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1.Relation and Function
hard
The function $f$ satisfies the functional equation $3f(x) + 2f\left( {\frac{{x + 59}}{{x - 1}}} \right) = 10x + 30$ for all real $x \ne 1$. The value of $f(7)$ is
A
$8$
B
$4$
C
$-8$
D
$11$
Solution
(b) $3f(x) + 2f\left( {\frac{{x + 59}}{{x – 1}}} \right) = 10x + 30$
For $x = 7$, $3f(7) + 2f(11) = 70 + 30 = 100$
For $x = 11$, $3f(11) + 2f(7) = 140$
$\frac{{f(7)}}{{ – 20}} = \frac{{f(11)}}{{ – 220}} = \frac{{ – 1}}{{9 – 4}}$ ==> $f(7) = 4$.
Standard 12
Mathematics
