k = 2

In ΔPAC and ΔPDB,
∠P = ∠P (Common)
∠PAC = ∠PDB (Exterior angle of a cyclic quadrilateral is ∠PCA = ∠PBD equal to the opposite interior angle)
∴ ΔPAC ∼ ΔPDB

( -7,9 )

The multiples of 4 between 10 and 250 be 12, 16, 20, ..., 248.
Here, a = 12, d = 16 - 12 = 4 and a_n = 248.
From formula, a_n = a + ( n - 1 )d, we get
12 + ( n - 1 )4 = 248
\(\Rightarrow\)  4 ( n - 1) = 248 - 12   \(\Rightarrow\)  4 ( n - 1 ) = 236
\(\Rightarrow\)  n - 1 = \({236 \over 4}\)  =  59   \(\Rightarrow\)  n = 59 + 1 = 60
\(\because\)         S_n = \({n\over2}\) ( a + l )    \(\Rightarrow\)   S₆₀ = \({60\over2}\) ( 12 + 248 ) = 30 x 260 = 7800

Simplify the given expression
7\(\times\)11\(\times\)13 + 2\(\times\)11 = 1001 + 22 = 1023
Prime factorisation of 1023 = 3\(\times\)11\(\times\)31. 
Thus, 1023 is divisible by 3, 11 and 31.
Since, a prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.
Therefore, 1023 is not a prime number as it has other divisor other than 1 and itself.
Hence,(7\(\times\)11\(\times\)13 + 2\(\times\)11) is not a prime number.

 0.4

PQ || MN
So, \(\frac { KP }{ PM } =\frac { KQ }{ QN } \)
\(\Rightarrow \quad \frac { KP }{ PM } =\frac { KQ }{ KN-KQ } \)
\(\Rightarrow \quad \frac { 4 }{ 13 } =\frac { KQ }{ 20.4-KQ } \)
\(\Rightarrow \) 4 x 20.4-4KQ=13 KQ
\(\Rightarrow \)17 KQ=4 x 20.4
\(\therefore \quad KQ=\frac { 20.4\times 4 }{ 17 } =4.8\quad cm,\)

True, since, \(-\frac { b }{ a } \)= sum of the zeroes <0, so \(\frac { b }{ a } \) >0. Also, the product of the zeroes =\(\frac { c }{ a } \) >0
∴ a,b and c all have same sign.

a_n=6-n
a₁=6-1=5
a₂=6-2=4
a₃=6-3=3
sequence is 5,4,3,.......

Infintely many solutions

Let us assume that \(\frac{2\sqrt 3}{5} \)is rational number. Then, it will be
of the form \( \frac{a}{b} \) where a, b are coprime integers and b \(\neq\)0 
Now \(\frac{2\sqrt 3}{5}\) \(​​\frac{a}{b}\)
On rearranging, we get \(\frac{5a -2b}{b}\) = \( \sqrt{3} \)
Since, 5 and  are rational. So, \(\frac{5a -2b}{b}\)  Will be rational
\(\therefore\) \( \sqrt{3} \) is rational.
But given that, \( \sqrt{3} \) is irrational So, this contradicts the fact that \( \sqrt{3} \) is irrational. Therefore, our assumption is wrong.
Hence,  \(\frac{ 2+ \sqrt 3}{5}\)is Irrational

In quadrilateral ABCD,
\( \frac { AO }{ BO } =\frac { CO }{ DO } \)
\(\Rightarrow \frac { AO }{ CO } =\frac { BO }{ DO } \)

In ABD, EO || AB (Construction)
(By BPT)    ...(ii)
From eqns. (i) and (ii),
\(\frac { AE }{ ED } =\frac { AO }{ CO } \)
In ADC,   \(\frac { AE }{ ED } =\frac { AO }{ CO } \)
\(\Rightarrow\)EO || DC (Converse of BPT)
EO || AB (Construction)
\(\therefore\) AB || DC
\(\Rightarrow\)In quad. ABCDm AB || DC
\(\Rightarrow\)ABCD is a trapezium.
Hence proved.

Given, \(\frac { 16 }{ x } -1=\frac { 15 }{ x+1 } \)
\(\frac { 16 }{ x } -\frac { 15 }{ x+1 } =1\\ \frac { 16(x+1)-15x }{ x(x+1) } =1 \)
16x+16-15x=x²+x
x+16=x²+x
x²=16
x²=\(\pm\)4
Hence, the roots are 4 and -4.

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.

