
= 3x² + 12x - x - 4
= 3x(x + 4) - 1(x + 4)
= (3x - 1)(x + 4)
So, zeroes are \(m=\frac{1}{3}\) and n = - 4
Now, \(\frac { m }{ n } +\frac { n }{ m } =\frac { \left( \frac { 1 }{ 3 } \right) }{ -4 } +\frac { -4 }{ \left( \frac { 1 }{ 3 } \right) } =\frac { 1 }{ -12 } -12\)
\(=\frac{-145}{12}\)

The cumulative frequency table for given data is

Here, \(\frac { n }{ 2 } =\frac { 67 }{ 2 } =33.5\)
The cumulative frequency just greater than 33.5 is 42 and the corresponding class is 125-145.
Thus, we have median class 125-145.
Also, the maximum frequency is of the class 125-145.
Therefore, the modal class is 125-145.
Difference of the upper limit of median class and the lower limit of modal class=145-125=20

0.35

\(1\over6\)

Since, the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
i.e  ㄥAOB + ㄥCOD = 180^o
⇒  ㄥCOD =180^o  - ㄥAOB
⇒ ㄥCOD =  = 180^o  - 125^o = 55^o

(i) (d):
Total number of participants = 60 + 84 + 108
= 252
(ii) (d):
60 = 22 x 3 x 5
84 = 22 x 3 x 7
108 = 22 x 33
LCM(60, 84, 108) = 22 x 33 x 5 x 7
= 3780
(iii) (a):
60 = 22 x 3 x 5
84 = 22 x 3 x 7
108 = 22 x 33
HCF(60, 84, 108) = 22 x 3
= 12
(iv) (c):
Minimum number of rooms required for all the participants = 252/12
= 21
(v) (d):
Minimum number of rooms required for all = 21 + 1 = 22

(i) (d): We know that,
\(\text { Class mark }=\frac{\text { Lower limit }+\text { Upper limit }}{2} \)
\(\Rightarrow m=\frac{\text { Lower limit }+b}{2} \Rightarrow \text { Lower limit }=2 m-b\)
(ii) (a):

\(\begin{aligned} &\therefore \quad \text { Average lifetime of a packet }\\ &=A+\frac{\sum f_{i} d_{i}}{\sum f_{i}}=325+\frac{6400}{400}=341 \mathrm{hrs} \end{aligned}\)

(iii) (b) : \(\text { Here, } N=400 \Rightarrow \frac{N}{2}=200\)
Also, cumulative frequency for the given distribution are 14, 70, 130,216,290,352,400
\(\therefore\) c.f just greater than 200 is 216, which is
corresponding to the interval 300-350.
l= 300, f=86, c.f = 130, h = 50
\(\therefore \quad \text { Median }=l+\left(\frac{\frac{N}{2}-c . f .}{f}\right) \times h=300+\left(\frac{200-130}{86}\right) \times 50\)
= 300 + 40.697 = 340.697 ""340 hrs (approx.)
(iv) (a) : We know that Mode = 3 Median - 2 Mean
= 3(340.697) -2(341)
= 1022.091 - 682 = 340.091 ""340 hrs
(v) (c): Since, minimum of mean, median and mode is approximately 340 hrs. So, manufacturer should claim that lifetime of a packet is 340 hrs.

(i) (b): OA²=AB²+OB²

\(\Rightarrow \quad O A=\sqrt{10^{2}+5^{2}}=5 \sqrt{5} \mathrm{~cm}\)
(ii) (a) : \(O A=\sqrt{O P^{2}-A P^{2}} \text { (Given) }\)
\(=\sqrt{17^{2}-15^{2}}=\sqrt{64}=8 \mathrm{~cm}\)
(iii) (b): Length of tangent \(A P=\sqrt{O P^{2}-O A^{2}} \text { (Given) }\)
\(=\sqrt{25^{2}-7^{2}}=\sqrt{576}=24 \mathrm{~cm}\)
(iv) (b):

\(\text { Since, } O P=\sqrt{(P T)^{2}+(O T)^{2}}=\sqrt{21^{2}+20^{2}}=29 \mathrm{~cm}\)
(v) (a): Since, OP² + PQ² = OQ²
\(\Rightarrow\) 8² + x² = (x + 2)²\(\Rightarrow\) 64 = 4x + 4\(\Rightarrow\) x = 15 cm
So, length of tangent, PQ = 15 cm.