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

Given, HCF of 306 and 657 = 9
We know that
LCM \(\times\) HCF = Product of two numbers
\(\Rightarrow\) LCM \(\times\) 9 = 306 \(\times\) 657
\(\Rightarrow\)   LCM = \( \frac{306 \times 657}{9}\)
= 34 \(\times\) 657 = 22338
\(\therefore\) LCM of 306 and 657 = 22338

P(x)=x²+ 7x+ 7
Here, a =1, b =7, c=7
∴ α, β are both zeroes of p(x)
\(∴\ \alpha+\beta={-b\over a}=-7\)
\(\alpha\beta={c\over a}=7\)
Now \({1\over \alpha}+{1\over \beta}-2\alpha\beta={\beta+\alpha\over \alpha\beta}-2\alpha\beta\)
\(={-7\over 7}-2\times7=-1-14=-15\)

Multiplying Equation (2) by \(\frac{5}{3},\) we get
5x – 8y + 1 = 0
But, this is the same as Equation (1). Hence the lines represented by Equations (1) and (2) are coincident. Therefore, Equations (1) and (2) have infinitely many solutions Plot few points on the graph and verify it yourself.

\(\triangle\)ABC~\(\triangle\)PQR   (Given)
\(\frac { AB }{ PQ } =\frac { BC }{ QR } =\frac { AC }{ PR } \)
\(\Rightarrow \frac { z }{ 3 } =\frac { 8 }{ 6 } =\frac { 4\sqrt { 3 } }{ y } \)
\(\Rightarrow \frac { z }{ 3 } =\frac { 8 }{ 6 } and\frac { 8 }{ 6 } =\frac { 4\sqrt { 3 } }{ y } \)
\(\Rightarrow z=\frac { 8\times 3 }{ 6 } and\quad y=\frac { 4\sqrt { 3 } \times 6 }{ 8 } \)
\(\therefore z=4 and\ y=3\sqrt { 3 } \)
\(\therefore y+z=4+3\sqrt { 3 } \)

The pair of equations y = 0 and y = -5 has no solution.

The required number is the LCM of 1, 2, 3, 4, 5, 6, 7, 8,9,10
∴  LCM = 2 x 2 x 3 x 2 x 3 x 5 x 7
 =2520

(i) (a) Pair of the equilateral triangles are similar figures.
(b) Pair of the squares are similar figures.
(ii) (a) A triangle and a quadrilateral form a pair of non-similar figures.
(b) A square and a circle form a pair of non-similar figures.

We have, last term, l = 184 and common difference, d = 10 - 7 = 3

\(\therefore\) 7th term from the end = l -(7-1)d
= \(184 - 6 \times 3\)
=  184 - 18 = 166

Let a be the first term and d be the common difference of given AP.
Now, according to the question, a₁₇ = a₁₀ + 7
\(\Rightarrow\)   a₁₇ - a₁₀ = 7
\(\Rightarrow\) [a + (17 - 1)d] - [a + (10 - 1)d] = 7
[\(\because\) a_n = a + (n - 1)d]
\(\Rightarrow\) (a + 16d) - (a + 9d) = 7
\(\Rightarrow\) 7d = 7 \(\Rightarrow\) d= 1
Hence, the common difference of this AP is 1.

Let P(x) = 2x⁴ - 9x³ + 5x² + 3x -1 
Since, \((2+\sqrt{3})\) and \((2-\sqrt{3})\) are the zeroes of P(x).
Therefore, product of the zeroes is the factors of P(x).
Now, \((x-(2+\sqrt{3}))(x-(2-\sqrt{3}))\) 
\(=x^{2}-x(2-\sqrt{3})-x(2+\sqrt{3})+\left(2^{2}-(\sqrt{3})^{2}\right.\) 
\(=x^{2}-2 x+x \sqrt{3}-2 x-x \sqrt{3}+(4-3)\) 
= x² - 4x + 1
Since,(x² - 4x + 1) is a factor of P(x). So, P(x) is divisible by (x² - 4x + 1).
By using long division,
 
\(\therefore \quad P(x)=\left(x^{2}-4 x+1\right)\left(2 x^{2}-x-1\right)\) 
Now, consider 2x²- x-1
= 2x² - (2 - 1)x - 1
[by splitting the middle term]
= 2x² - 2x + x-1= 2x(x -1) +1(x - 1)
= (x -1)(2x + 1
For zeroes (x -1)(2x +1)= 0 and \(2 x+1=0 \Rightarrow\) \(x=-\frac{1}{2}\) 
when \(x-1=0 \Rightarrow x=1\) 
So all zeroes are,\(1,-\frac{1}{2}, 2+\sqrt{3}, 2-\sqrt{3}\) 

