\(-\frac{1}{4}, \frac{1}{4}\) 
Let the quadratic polynomial be ax² + bx + c, and its zeroes be α + ß
Given  \(\alpha+\beta=-\frac{1}{4}=-\frac{b}{a}\) 
\(\alpha \beta=\frac{1}{4}=\frac{c}{a}\) 
If a = 4, b = 1, c = 1
Therefore, the quadratic polynomial is 4x² + x +1.

In \(\triangle\)AOB and \(\triangle\)COD,
AB || CD

OAB = DCO and OBA = ODC (Alternate angles)
\(\triangle\)AOB ~\(\triangle\)COD    (AA similarity)
\(\therefore \frac { AO }{ BO } =\frac { CO }{ DO } .\\ \) (Corresponding sides of similar triangles)
or \(\frac { AO }{ CO } =\frac { BO }{ DO } \)

Given HCF (306, 1,314) =18
LCM (306, 1,314) = ?
Let  a = 306
b = 1,314
We know that
a x b = LCM (a, b) x HCF (a, b)
⇒ 306 x 1,314 = LCM (a, b) x 18
\(\Rightarrow \quad LCM(a,b)=\frac { 306\times 1,314 }{ 18 } \)
∴ LCM (306, 1,314) = 22,338

For parallel lines, \(\frac { { a }_{ 1 } }{ { a }_{ 2 } } =\frac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \frac { { c }_{ 1 } }{ { c }_{ 2 } } \)
k=-6

\(AB=\sqrt { (4-{ 2) }^{ 2 }+(6-3)^{ 2 } } =\sqrt { 13 } ;\)
\(BC=\sqrt { (6-{ 4) }^{ 2 }+(9-6)^{ 2 } } =\sqrt { 13 } \)
\(AC=\sqrt { (6-{ 2) }^{ 2 }+(9-3)^{ 2 } } =\sqrt { 52 } =2\sqrt { 13 } \)
Here AB+BC=AC
∴ AB, BC and AC cannot be the side of any triangle.
( ∵ Sum of two sides of a triangle is always more than the third side)
∴  A, B and C are not vertices of any triangle. A, B and C are collinear.

In right ΔABC
\(\frac { BC }{ AC } \) = cos 60\(\unicode{xb0} \)
⇒ \(\frac { 2.5 }{ AC } =\frac { 1 }{ 2 } \)
⇒ AC = 2.5 x 2 = 5 m

Let one number be x,
Then, another number be 27 - x
[\(\because\) sum of two numbers = 27]
According to the question,
Product of these two numbers = 182
\(\therefore\) x(27-x)=182
\(\Rightarrow\) 27x-x² = 182
\(\Rightarrow\) x²-27x+182=0 
\(\Rightarrow\) x²-14x-13x+182=0
[\(\because\) (-14) \(\times\) (-13) =182 and -14-13=-27
\(\Rightarrow\) x(x-14)-13(x-14)=0 \(\Rightarrow\) (x-14)(x-13)=0
\(\Rightarrow\) x-14 = 0 or x-13 = 0
\(\Rightarrow\) x=14 or x=13
If x = 14, then 27-x= 27-14 =13 and if x=13, then 27-x= 27-13 = 14 Hence, in both cases, the numbers are 13 and 14.

\(\therefore \quad \frac{n}{2}=\frac{250}{2}=125 \text {, }\) which lies in the interval 2-3.
\(\begin{aligned} \therefore \text { Median } & =l+\frac{\left(\frac{n}{2}-c f\right)}{f} \times h \end{aligned}\)
\(\begin{aligned} =2+\frac{125-120}{62} \times 1 \end{aligned}\)
\(\begin{aligned} =2+\frac{5}{62}=2+0.08=2.08 \mathrm{~m} \end{aligned}\)
Hence, we observe that 50% of students jump below 2.08 m and 50% above it.

Here, the maximum frequency is 10 and the corresponding class is 30-40. So, this is the modal class.
We have, 1= 30, h = 10, f = 16, f₁ = 10 and f₂ = 12
\(\therefore \quad \text { Mode } =l+\frac{f-f_{1}}{2 f-f_{1}-f_{2}} \times h \)
\(=30+\frac{16-10}{2(16)-10-12} \times 10 \)
\(=30+\frac{6}{32-22} \times 10 \)
\(=30+\frac{6}{10} \times 10 \)
\(=30+6\)
Mode = 36

3x² – x – 4 = 3x² – 4x + 3x – 4
= x(3x - 4) +1(3x - 4)
=(3x - 4)(x +1)
The zeroes of the polynomial are {4/3, -1}
Relationship between the zeroes and the coefficient of the polynomial:
Also sum of the zeroes = \(\frac{4}{3}-1=\frac{4-3}{3}=\frac{1}{3}\) 
Also product of the zeroes = \(\frac{4}{3} x-1=-\frac{4}{3}\) 
Hence verified.

