Question

For x(-R. the function f(x) satisfies 2f(x) + f(1-x) = x^2. The value of f(4) is equal to​

Answer 1

Answer:

15

see the attachment

for the detailed explanation

Answer 2

SOLUTION

TO DETERMINE

Let f : R → R be a real valued function such that 2f(x) + f(1 – x) = x² then f(4) is equal to

EVALUATION

Here it is given that f : R → R be a real valued function such that

$\displaystyle \sf{ 2f(x) + f(1 - x) = {x}^{2} \: \: \: \: - - - - (1)}$

Replacing x by 1 - x in both sides we get

$\displaystyle \sf{ 2f(1 - x) + f(x) = {(1 - x)}^{2} }$

$\displaystyle \sf{ \implies f(x) +2f(1 - x) = {(x - 1)}^{2} \: \: \: \: - - - - (2)}$

2 × Equation 1 - Equation 2 gives

$\displaystyle \sf{ 3f(x) =2 {x}^{2} - {(x - 1)}^{2} }$

$\displaystyle \sf{ \implies 3f(x) =2 {x}^{2} - {x}^{2} + 2x - 1 }$

$\displaystyle \sf{ \implies 3f(x) = {x}^{2} + 2x - 1 }$

$\displaystyle \sf{ \implies f(x) = \frac{ {x}^{2} + 2x - 1 }{3} }$

Putting x = 4 we get

$\displaystyle \sf{ f(4) = \frac{ {4}^{2} + (2 \times 4) - 1 }{3} }$

$\displaystyle \sf{ \implies f(4) = \frac{ 16 + 8 - 1 }{3} }$

$\displaystyle \sf{ \implies f(4) = \frac{ 23 }{3} }$

FINAL ANSWER

$\displaystyle \sf{ f(4) = \frac{ 23 }{3} }$

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