The function f satisfies the functional equation 3 f ( x ) + 2 f ( x + 59 x - 1 ) = 10 x + 30 for all real x 1 . The value of f ( 7 ) is

Question

The function $f$ satisfies the functional equation $3 f (x) + 2 f (\frac{x + 59}{x - 1}) = 10 x + 30$ for all real $x \neq 1$ . The value of $f (7)$ is

Smart Achievers · Accepted Answer

(2)

$Given functional equation is$

$3 f (x) + 2 f (\frac{x + 59}{x - 1}) = 10 x + 30$ ...(i)

$Put x = \frac{x + 59}{x - 1}, we get$

$3 f (\frac{x + 59}{x - 1}) + 2 f [\frac{\frac{x + 59}{x - 1} + 59}{\frac{x + 59}{x - 1} - 1}] = 10 (\frac{x + 59}{x - 1}) + 30$

$\Rightarrow$ $3 f (\frac{x + 59}{x - 1}) + 2 f (x) = \frac{40 x + 560}{x - 1}$ ...(ii)

$Multiply (i) by 3 and (ii) by 2 and then subtract, we get$

$9 f (x) - 4 f (x) = (30 x + 90) - (\frac{80 x + 1120}{x - 1})$

$\Rightarrow$ $5 f (x) = 10 [(3 x + 9) - (\frac{8 x + 112}{x - 1})]$

$\Rightarrow$ $f (x) = 2 [(3 x + 9) - (\frac{8 x + 112}{x - 1})]$

$Putting x = 7, we get$

$f (7) = 2 [30 - \frac{56 + 112}{6}] = 2 (30 - 28) = 4$

Solution

Q.
The function $f$ satisfies the functional equation $3 f (x) + 2 f (\frac{x + 59}{x - 1}) = 10 x + 30$ for all real $x \neq 1$ . The value of $f (7)$ is

1 8

2 4

3 - 8

4 11

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Solution

Q. The function f satisfies the functional equation 3f(x)+2f(x+59x-1)=10x+30 for all real x≠1. The value of f(7) is

1 8

2 4

3 - 8

4 11

Want Us to Email you About Special Offers & Updates?

Q.
The function $f$ satisfies the functional equation $3 f (x) + 2 f (\frac{x + 59}{x - 1}) = 10 x + 30$ for all real $x \neq 1$ . The value of $f (7)$ is