Given,  \(\triangle ABE\cong \triangle ACD\)
\(\Rightarrow \)   AB = AC and AE = AD    [by CPCT]
\(\Rightarrow \)   \(\frac{AB}{AC} =1\)
and \(\frac{AD}{AE}=1\)
\(\Rightarrow\frac { AB }{ AC } =\frac { AD }{ AE } \)    .... (i)
In \(\triangle ADE\) and \(\triangle ABC\) , we have
\(\frac{AD}{AE}=\frac{AB}{AC}\)  [from Eq.(i)]
\(\Rightarrow\frac{AD}{AB}=\frac{AE}{AC}\)
and \(\angle DAE=\angle BAC\)   [common angle]
\(\therefore \) \(\triangle ADE\sim \triangle ABC\)  [by SAS similarity criterion]
Hence proved.

Given pair of linear equations is
41x+53y=135   ..(i)
and 53x+41y=147  ...(ii)
On adding Eqs. (i) and (ii), we get
94x+94y=282
\(\Rightarrow\) x+y=3 [dividingboth sides by 94] ...(iii)
On subtracting Eq. (i) from Eq. (ii), we get
12x-12y=12
\(\Rightarrow\) x-y=1 [dividing both sides by 12]   ...(iv)
Now, on adding Eqs.(iii) and (iv), we get
2x=4 \(\Rightarrow\) x=2
On substituting x=2 in Eq. (iii), we get
y=3-2=1
Hence, x=2 and  y=1 is the required solution.

Given numbers are 156, 208 and 260.
Here, 260 > 208 > 156
Let us first find the HCF of 260 and 208
By using Euclid's division lemma for 260 and 208, 
we get 
260 = (208 x 1) + 52
Here, remainder = 52 \(\neq \) 0
On taking 208 as new dividend and 52 as new divisor and then apply Euclid's division lemma, we get
208 = (52 x 4) + 0
Here, the remainder is zero and  the divisor is 52.
So, 52 is the HCF of 208 and 260.
Now, 156 > 52
Let us find the HCF of 52 and 156. By using Euclid's division lemma, we get
156 = (52 x 3) + 0
Here, the remainder is zero and the divisor is 52.
So, 52 is the HCF of 52 and 156.
Thus, HCF of 156, 208 and 260 is 52.
Hence, the minimum number of buses
\(=\frac { 156 }{ 52 } +\frac { 208 }{ 52 } +\frac { 260 }{ 52 } \)
= 3 + 4 + 5 = 12

Point P(k-1,2) is equidistant from the points A(3,k) and B(k,5).
∴ AP = BP
⇒ AP²=BP²
⇒ (k-1-3)²+(2-k)² = (k-1-k)²+(2-5)²
⇒ (k-4)²+(2-k)²=(-1)²+(-3)²
⇒ k²-8k+16+4-4k+k² =1+9
⇒ 2k²-12k+10=0
⇒ k²-6k+5=0
⇒ k²-5k-k+5=0
⇒ k(k-5)-1(k-5)=0
⇒ (k-1)(k-5)=0
⇒ k-1=0 or k-5=0
⇒ k=1,5

Let AB is statue, BC is pedestal and BC=x m, CD=y m.
In right ΔBCD, ΔBCD, \(\frac { BC }{ CD } \)=tan 45^o
⇒, \(\frac { x }{ y } \)=1 ⇒ x=y     .....(i)
In right ΔACD, \(\frac { AC }{ CD } \)=tan 600
⇒ \(\frac { x+1.46 }{ y } =\sqrt { 3 } \)
⇒ \(\frac { x+1.46 }{ x } \)=1.73    [Using (i)]
⇒ x+1.46=1.73x ⇒ 0.73x=1.46 ⇒ x=2

