Question

A function satisfies
$\sf{3f(x) +2f(1-x)=5x}$
Find the value of a such that
$\sf{f(a)=13}$
I don't understand the method in my book.
For correct methods and answers I will mark as ☆brainliest☆
Explain it well!​

Answer 1

Answer:-

$\red{\bigstar}$$\large\boxed{\rm\blue{a = 3}}$

Given:-

$\sf 3f(x) + 2f(1-x) = 5x$

$\sf f(a)= 13$

Solution:-

• Substituting x = a

Hence,

→ $\sf 3f(a) + 2f(1-a) = 5a$

→ $\sf 3 \times 13 + 2f(1-a) = 5a$

→ $\sf 2f(1-a) = 5-39$

→ $\bf\green{f(1-a) = \dfrac{5-39}{2}}$

• Now, Substituting x = 1-a

Hence,

$\sf 3f(x) + 2f(1-x) = 5x$

→ $\sf 3f(1-a) + 2f(a) = 5(1-a)$

→ $\sf 3f(1-a) + 2 \times 13 = 5 - 5a$

→ $\sf 3f(1-a) = 5-5a -26$

→ $\sf 3f(1-a) = -5a - 21$

→ $\bf\green{f(1-a) = \dfrac{-5a-21}{3}}$

Now,

→ $\sf \dfrac{5a-39}{2} = \dfrac{-5a-21}{3}$

→ $\sf 3(5a-39) = 2(-5a-21)$

→ $\sf 15a - 117 = -10a - 42$

→ $\sf 15a + 10a = -42 + 117$

→ $\sf 25a = 75$

→ $\sf a = \dfrac{75}{25}$

→ $\large\boxed{\green{\bf a = 3}}$

Answer 2

★ \large\boxed{\rm\blue{a = 3}}

a=3

Given:-

\sf 3f(x) + 2f(1-x) = 5x3f(x)+2f(1−x)=5x

\sf f(a)= 13f(a)=13

Solution:-

• Substituting x = a

Hence,

→ \sf 3f(a) + 2f(1-a) = 5a3f(a)+2f(1−a)=5a

→ \sf 3 \times 13 + 2f(1-a) = 5a3×13+2f(1−a)=5a

→ \sf 2f(1-a) = 5-392f(1−a)=5−39

→ \bf\green{f(1-a) = \dfrac{5-39}{2}}f(1−a)=

2

5−39

• Now, Substituting x = 1-a

Hence,

\sf 3f(x) + 2f(1-x) = 5x3f(x)+2f(1−x)=5x

→ \sf 3f(1-a) + 2f(a) = 5(1-a)3f(1−a)+2f(a)=5(1−a)

→ \sf 3f(1-a) + 2 \times 13 = 5 - 5a3f(1−a)+2×13=5−5a

→ \sf 3f(1-a) = 5-5a -263f(1−a)=5−5a−26

→ \sf 3f(1-a) = -5a - 213f(1−a)=−5a−21

→ \bf\green{f(1-a) = \dfrac{-5a-21}{3}}f(1−a)=

3

−5a−21

Now,

→ \sf \dfrac{5a-39}{2} = \dfrac{-5a-21}{3}

2

5a−39

=

3

−5a−21

→ \sf 3(5a-39) = 2(-5a-21)3(5a−39)=2(−5a−21)

→ \sf 15a - 117 = -10a - 4215a−117=−10a−42

→ \sf 15a + 10a = -42 + 11715a+10a=−42+117

→ \sf 25a = 7525a=75

→ \sf a = \dfrac{75}{25}a=

25

75

→ \large\boxed{\green{\bf a = 3}}

a=3

Hope it's helpful for you...