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Let $\mathrm{f(x) \text { and } g(x)}$ be two real polynomials of degree $\mathrm{2\: and\: 1}$ respectively. If $\mathrm{f(g(x))=8 x^{2}-2 x, \text { and } g(f(x))=4 x^{2}+6 x+1}$ , then the value of $\mathrm{f(2)+g(2)}$ is ______.
Option: 1
18

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{f(g(x))=8 x^{2}-2 x} \\$

$\mathrm{g(f(x))=4 x^{2}+6 x+1} \\$

$\mathrm{\text { Let } f(x)=a x^{2}+b x+c \text { and } g(x)=d x+e}$

$\mathrm{f(g(x)) =a(d x+e)^{2}+b(d x+e)+c} \\$

$\mathrm{=a d^{2} x^{2}+2 a d e x+a e^{2}+b d x+b e+c} \\$

$\mathrm{=a d^{2} x^{2}+(2 a d e+b d) x+\left(a e^{2}+b e+c\right)}$

$\mathrm{g(f(x)) =d\left(a x^{2}+b x+c\right)+e }\\$

$\mathrm{=a d x^{2}+b d x+(c d+e)}$

Comparing

$\mathrm{a d=4, a d^{2}=8}\\$

$\mathrm{Dividing\: d=2 \Rightarrow a=2}\\$

$\mathrm{And \: b d=6 \Rightarrow b=3}\\$

$\mathrm{And \: 2 a d e+b d=-2}\\$

$\mathrm{8 e+6=-2 \Rightarrow e=-1}\\$

$\mathrm{And\: c d+e=1 \Rightarrow 2 c-1=1 \Rightarrow c=1}$

$\mathrm{\therefore f(x) =2 x^{2}+3 x+1, g(x)=2 x-1 }\\$

$\mathrm{\therefore f(2)+g(2) =(8+6+1)+(4-1)} \\$

$\mathrm{=15+3} \\$

$\mathrm{=18}$

Hence answer is $\mathrm{18}$

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