Let speed of the stream be x km/h
Speed of the boat in still water = I I km/h 
\(\therefore \) Upstream speed be (11-x) km/h and downstream speed be (11+x) km/h
Distance = 12 km
Time taken for downstream direction = \(\frac { 12 }{ 11+x } \) hours
Time taken for upstream direction = \(\frac { 12 }{ 11-x } \) hours
ATQ \(\frac { 12 }{ 11+x } +\frac { 12 }{ 11-x } =2\frac { 3 }{ 4 } \Rightarrow \frac { 12(11-x)+12(11+x) }{ (11+x)(11-x) } =\frac { 11 }{ 4 } \)
\(\Rightarrow \frac { 132+132 }{ 121-x^{ 2 } } =\frac { 11 }{ 4 } \Rightarrow 4\times 264=11(121-x^{ 2 })\)
\(\Rightarrow \frac { 4\times 264 }{ 11 } =121-x^{ 2 }\Rightarrow 4\times 24=121-x^{ 2 }\)
\(\Rightarrow x^{ 2 }=25\Rightarrow x=\pm 5\)
Hence speed of the stream is x=5 km/h

\(\because\)  AB II PQ
\(​​\angle\)ABQ = \(​​\angle\).PQD (Corresponding angles) 
In \(\triangle\)ADB and \(\triangle\)PDQ,
\(​​\angle\)ADB = \(​​\angle\).PDQ (Common)
\(​​\angle\)ABQ = L\(​​\angle\)PQD (Proved above)
By AAsimilarity,
\(\triangle\)PDB~\(\triangle\)PDQ
\(\frac { DQ }{ DB } =\frac { PQ }{ AB } \)
\(\Rightarrow \frac { DQ }{ DB } =\frac { z }{ x } ...(i)\)
Similarly, \(\triangle\)PBQ~\(\triangle\)CBD
\(\therefore \frac { BQ }{ DB } =\frac { z }{ y } ...(ii)\)
Adding (i) and (ii), we get
\(\frac { z }{ x } +\frac { z }{ y } =\quad \frac { DQ+BQ }{ DB } =\frac { BD }{ BD } \)
\(\Rightarrow \frac { z }{ x } +\frac { z }{ y } =1\quad \)
\(\therefore \frac { 1 }{ x } +\frac { 1 }{ y } =\frac { 1 }{ z } .\)

\(\frac { x-3 }{ x-4 } +\frac { x-5 }{ x-6 } =\frac { 10 }{ 3 } \)
\(\frac { (x-3)(x-6)+(x-4)(x-5) }{ (x-4)(x-6) } =\frac { 10 }{ 3 } \)
\(\frac { (x-3)(x-6)+(x-4)(x-5) }{ (x-4)(x-6) } =\frac { 10 }{ 3 } \)
\(3(2x^{ 2 }-18x+38)=10x^{ 2 }-100x+240\)
\(6x^{ 2 }-54x+114=10x^{ 2 }-100x+240\)
4x²-46x+126=0
2x²-14x-9x+63=0
2x(x-7)-9(x-7)=0
(2x-9)(x-7)=0
2x-9=0,x-7=0
x=9/2 , x=7

Given, a pair of linear equations is :
2x + 3y = 11     ......(i)
and 2x - 4y = - 24  ......(ii)
From eqn. (ii), 4y = 2x + 24
\(\\ \Rightarrow \quad y=\frac { x+12 }{ 2 } \)  ......(iii)
On substituting y from eqn. (iii) in eqn. (i), we get
2x + 3 \(\left( \frac { x+12 }{ 2 } \right) \)= 11
\(\Rightarrow\) 4x + 3(x + 12) = 11 \(\times\) 2
\(\Rightarrow\) 4x + 3x + 36 = 22
\(\Rightarrow\) 7x=22 - 36
\(\Rightarrow\) 7x = -14
\(\therefore\) x = -2
From eqn. (iii), \(y=\frac { -2+12 }{ 2 } \)
\(\therefore\)   y = 5
On substituting x = - 2 and y = 5 in the equation
y = mx + 3,
5 = m \(\times\) (- 2) + 3
\(\Rightarrow\) 5 = -2m + 3
\(\Rightarrow\) 2m  = 3 - 5 = -2
\(\therefore\)  m = -1

Let P(x₁ , y₁) and Q(x₂ , y₂) are two points which divide AB in three equal parts.
By section formula
\(P\left( { x }_{ 1 }{ y }_{ 1 } \right) =\left( \frac { 1\times \left( -4 \right) +2\times \left( 2 \right) }{ 1+2 } ,\frac { 1\times \left( -6 \right) +2\times \left( -3 \right) }{ 1+2 } \right) \)
\(=\left( \frac { -4+4 }{ 3 } ,\frac { -6+(-6) }{ 3 } \right) \)
= (0, -4)
\(Q\left( { x }_{ 2 }{ y }_{ 2 } \right) =\left( \frac { 2\times \left( -4 \right) +1\times \left( 2 \right) }{ 2+1 } ,\frac { 2\times \left( -6 \right) +1\times \left( -3 \right) }{ 2+1 } \right) \)
\(=\left( \frac { -8+2 }{ 3 } ,\frac { -12+(-3) }{ 3 } \right) \)
= (- 2, - 5)

