Question

If $f( {\rm{x}})=\rm{6x}$ and b = $\rm{\dfrac{a+c}{2}}$ then ,

a] $f( {\rm{a - 3)}} + f {\rm{(c - 3)}}= 2f {\rm{(b - 3)}}$

b] $\dfrac{ f{\rm{(a - 3)}} - f \rm{{(b - 3)}}}{f {\rm{(b - 3)}} - f {\rm{(c - 3)}}} = 1$

c] $\dfrac{ f{\rm{(a - 3)}} - f \rm{{(c - 3)}}}{f {\rm{(b - 3)}} - f {\rm{(a - 3)}}} = 2$

d] $f({\rm{a - 3)}} - f {\rm{(c - 3)}} = \rm 6(a - c)$

NOTE : This question has more than one Correct answer.​

Answer 1

Given :

$\sf f(x)=6x\ ,\ b=\dfrac{a+c}{2}$

$\to \sf c=2b-a$

Note : Make sure that every function is in terms of a and b variables only except for exceptions

Explanation :

════════════════════════════════

$\sf f(a-3)$

• f(x) = 6x

$\to \sf 6(a-3)\\\to \sf 6a-18$

$\sf f(b-3)\\\to \sf 6(b-3)\\\to \sf 6b-18$

$\sf f(c-3)\\\to \sf f(2b-a-3)\\\to \sf 6(2b-a-3)\\\to \sf 12b-6a-18$

════════════════════════════════

• Let's solve for a] ,

$\implies \sf{\bf{f(a-3)+f(c-3)}}$

$\to \sf 6a-18+12b-6a-18$

$\to \sf 12b-36$

$\to \sf2(6b-18)$

$\to \sf 2[6(b-3)]$

$\leadsto \sf{\bf{2f(b-3)}}$

Hence option a] is correct

________________________

• Let's solve for b] ,

$\implies \bf \dfrac{f(a-3)-f(b-3)}{f(b-3)-f(c-3)}$

$\to \sf \dfrac{6a-18-(6b-18)}{6b-18-(12b-6a-18)}$

$\to \sf \dfrac{6a-6b}{6b-12b+6a}$

$\to \sf \dfrac{6a-6b}{6a-6b}$

$\leadsto \bf\ \; 1$

Hence option b] is correct

________________________

• Let's solve for c] ,

$\implies \bf \dfrac{f(a-3)-f(c-3)}{f(b-3)-f(a-3)}$

$\to \sf \dfrac{6a-18-(12b-6a-18)}{6b-18-(6a-18)}$

$\to \sf \dfrac{12a-12b}{6b-6a}$

$\to \sf \dfrac{-12(b-a)}{6(b-a)}$

$\leadsto \bf\ \; -2$

Hence option c] is incorrect

________________________

• Let's solve for d] ,

Note : Function [LHS and RHS] is in terms of a  and c , so proceed through it

$\implies \bf f(a-3)-f(c-3)$

$\to \sf 6a-18-(6c-18)$

$\to \sf 6a-6c$

$\leadsto \bf 6(a-c)$

Hence option d] is correct

________________________

Correct Options : a , b , d

Incorrect Options: c

Answer 2

• $\bf {f( {x})= {6x}}$

• And $\bf{b\:=\:\dfrac{a+c}{2}}$

➪ c = 2b - a

__________________________

a] $\sf {f( {\rm{a - 3)}} + f {\rm{(c - 3)}}= 2f {\rm{(b - 3)}}}$

L.H.S ;-

$\huge{\color{aqua}\checkmark}$ $\bf \orange {f( {\rm{a - 3)}} + f {\rm{(c - 3)}}}$

• Given, f(x) = 6x

• ⇛ f (a - 3) = 6 (a - 3)

And

• ⇛ f (c - 3) = 6 (c - 3)

$\\ \bf{\implies\:6\:(a\:-\:3)\:+\:6\:(c\:-\:3)\:}$

$\\ \bf{\implies\:6a\:-\:18\:+\:6c\:-\:18\:}$

$\\ \bf{\implies\:6a\:+\:6c\:-\:36\:}$

R.H.S ;-

➪ $\bf \green {2f {\rm{(b - 3)}}}$

$\bf{\implies\:2\:\Big(\:f\:(b\:-\:3)\:\Big)\:}$

• Given, f(x) = 6x

• ⇛ f (b - 3) = 6 (b - 3)

$\\ \bf{\implies\:2\:\Big(\:6\:(b\:-\:3)\:\Big)\:}$

$\\ \bf{\implies\:2\:\Big(\:6b\:-\:18\:\Big)\:}$

$\\ \bf{\implies\:2\:\Big(\:6\times{\dfrac{a\:+\:c}{2}}\:-\:18\:\Big)\:}$

$\\ \bf{\implies\:2\:\Big(\:3\times(a\:+\:c)\:-\:18\:\Big)\:}$

$\\ \bf{\implies\:2\:\Big(\:3a\:+\:3c\:-\:18\:\Big)\:}$

$\\ \bf{\implies\:6a\:+\:6c\:-\:36\:}$

$\huge\therefore$ L.H.S = R.H.S . Hence it is a correct option .

