Question

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

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

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

$c] \dfrac{ f{\rm{(a - 3)}} - f \rm{{(c - 3)}}}{f {\rm{(b - 3)}} - f {\rm{(a - 3)}}} = 2$
$d] f({\rm{a - 3)}} - f {\rm{(c - 3)}} = \rm 6(a - c)f(a−3)−f(c−3)=6(a−c)$


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

Answer 1

Answer:

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Answer–}}}}$

Given:

$→f(x) = 6x \: \: \: \: \: \: \: \: →b = \frac{a + c}{2} \\ \\→ c = 2b - a$

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

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$⟹f(a - 3) \\ →f(x) = 6x \\ \\ →6(a - 3) \\ →6a - 18 \\ \\ →f(b - 3) \\ →6(b - 3) \\→ 6b - 18 \\ \\ →f(c - 3) \\→ f(2b - a - 3) \\→12b - 6a - 18$

Let's solve for a,

$⟹f(a - 3) + f(c - 3) \\ →6a - 18 + 12b - 6a - 18 \\ →12b - 36 \\→ 2(6b - 18) \\ →2f(b - 3) \\ \\ \\ Hence \: Option \: a \: is \: Correct$

Let's solve for b,

$\frac{⟹ f(a−3)−f(b−3)}{f(b−3)−f(c−3)} \\ \\ \begin{gathered}\to \sf \dfrac{6a-18-(6b-18)}{6b-18-(12b-6a-18)}\\\\\end{gathered} \\ \begin{gathered}\to \sf \dfrac{6a-6b}{6b-12b+6a}\\\\\end{gathered} \\ \frac{→6a−6b}{6a−6b} \\ \leadsto \bf\ \; 1 \\Hence  \: Option \: b \: is \: Correct$

Let's solve for d,

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

$⟹f(a−3)−f(c−3) \\ \\ \to \sf 6a-18-(6c-18) \\ \\ \to \sf 6a-6 c\\ \\ \leadsto \bf 6(a-c) \\ \\Hence  \: Option \: \: d \: \: is \: \: Correct$

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Correct Options : a , b , d

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Answer 2

Answer:

Given :

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

2

a+c

\to \sf c=2b-a→c=2b−a

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

Explanation :

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\sf f(a-3)f(a−3)

f(x) = 6x

\begin{gathered}\to \sf 6(a-3)\\\to \sf 6a-18\end{gathered}

→6(a−3)

→6a−18

\begin{gathered}\sf f(b-3)\\\to \sf 6(b-3)\\\to \sf 6b-18\end{gathered}

f(b−3)

→6(b−3)

→6b−18

\begin{gathered}\sf f(c-3)\\\to \sf f(2b-a-3)\\\to \sf 6(2b-a-3)\\\to \sf 12b-6a-18\end{gathered}

f(c−3)

→f(2b−a−3)

→6(2b−a−3)

→12b−6a−18

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Let's solve for a] ,

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

\begin{gathered}\to \sf 6a-18+12b-6a-18\\\end{gathered}

→6a−18+12b−6a−18

\to \sf 12b-36→12b−36

\to \sf2(6b-18)→2(6b−18)

\to \sf 2[6(b-3)]→2[6(b−3)]

\leadsto \sf{\bf{2f(b-3)}}⇝2f(b−3)

Hence option a] is correct

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Let's solve for b] ,

\begin{gathered}\implies \bf \dfrac{f(a-3)-f(b-3)}{f(b-3)-f(c-3)}\\\\\end{gathered}

⟹

f(b−3)−f(c−3)

f(a−3)−f(b−3)

\begin{gathered}\to \sf \dfrac{6a-18-(6b-18)}{6b-18-(12b-6a-18)}\\\\\end{gathered}

→

6b−18−(12b−6a−18)

6a−18−(6b−18)

\begin{gathered}\to \sf \dfrac{6a-6b}{6b-12b+6a}\\\\\end{gathered}

→

6b−12b+6a

6a−6b

\begin{gathered}\to \sf \dfrac{6a-6b}{6a-6b}\\\\\end{gathered}

→

6a−6b

6a−6b

\leadsto \bf\ \; 1⇝ 1

Hence option b] is correct

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Let's solve for c] ,

\begin{gathered}\implies \bf \dfrac{f(a-3)-f(c-3)}{f(b-3)-f(a-3)}\\\\\end{gathered}

⟹

f(b−3)−f(a−3)

f(a−3)−f(c−3)

\begin{gathered}\to \sf \dfrac{6a-18-(12b-6a-18)}{6b-18-(6a-18)}\\\\\end{gathered}

→

6b−18−(6a−18)

6a−18−(12b−6a−18)

\begin{gathered}\to \sf \dfrac{12a-12b}{6b-6a}\\\\\end{gathered}

→

6b−6a

12a−12b

\begin{gathered}\to \sf \dfrac{-12(b-a)}{6(b-a)}\\\\\end{gathered}

→

6(b−a)

−12(b−a)

\leadsto \bf\ \; -2⇝ −2

Hence option c] is incorrect

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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)⟹f(a−3)−f(c−3)

\to \sf 6a-18-(6c-18)→6a−18−(6c−18)

\to \sf 6a-6c→6a−6c

\leadsto \bf 6(a-c)⇝6(a−c)

Hence option d] is correct

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Correct Options : a , b , d

Incorrect Options: c