It can be observed that class intervals are not continuous. There is a gap of 1 between two class intervals. Therefore,1/2  has to be added to the upper class limit and1/2 has to be subtracted from the lower class limit of each interval.
Class mark (x_i) can be obtained by using the following relation.
\(x_{i}=\frac{\text { Upper class limit }+\text { Lower class limit }}{2}\)
Class size (h) of this data = 3
Taking 57 as assumed mean (a), d_i, u_i, f_iu_i are calculated as follows.

It can be observed that
\(\sum f_{i}=400 \)
\(\sum f_{i} u_{i}=25 \)
\(\text { Mean } \bar{x}=a+\left(\frac{\sum f_{1} u_{i}}{\sum f_{i}}\right) x h \)
\(=57+\left(\frac{25}{400}\right) \times 3\)
= 57+3/16 =  57+ 0.1875
= 57.1875
= 57.19
Mean number of mangoes kept in a packing box is 57.19.
Step deviation method is used here as the values of f_i, d_i are big and also, there is a common multiple between all d_i.

Let CD = h m be the height of the tower. At point D of the tower, a man is standing and observes the car at an angle of depression of 30°. After six seconds, the angle of depression of the car is 60°.
i.e. \(\angle\)ODA = 30° and \(\angle\)ODB = 60°
\(\Rightarrow\) \(\angle\)DAC = \(\angle\)ODA = 30°  [alternate angles]
and \(\angle\)DBC = \(\angle\)ODB = 60°  [alternate angles]
Let AB = y m and BC = x m
In right angled \(\Delta\)BCD,

\(\begin{array}{rlrl} \tan 60^{\circ} & =\frac{P}{B}=\frac{C D}{B C} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad \sqrt{3} & =\frac{h}{x} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad h & =\sqrt{3} x & {\left[\because \tan 60^{\circ}=\sqrt{3}\right]} \end{array}\)
\(\Rightarrow \quad h = \sqrt3 x\)....(i)
In right angled \(\Delta\)ACD,
\(\begin{array}{rlrl} \tan 30^{\circ} & =\frac{C D}{A C}=\frac{C D}{A B+B C} & & {[\because A C=A B+B C]} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad \frac{1}{\sqrt{3}} & =\frac{h}{x+y} & & {\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right]} \\ \end{array}\)
\(\begin{array}{rlrl} \Rightarrow \quad x+y & =h \sqrt{3} & \end{array}\)
\(\Rightarrow \quad x + y=\sqrt3 x(\sqrt3)\) [from Eq. (i)]
\(\Rightarrow\) x + y = 3x  .....(ii)
It is given that a car moves from point A to B in six seconds. Let its speed be k km/s.
\(\therefore \quad \text { Time }=\frac{\text { Distance }}{\text { Speed }}\)
\(\Rightarrow \quad 6=\frac{y}{k} \Rightarrow y=6 k\)
On putting y = 6k in Eq. (ii), we get
x + 6k = 3x \(\Rightarrow\) 6k - 2x \(\Rightarrow\) x = 3k
\(\therefore \quad \text { Time }=\frac{\text { Distance }}{\text { Speed }}=\frac{x}{k}=\frac{3 k}{k}=3 \mathrm{~s}\)
Hence, the car moves from point B to point C in 3s.

(a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

(d) If Assertion is incorrect but Reason is correct.