__________________________

b] $\sf {\dfrac{ f{\rm{(a - 3)}} - f \rm{{(b - 3)}}}{f {\rm{(b - 3)}} - f {\rm{(c - 3)}}} = 1}$

L.H.S ;-

$\huge{\color{aqua}\checkmark}$ $\bf \orange {\dfrac{ f {\rm{(a - 3)}} - f {\rm{(b - 3)}}}{f {\rm{(b - 3)}} - f {\rm{(c - 3)}}}}$

$\\ \bf{\implies\:{\dfrac{6\:{(a\:-\:3)}\:-\:6\:{(b\:-\:3)}}{6\:{(b\:-\:3)}\:-\:6\:{(c\:-\:3)}}}\:}$

$\\ \bf{\implies\:{\dfrac{6a\:-\:18\:-\:6b\:+\:18}{6b\:-\:18\:-\:6c\:+\:18}}\:}$

$\\ \bf{\implies\:{\dfrac{6a\:-\:6b}{6b\:-\:6c}}\:}$

$\\ \bf{\implies\:{\dfrac{6a\:-\:6b}{6b\:-\:6\:(2b\:-\:a)}}\:}$

$\\ \bf{\implies\:{\dfrac{6a\:-\:6b}{6b\:-\:12b\:+\:6a}}\:}$

$\\ \bf{\implies\:{\dfrac{6a\:-\:6b}{-\:6b\:+\:6a}}\:}$

$\\ \bf{\implies\:{\dfrac{6a\:-\:6b}{6a\:-\:6b}}\:}$

$\\ \bf{\implies\:1}$

R.H.S ;-

➪ $\bf\green{1}$

$\huge\therefore$ L.H.S = R.H.S . Hence it is a correct option .

__________________________

c] $\sf {\dfrac{ f{\rm{(a - 3)}} - f \rm{{(c - 3)}}}{f {\rm{(b - 3)}} - f {\rm{(a - 3)}}} = 2}$

L.H.S ;-

$\huge{\color{aqua}\checkmark}$$\bf \orange {\dfrac{ f {\rm{(a - 3)}} - f {\rm{(c - 3)}}}{f {\rm{(b - 3)}} - f {\rm{(a - 3)}}}}$

$\\ \bf{\implies\:{\dfrac{6\:{(a\:-\:3)}\:-\:6\:{(c\:-\:3)}}{6\:{(b\:-\:3)}\:-\:6\:{(a\:-\:3)}}}\:}$

$\\ \bf{\implies\:{\dfrac{6a\:-\:18\:-\:6c\:+\:18}{6b\:-\:18\:-\:6a\:+\:18}}\:}$

$\\ \bf{\implies\:{\dfrac{6a\:-\:6c}{6b\:-\:6a}}\:}$

$\\ \bf{\implies\:{\dfrac{6a\:-\:6\:(2b\:-\:a)}{6b\:-\:6a}}\:}$

$\\ \bf{\implies\:{\dfrac{6a\:-\:12b\:+\:6a}{6b\:-\:6a}}\:}$

$\\ \bf{\implies\:{\dfrac{12a\:-\:12b}{6b\:-\:6a}}\:}$

$\\ \bf{\implies\:{\dfrac{12\:(a\:-\:b)}{6\:(b\:-\:a)}}\:}$

$\\ \bf{\implies\:{\dfrac{2\:(a\:-\:b)}{b\:-\:a}}\:}$

$\\ \bf{\implies\:{\dfrac{-\:2\:(b\:-\:a)}{b\:-\:a}}\:}$

$\\ \bf{\implies\:-\:2\:}$

R.H.S ;-

➪ $\bf\green{2}$

$\huge\therefore$ L.H.S ≠ R.H.S . Hence it is a incorrect option .

__________________________

d] $\sf {f({\rm{a - 3)}} - f {\rm{(c - 3)}} = \rm 6(a - c)}$

L.H.S ;-

$\huge{\color{aqua}\checkmark}$$\bf \orange {f {(a - 3)} - f {(c - 3)}}$

$\\ \bf{\implies\:6\:(a\:-\:3)\:-\:6\:(c\:-\:3)\:}$

$\\ \bf{\implies\:6a\:-\:18\:-\:6c\:+\:18\:}$

$\\ \bf{\implies\:6a\:-\:6c\:}$

R.H.S ;-

➪ $\bf \green {6 {(a - c)}}$

$\bf{\implies\:6a\:-\:6c\:}$

$\huge\therefore$ L.H.S = R.H.S . Hence it is a correct option .

__________________________

$\huge\red\therefore$ The correct options are (a) , (b) & (d).