A is point of intersection of line x=2 and 4x+5y=20
⇒ 4 x 2 + 5y =20  ⇒ \(y={12\over 5}\)
Coordinates of A are \(\left(2, {12\over 5}\right)\)
B is point of intersection of x=2 and y=0
Coordinates of A are (2, 0)
C is point of intersection y=0 and 4x+5y=20
⇒ 4x+5x0=20 ⇒ x=5
⇒ Coordinates of C are (5, 0)
Area ΔABC
\(={1\over 2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|\)
\(={1\over 2}\left|2(0-0)+2\left(0-{12\over 5}\right)+5\left({12\over 5}-0\right)\right|\)
\(={1\over 2}\left| {-24\over 5}+12\right|={1\over2}\times{36\over 5}={18\over5}=\)sq.units= 3.6sq. units

Let p(x) = ax³ + bx² + cx + d
Given that, one of the zeroes of the cubic polynomial p(x) is zero.
Let \(\alpha,\beta\) and \( \gamma\) arethe zeroes of cubic polynomial p (x),
where a = 0.
We know that,
Sum of product of two zeroes at a time =\(\frac{c}{a}\)
\(​​​​\alpha\beta +\beta\gamma+\beta​​​​\alpha=\frac{c}{a}\)
\(​​​​\alpha\beta +\beta\gamma+\gamma*0=\frac{c}{a}[\because \alpha=0,given]\)
  \(0+\beta\gamma+0=\frac{c}{a}\Rightarrow\beta\gamma=\frac{c}{a}\)
Hence, product of other two zeroes =\(\frac{c}{a}\)

(i) Number of throws on first day a=40
Number of throws on second day =40+12=52
AP formed by number of throws is
40,52,(52+12)… 
(I day) (II day)  (III day) 
Here, common difference, d=12
Number of throws she practiced on 11th day
a₁₁=a+(11−1)d
=40+10×12=160
(ii) Distance of Sanjitha's throw on first day, a=756 m Distance of Sanjitha throw after a week
=(7.56+0.09)m=7.65 m
∴ AP formed by the distances is
7.56,7.65,7.65+(0.09),…… 
(I week) (II week) (III week) 
Here, common difference, d=0.09 m
Sanjitha's throw distance at the end of 6 months (i.e. 26 weeks)
a₂₆=a+(26−1)d
=7.56+25×0.09
=7.56+2.25=9.81 m 
Or 
\( \text { Here, } a_n=11.16=a+(n-1) d\)
\(\Rightarrow 7.56+(n-1) 0.09=11.16 \)
\( \Rightarrow \quad n-1=\frac{3.6}{0.09} \Rightarrow n=41\)
∴ On 41st week, she will be able to achieve a throw of 11.16 m .
(iii) Total throws done by her during the entire camp of 15 days \(=S_n=\frac{n}{2}[2 a+(n-1) d]\)
where, n is total number of terms
\(\Rightarrow \quad S_{15}  =\frac{15}{2}[2 \times 40+14 \times 12]\)
=15[-10+8.1] 
\( =15 \times 124=1860\)

(i) (d)
(ii) (b)
(iii) (a): x(x + 3) + 7 = 5x - 11
\(\Rightarrow x^{2}+3 x+7=5 x-11\)
\(\Rightarrow x^{2}-2 x+18=0 \) is a quadratic equation.
\((b) (x-1)^{2}-9=(x-4)(x+3)\)
\(\Rightarrow x^{2}-2 x-8=x^{2}-x-12\)
\(\Rightarrow x-4=0\) is not a quadratic equation.
\((c) x^{2}(2 x+1)-4=5 x^{2}-10\)
\(\Rightarrow 2 x^{3}+x^{2}-4=5 x^{2}-10\)
\(\Rightarrow 2 x^{3}-4 x^{2}+6=0\) is not a quadratic equation.
\((d) x(x-1)(x+7)=x(6 x-9)\)
\(\Rightarrow x^{3}+6 x^{2}-7 x=6 x^{2}-9 x\)
\(\Rightarrow x^{3}+2 x=0\)   is not a quadratic equation.
(iv) (d)
(v) (d)

(i) (b): Given, a and \(\beta\) are the zeroes of 
\(p(x)=x^{2}-24 x+128\)
\(\text { Putting } p(x)=0 \text { , we get }\)
\( x^{2}-8 x-16 x+128=0 \)
\(\Rightarrow x(x-8)-16(x-8)=0 \)
\(\Rightarrow (x-8)(x-16)=0 \Rightarrow x=8 \text { or } x=16 \)
\(\therefore \alpha=8, \beta=16\)
(ii) (c) : \(\alpha+\beta+\alpha \beta =8+16+(8)(16) =24+128=152 \)
(iii) (d) : \(p(2)=2^{2}-2 4(2)+128=4-48+128=84\)
(iv) (a): Since a and \(\beta\) are zeroes of \(x^{2}+x-2\)
\(\therefore \quad \alpha+\beta=-1 \text { and } \alpha \beta=-2 \)
\(\text { Now, } \frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha \beta}=\frac{-1}{-2}=\frac{1}{2}\)
(v) (c): Sum of zeroes \(=\frac{-2}{k}\)
Product of zeroes \(=\frac{3 k}{k}=3\)
According to question, we have \(\frac{-2}{k}=3\)
\(\Rightarrow \quad k=\frac{-2}{3}\)