Given OB = x, OD = x + 3, OC = 3x + 19 and OA = 3x + 4

and AC || BD
To prove \(\triangle O A C \sim \triangle O B D\)
Proof In \(\triangle O A C \text { and } \triangle O B D\)
\(\angle D O B=\angle C O A\)
                         [vertically opposite angles]
As, AC is parallel to BD, AB and DC are transversal
\(\begin{aligned} & \angle B A C=\angle A B D \end{aligned}\)
and   \(\begin{aligned} & \angle C D B=\angle D C A \end{aligned}\)
                              [\(\because\)alternate interior angles]
\(\triangle O A C \sim \triangle O B D\)
[by AA similarity criteria]
                               Hence proved.
(ii) To prove \(\frac{O A}{A C}=\frac{O B}{B D}\)
Proof From part (i) we know that
\(\triangle O A C \sim \triangle O B D\)
Corresponding sides of \(\Delta\)OAC and \(\Delta\)OBD must be proportional
\(\begin{aligned} \frac{O A}{O B} & =\frac{A C}{B D} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad & \frac{O A}{A C}=\frac{O B}{B D} \end{aligned}\)        Hence proved.
(iii) (a) From part (i)
\(\begin{aligned} \triangle O A C & \sim \triangle O B D \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{O A}{O B} & =\frac{O C}{O D} \Rightarrow \frac{3 x+4}{x}=\frac{3 x+19}{x+3} \end{aligned}\)
[given OA = 3x +4, OB = x, OC = 3x+19, OD = x+3]
\(\Rightarrow\) (3x + 4)(x + 3) = (3x + 19)x
\(\Rightarrow\) 3x² + 9x + 4x + 12 = 3x² + 19x
\(\Rightarrow\)      13x + 12 = 19x \(\Rightarrow\) x = 2
Given,       OC = 3x + 19
On substituting x = 2, we get
OC = 3 \(\times\) 2 + 19 = 6 + 19 = 25
Therefore, OC = 25
(b) From part (ii), we have
\(\begin{aligned} & \frac{O A}{A C}=\frac{O B}{B D} \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad \frac{B D}{A C}=\frac{O B}{O A} \\ \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad \frac{B D}{A C}=\frac{x}{3 x+4} \end{aligned}\)
                       [given, OB = x and OA = 3x + 4]
On substituting x = 2, we get
\(\frac{B D}{A C}=\frac{2}{3 \times 2+4}=\frac{2}{10} \Rightarrow \frac{B D}{A C}=\frac{1}{5}\)
Therefore, \(\frac{B D}{A C}=\frac{1}{5}\)

(i) (a) Since, Ajay's car travels a distance in one hour is (x + 5) km. Therefore, Ajay's car travels a distance two hours is 2(x + 5) km.
(ii) (c) \(\because \text { Time }=\frac{\text { Distance }}{\text { Speed }}\)
Time taken by Ajay and Raj to complete the 400 km journey,
\(t_1=\frac{400}{x+5} \mathrm{~h} \)
and \(t_2=\frac{400}{x} \mathrm{~h}\)
According to the question,
t₂ = t₁ + 4
\( \therefore \frac{400}{x}=\frac{400}{x+5}+4\)
\(\Rightarrow \frac{100}{x}=\frac{100}{x+5}+1\) [dividing by 4]
\(\Rightarrow\) 100(x + 5) = 100x + x(x + 5)
\(\Rightarrow\) 100x + 500 = 100x + x² + 5x
\(\Rightarrow\) x²  +5x - 500 = 0
(iii) (a) Consider the quadratic equation x² + 5x - 500 = 0
On comparing with ax² + bx + c =0, we get
a = 1, b = 5 and c = -500
\(\begin{aligned} \because \quad x & =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{-5 \pm \sqrt{(5)^2-4 \times(1)(-500)}}{2 \times 1} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{-5 \pm \sqrt{25+2000}}{2}=\frac{-5 \pm \sqrt{2025}}{2} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{-5 \pm 45}{2}=\frac{-50}{2}, \frac{40}{2}=-25,20 \end{aligned}\)
Since, speed cannot be negative, so we consider only x = 20.
Hence, the speed of Raj's car is 20 km/h.
(iv) (d) To travel 400 km, time taken by Ajay
\(\begin{aligned} t_1 & =\frac{400}{(x+5)} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{400}{20+5}=\frac{400}{25}=16 \mathrm{~h} \end{aligned}\)

(i) For batch I,
20x + 5y = 9000
For batch II,
5x + 25y = 26000
(ii) We have,
20x + 5y = 9000
\(\Rightarrow\) 4x + y = 1800  ...(i)
and 5x + 25y = 26000
\(\Rightarrow\)  x + 5y = 5200 ...(ii)
Multiplying Eq. (i) by 5 and subtracting Eq. (ii) from it.
5(4x + y) - (x + 5y) = 5 \(\times\) 1800 - 5200
\(\Rightarrow\)  19x = 3800
\(\Rightarrow\) x = 200
Or
On substituting x = 200 in Eq. (i), we get
4(200) + y = 1800
\(\Rightarrow\) 800 + y = 1800
\(\Rightarrow\) y = 1000
\(\therefore\) Difference in the monthly fee paid by a poor child and a rich child = y - x = Rs (1000 - 200) = Rs 800
(iii) Total monthly collection of fees, if there are 10 poor and 20 rich children = 10x + 20 y
= 10 \(\times\) 200 + 20 \(\times\)1000
= 2000 + 20000
= Rs 22000