Question

if f(x)=x^3-3x+1 and f(2a)=2f(a) find a.​

Answer 1

$\rm\red{Answer - }$

$\rm\pink{Given -}$

$\rm f(x) = x^3 - 3x + 1$

$\rm f(2a) = 2f(a)$

$\rm\pink{To \:find - }$

Value of a

$\rm\red{Solution - }$

$\rm f(x) = x^3 - 3x + 1$

$\bf Calculating \:f(2a)$

$\rm f(2a) = (2a)^3 - 3(2a) + 1$

$\rm f(2a) = 8a^3 - 6a + 1 \: \:\:\:\: \:\longrightarrow\:\:\: eq 1$

$\bf Calculating \:2f(a)$

$\rm 2 f(a) = 2 [ a^3 - 3a + 1 ]$

$\rm 2 f(a) = 2a^3 - 6a + 2 \: \:\:\: \:\longrightarrow\:\:\: eq 2$

$\bf Equating \:the \:equations -$

$\rm 8a^3 - 6a + 1 = 2a^3 - 6a + 2$

$\rm 8a^3 - 6a + 1 - ( 2a^3 - 6a + 2 ) = 0$

$\rm 6a^3 - 6a + 6a + 1 - 2 = 0$

$\rm 6a^3 - 1 = 0$

$\rm a = \sqrt[3]{\dfrac{1}{6}}$

$\boxed{\rm\red{a = \dfrac{1}{ \sqrt[3]{6} }}}$

Answer 2

Step-by-step explanation:

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