f(3) = 6, f(-2) = Not defined, f(0) = Not defined,
\(f(\frac { 1 }{ 2 } )=\frac { -3 }{ 4 } ;\ f(2-h)={ h }^{ 2 }-h+3,f(-1+h)=h\)

Domain = R - {2, 6}

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. \(\lim _{ x\rightarrow 0 }{ f\left( x \right) } =3\quad and\quad \lim _{ x\rightarrow 1 }{ f\left( x \right) } =6\)

(a) Thousand's place can be filled in 2 ways(5 or 7).
(b) The remaining 3 places can be filled in 4 x 3 x 2 ways.
Number of 4digits numbers =  2 x 4 x 3 x 2
(c) If the number is divisible by 5 then its units place is either 0 or 5.
(d) If units place is 0, then thousands, place can be filled in 2 ways and other two places can be filled in 3 x 2 ways.
(e) If units place is 5, then thousands place can be filled in 1 way and other two places can be filled in 3 x 2 ways.
Number of 4digits numbers divisible by 3 
=   2 x 3 x 2 +1 x 3 x 2
\(=\frac { 3 }{ 8 } \)