Let P(h,k) be the moving and A(0,2) and B(0,-2) be
the given points.By the given condition,PA + PB = 6
\(\Rightarrow \sqrt { { (h-0) }^{ 2 }+{ (k-2) }^{ 2 } } +\sqrt { { (h-0) }^{ 2 }+{ (k+2) }^{ 2 } } =6\)
\(\Rightarrow (2k+9)=3\sqrt { { h }^{ 2 }+{ (k+2) }^{ 2 } } \)
On squaring both sides, we get
(2k + 9)² = 9[h² + (k+2)²] \(\Rightarrow \) 9h² + 5k² = 45
9x² + 5y² = 45

\(6\times 0.1+6\times 0.11+6\times 0.111+....\)n terms
\(=\frac { 6 }{ 9 } [0.9+0.99+0.999+\)...n terms]
\(=\frac { 2 }{ 3 } \left[ \frac { 9 }{ 10 } +\frac { 99 }{ 100 } +\frac { 999 }{ 1000 } +...\quad n\quad terms \right] \)
\(=\frac { 2 }{ 3 } \left[ \left( 1-\frac { 1 }{ 10 } \right) +\left( 1-\frac { 1 }{ 100 } \right) +\left( 1-\frac { 1 }{ 1000 } \right) +...n\quad terms \right] \)
\(=\frac { 2 }{ 3 } \left[ (1+1+1+...n\quad terms)-(\frac { 1 }{ 10 } +\frac { 1 }{ { 10 }^{ 2 } } +\frac { 1 }{ { 10 }^{ 3 } } +...n\quad terms) \right] \) 
Ans.\(\frac { 2 }{ 3 } n-\frac { 2 }{ 27 } (1-10^{ -n })\)

\(\left( 53 \right) ^{ 7 }-\left( 47 \right) ^{ 7 }=\left( 50+3 \right) ^{ 7 }+\left( 50-3 \right) ^{ 7 }\\ \left( 50+3 \right) ^{ 7 }=^{ 7 }{ C }_{ 0 }{ \left( 50 \right) }^{ 7 }+^{ 7 }{ C }_{ 1 }{ \left( 5 \right) }^{ 6 }{ \left( 3 \right) }^{ 1 }+^{ 7 }{ C }_{ 2 }{ \left( 50 \right) }^{ 5 }{ \left( 2 \right) }^{ 2 }.........(i)\\ \left( 50-3 \right) ^{ 7 }=^{ 7 }{ C }_{ 0 }{ \left( 50 \right) }^{ 7 }-^{ 7 }{ C }_{ 1 }{ \left( 50 \right) }^{ 6 }{ \left( 3 \right) }^{ 1 }+^{ 7 }{ C }_{ 2 }{ \left( 3 \right) }^{ 2 }.........(ii)\\ Sub\quad equ(i)\quad and\quad (ii)\\ \left( 50+3 \right) ^{ 7 }-\left( 50-3 \right) ^{ 7 }=2\left[ ^{ 7 }{ C }_{ 1 }{ \left( 50 \right) }^{ 6 }{ \left( 3 \right) }^{ 1 }+^{ 7 }{ C }_{ 3 }{ \left( 50 \right) }^{ 4 }{ { \left( 3 \right) }^{ 3 }+^{ 7 }{ C }_{ 5 }{ \left( 50 \right) }^{ 2 }{ { \left( 3 \right) }^{ 5 }+^{ 7 }{ C }_{ 7 }{ { \left( 3 \right) }^{ 7 } } } } \right] \\ =668088019374\)

Thousand's place can be filled either by 1 or 2 or 3, i.e. thousand's place can be filled in 3 ways.
Ans. (i) 375
(ii) 72

Less than or equal to 4.