Here, \(\frac { AB }{ DF } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } ,\frac { BC }{ EF } =\frac { 5 }{ 7.5 } =\frac { 2 }{ 3 } ,\frac { AC }{ DE } =\frac { 3 }{ 4.5 } =\frac { 2 }{ 3 } \)
As, \(\frac { AB }{ DF } =\frac { BC }{ EF } =\frac { AC }{ DE } \)
So, \(\triangle ABC\sim \triangle DFE\)  [by SSS similarity criterion]
Hence, figures (i) and (ii) are similar triangles, but no other pairs of triangles in the given figure are similar.

In \(\triangle ABC,\frac { AD }{ DC } =\frac { BE }{ EC } \)
\(\Rightarrow DE\parallel AB\)  [by converse of BPT]
\(\therefore \angle CDE=\angle A\quad \angle CED=\angle B\)   [corresponding angles]
Since, \(\angle CDE=\angle CED\)
\(\angle A=\angle B\)
CB = CA
[since, sides opposite to equal angles are also equal]
\(\therefore \triangle CAB\) is an isosceles triangle.

Given, rational number is \(\frac{1}{13}\).
Here, prime factors of 13 are not of the form 2ⁿ 5^m.
So, it will not have a terminating decimal expansion.
[ by using theorems 3 and 4]
Now, \(\frac { 1 }{ 13 } =0.076923076923...=\overline { 0.076923 } \)
Thus, \(\frac{1}{13}\) has non-terminating repeating decimal expansion.

Two digits numbers are 10,11,12,13,14,15,...,97,98,99 in which only 14,21,28,..., 98 are divisible by 7.
Here, 21 - 14 = 28 - 21 = 7.
So, this list of numbers is an AP, whose first term (a) = 14, common difference (d) = 7 and nth term = 98.
\(\therefore\)     a  +( n- d ) d = 98
\(\Rightarrow\)   14 + (n - 1) 7 = 98
\(\Rightarrow\)    14 + 7n - 7 = 98      \(\Rightarrow\)     7n=91
\(\Rightarrow\)         \(n=\frac{91}{7}=13\)
Hence, 13 numbers of two digits are divisible by 7.

D \(=[2(k-12)]^{ 2 }-4(k-12)\times 2\)
\( =4(k-12)^{ 2 }-8(k-12)\quad \)
For equal and real roots, D = 0
\(\Rightarrow \ 4(k-12)^{ 2 }-8(k-12)=0\)
\(\Rightarrow 4(k-12)(k-12-2)=0\)
\(\Rightarrow 4(k-12)(k-14)=0\)
\(\Rightarrow k=12\) or k=14 \(\because a\neq 0\)
\(\Rightarrow d k\neq 12;\ \therefore k=14.\)

Let present age of Aftab = x years and present age of Aftab's daughter = y years. 

1st Condition 
x-7=7(y-7)
\(\Rightarrow x-7=7 y-49\) 
\(\Rightarrow x-7 y=-42\) 
table

2nd condition :
Three years later 
x+3=3(y+3)
x+3=3y+9
x-3y=6
Table

Thus,the algebraic eqation are
x-7y+42=0 and x-3y-6=0

(i) Let AB and CD be the two trees of height a and b metre such that the trees are p metre apart i.e., AC = p. Let the lines AD and BC meet at O such that OL = hm.

