We have, \({ z }^{ 2 }=\overset { \_ }{ z } \quad \Rightarrow { (x+iy) }^{ 2 }\quad =x-iy\)
\(\Rightarrow { x }^{ 2 }+{ (iy) }^{ 2 }+2xyi=x-iy\)
\([\because { ({ z }_{ 1 }+{ z }_{ 2 }) }^{ 2 }={ z }_{ 1 }^{ 2 }+{ z }_{ 2 }^{ 2 }+2{ z }_{ 1 }{ z }_{ 2 }]\)
\( \Rightarrow { x }^{ 2 }-{ y }^{ 2 }+2xyi=x-y \quad [\because { i }^{ 2 }=-1]\)
On equating  real and imaginary parts, we get 
 \({ x }^{ 2 }-{ y }^{ 2 }=x\)  ... (i)  and
 \(2xy=-y\)  ...(ii)
From Eq. (i) and (ii), we have 
 2xy+y = 0 \(​​\Rightarrow y(2x+1)=0\)
\(\Rightarrow y=o\ or \ x= -\frac { 1 }{ 2 } \)
Case I When y=0.
In this case, we have \({ x }^{ 2 }=x\)   [From Eq. (i)]
\(\Rightarrow { x }^{ 2 }-x=0\ \Rightarrow x(x-1)=0\Rightarrow x=0\quad or\quad x=1\)
\(\because \ z=0+oi\ or\ z=1+oi\)
Case II When x = -\(\frac { 1 }{ 2 } \)
In this case, we have
\(\frac { 1 }{ 4 } -{ y }^{ 2 }=-\frac { 1 }{ 2 } \)    [from Eq. (i)]
\(\Rightarrow { y }^{ 2 }=\frac { 1 }{ 4 } +\frac { 1 }{ 2 } =\frac { 3 }{ 4 } \Rightarrow y=\pm \frac { \sqrt { 3 } }{ 2 } \)
\(\because \ z=-\frac { 1 }{ 2 } \pm i\frac { \sqrt { 3 } }{ 2 } \)
Hence, the solution of given equation are 0+0i, 1+0i, \(-\frac { 1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \quad and\quad -\frac { 1 }{ 2 } -i\frac { \sqrt { 3 } }{ 2 } \)

Now solving  \(\frac{^nCr-1}{^nCr} \)= \(\frac{36}{34}\)
and \(\frac{^nCr} {^nCr-1}\)  = \(\frac{84}{126} \) ,  we get 
r = 3
\(\therefore\)  ^rC₂ = ³C₂=3  
Ans. 3

At \(x=0,LHL=\lim _{ x\rightarrow { 0 }^{ - } }{ f\left( x \right) } =\lim _{ h\rightarrow 0 }{ \quad f\left( 0-h \right) } \)
\(\lim _{ h\rightarrow 0 }{ 2(0-h) } =3=3\)
\(RHL=\lim _{ x\rightarrow 0 }{ f\left( x \right) } =\lim _{ h\rightarrow 0 }{ 2(0-h) } =\lim _{ h\rightarrow 0 }{ 3(0+h+1)=3 } \)
\(LHL=\lim _{ x\rightarrow { 1 }^{ - } }{ f\left( x \right) } =\lim _{ h\rightarrow 0 }{ f(1-h) } =\lim _{ h\rightarrow 0 }{ 3(1+h+1)=6 } \)
\(At\quad x=1,RHL=\lim _{ x\rightarrow { 1 }^{ + } }{ f\left( x \right) } =\lim _{ h\rightarrow 0 }{ f(1+h) } =\lim _{ h\rightarrow 0 }{ 3(1+h+1)=6 } \)
Ans. \(\lim _{ x\rightarrow 0 }{ f\left( x \right) } =3\quad and\quad \lim _{ x\rightarrow 1 }{ f\left( x \right) } =6\)

\(\frac { 28363 }{ 36315 } \)

The four digit numbers greater than 5000are randomly formed from the digits 0,1,3,5,7.
(i) Given, the digits are repeated.
The number formed is greater than 5000, the leftmost digit is either 7or 5. The remaining three places can be filled by any of the digits 0,1,3,5or 7.
Total four digit number formed greater than 5000 is 2×5×5×5−1=249.
If unit digit of a number is 0 or 5 then number is divisible by 5.
Total numbers that are divisible by 5is 2×5×5×2−1=99.
Assume E be the event when number is divisible by 5and digits are repeating.
The probability is, P( E )= 99 249 = 33 249
Thus, the probability of number divisible by 5when digits repeated is 33 249 .
(ii) Given, repetition of digit is not allowed. Then the thousands place can be filled with digits either 5or 7and remaining three places can be filled with any of the remaining four digits.
Total four digit number formed greater than 5000is 2×4×3×2=48.
If the digit at thousand place is 5 then unit place can be filled with 0 and tens and hundreds places can be filled with any of the remaining three digits.
The four digits number formed started with 5and divisible with 5is 3×2=6.
If the digit at thousand place is 7 then unit place can be filled with 0or 5.
The four digit numbers starting with 7and divisible by 5is 1×2×3×2=12.
Total number greater than 5000and divisible by 5is, 6+12=18
The probability of a number formed is divisible by 5 with repetition of digits is, 18 48 = 3 8
Thus, the required probability is 3 8 .