4

1, \(\frac { 9 }{ 2 }\), 14, \(\frac { 65 }{ 2 }\), 63

Given expression is (1-z)^4. Here, n= 4
\(\therefore \) The number of terms in the expansion is (n+1)
i.e. 4+1 = 5

First, flag can be selected in 5 ways and the second flag can be selected in 4 ways.
Ans. 20 

We have (3x² - 2ax + 3a²)³
=[(3x²- 2ax) + 3a²)]³
=³C₀(3x²- 2ax)³ + ³C₁(3x² - 2ax)² (3a²)+³C₂(3x²- 2ax) (3a²)²+ ³C₃(3a²)³
= (3x² - 2ax)³ + 3 x 3a² (3x² - 2ax)² + 3x 9a⁴ (3x² - 2ax) + 27a⁶
= (27x⁶ - 8a³x³ - 54ax⁵ + 36a²x⁴) + 9a² (9x⁴+ 4a²x² - 12ax³) + 27a⁴ (3x² - 2ax) + 27a⁶
= 27x⁶ - 108a³x³- 54ax⁵ + 36a²x⁴ + 81a²x⁴+ 36a⁴x² - 108a³x³ + 81a⁴x² - 54a⁵x + 27a⁶
= 27x⁶ - 54ax⁵ + 117a²x⁴ - 116a³x³
+ 117a⁴x² - 54a⁵x + 27 a⁶

Let m be the slope of required line
which passes through point (3, 2). Then equation of required line is
y - 2 = m(x - 3)... (i)
The equation of given line is x - 2y = 3
\(\Rightarrow y=\frac{x}{2}-\frac{3}{2}\)....(ii)
\(\therefore\)   Slope of given line is \(\frac{1}{2}\)
It is given that line (i) and (ii) make an angle of 45^o.
\(\therefore \tan 45^o=\left| \frac { m-\frac { 1 }{ 2 } }{ 1+\frac { m }{ 2 } } \right| \)
\(\therefore 1=|\frac{2m-1}{2+m}|\)
\(\Rightarrow \frac{2m-1}{2+m}=\pm 1\)
When \(\frac{2m-1}{2+m}=1\)
\(\Rightarrow\) 2m - 1 = 2 + m \(\Rightarrow\) m = 3
Then equation of required line is
y - 2 = 3(x - 3)
\(\Rightarrow\) y - 2 = 3x - 9 \(\Rightarrow\) 3x - y - 7 = 0
When \(\frac{2m-1}{2+m}=-1\Rightarrow\) 2m - 1 = -2 -m
\(\Rightarrow\)  3m = -1 \(\Rightarrow\) m = \(\frac{-1}{3}\)
Then the equation of required line is
\(y-2=\frac{-1}{3}(x-3)\)
\(\Rightarrow\)  3y - 6 = -x + 3 \(\Rightarrow\) x + 3y - 9 = 0.

Let S_n = 8 + 88 + 888 + 8888 + .... upto n terms
= 8 [1 + 11 + 111 + 1111 +.... upto n terms]
= \(\frac { 8 }{ 9 } \) [9 + 99 + 999 + 9999 + ... upto n terms]
= \(\frac { 8 }{ 9 } \) [(10 - 1)+(10² - 1) + (10³ - 1)] +....upto n terms
= \(\frac { 8 }{ 9 } \) [(10 + 10² +10³ + ...n terms) - (1+1+...n terms)]
= \(\frac { 8 }{ 9 } \left[ \frac { 10.\left( { 10 }^{ n }-1 \right) }{ \left( 10-1 \right) } -n \right] \)
= \(\frac { 8 }{ 9 } \left[ \frac { 10 }{ 9 } \left( { 10 }^{ n }-1 \right) -n \right] \)
= \(\frac { 80 }{ 81 } \left[ { 10 }^{ n }-1 \right] -\frac { 8 }{ 9 } n\)

(i) (a)
(ii) (c)
(iii) (b)
(iv) (a)
(v) (b)

(i) (b)
(ii) (d)
(iii) (a)
(iv) (a)
(v) (b)