Let CL = x and LA = y,
then x + y = p
In \(\Delta\)ABC and \(\Delta\)LOC,
\(\angle\)CAB = \(\angle\)CLO (each 90°)
\(\angle\)C= \(\angle\)C  (Common)
\(\therefore \Delta CAB\sim \Delta CLO\)  (AA Similarity)
\(\Rightarrow \frac { CA }{ CL } =\frac { AB }{ LO } \quad \)
\(\Rightarrow \frac { p }{ x } =\frac { a }{ h } \quad \)
\(\Rightarrow x=\frac { ph }{ a } ....(i)\)
In \(\Delta\)ALO and \(\Delta\)ACD,
\(\angle\)ALO = \(\angle\)ACD  (each 90°)
\(\angle\)A = \(\angle\)A   (Common)
\(\therefore \angle ALO\sim \angle ACD\ (AA \quad similarity)\)
\(\Rightarrow \frac { AL }{ AC } =\frac { OL }{ DC } \)
\(\Rightarrow \frac { y }{ p } =\frac { h }{ b } \)
\(\\ \\ \Rightarrow \quad y=\frac { ph }{ d } \)    .....(ii)
Adding eqns. (i) and (ii),
\(x+y=\frac { ph }{ a } +\frac { ph }{ b }\)
\(\Rightarrow p=ph\left( \frac { 1 }{ a } +\frac { 1 }{ b } \right) \)
\(\Rightarrow \frac { 1 }{ h } =\frac { 1 }{ a } +\frac { 1 }{ b } \)
\(\therefore h=\frac { ab }{ a+b } m.\)
(ii) Triangle. 
(iii) Trees are helpful to keep our life in this world. They should be saved at any cost.

The sequence is 103, 110, , 999
\(\therefore\) 999 = 103 + (n - 1) \(\times\) 7
\(\Rightarrow\) n = 129
Therefore \(\frac { 129+1 }{ 2 } \) = 65^th term is the middle term 
Middle term = 103 + (64)\(\times\)7 = 551
Sum of first 64 terms = 32 [206 + 63\(\times\)7]
= 20704
Sum of last 64 terms = 32 [1116 + 63 \(\times\) 7]
= 49824

2a-b, 2a+b

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

(a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion

Number on n^th spot = 20 + 4n i.e., tn = 20 + 4n
(i) (b): Number on 1^st spot = t₁ = 20 + 4(1) = 24
(ii) (c): Number on (n - 2)^th spot = t_{n - 2}
= 20 + 4 (n - 2)
= 20 + 4n - 8 = 12 + 4n
(iii) (a): Number on 34^th spot = t₃₄ = 20 + 4(34) = 156
(iv) (b): Here a = t₁ = 24
Now, t₂ = 20 + 4 (2) = 20 + 8 = 28
\(\therefore\) d = t₂ - t₁ = 4
So, required sum \(=S_{10}=\frac{10}{2}[2(24)+9(4)]=420\)
(v) (d): Let nth spot is numbered as 116.
\(\therefore\) t_n = 116
\(\Rightarrow 20+4 n=116 \Rightarrow 4 n=96 \Rightarrow n=24\)

(i) (b):Piyush sells a saree at 8% profit + sells a sweater at 10% discount = Rs 1008
\(\Rightarrow\) (100 + 8)% of x + (100 - 10)% of y = 1008
\(\Rightarrow\) 108% of x + 90% of y = 1008
\(\Rightarrow\) 1.08x + 0.9 y = 1008 ...(i)
(ii) (c): Piyush sold the saree at 10% profit + sold the sweater at 8% discount = Rs 1028
\(\Rightarrow\) (100 + 10)% of x + (100 - 8)% of Y = 1028
\(\Rightarrow\) 110%ofx+92%ofy= 1028
\(\Rightarrow\) 1.1 x + 0.92y = 1028 ...(ii)
(iii) (d): At x-axis, y = 0
\(\Rightarrow 1.08 x=1008 \Rightarrow x=\frac{1008}{1.08}=\frac{2800}{3}\)
(iv) (a): At y-axis, x = 0
\(\Rightarrow 0.92 y=1028 \Rightarrow y=\frac{1028}{0.92}=\frac{25700}{23}\)
(v) (b): Solving equations (i) and (ii), we get x = 600 and y = 400
Hence both linear equations intersect at (600, 400).

(i) (b): LCM of x and y = p³q³ and HCF of x and y = p²q Also, LCM = pq² x HCF.
(ii) (d): Number of marbles = 5m + 2 or 6n + 2.
Thus, number of marbles, p = (multiple of 5 x 6) + 2
= 30k + 2 = 2(15k + 1)
= which is an even number but not prime
(iii) (d): Here, required numbers
= HCF (398 - 7, 436 - 11,542 -15)
= HCF (391,425,527) = 17
(iv) (b): LCMof126and600 = 2 x 3 x 21 x 100= 12600 The least positive integer which on adding 1 is exactly divisible by 126 and 600 = 12600 - 1 = 12599
(v) (a): Here 8SC - 340A = 109 and 425A + 85B = 146 On adding them, we get 85A + 85B + 85C = 255 ~ A + B + C = 3, which is divisible by 3.