Question

If , f(x) = 6x³ - ax² + x - 3 , find 'a' at x = -1/2

Answer 1

$\huge\mathfrak\blue{answer}$

$we \: have \: {6}^{3} - a {x }^{2} + x - 3 \\ on \: putting \: x = { \frac{1}{2} }^{3} in\:f(x ) \\ \\ \\ we \: get f( \frac{ - 1}{2}) = 6 { \frac{ - 1}{2} }^{3} - a { \frac{ - 1}{2} }^{2} + \frac{ - 1}{2} - 3 \\ 6 { \frac{ - 1}{2} }^{3} - a { \frac{ - 1}{2} }^{2} + \frac{ - 1}{2} - 3\\ \implies 6 { \frac{ - 1}{8} } - 1 (\frac{a}{4} ) - 1( \frac{1}{2}) - 1( \frac{3}{1} ) = 0 \\ \implies \frac{ - 3}{4} -1 (\frac{a}{4}) - 1( \frac{1}{2}) - 1( \frac{3}{1} ) = 0 \\ \implies \frac{ - 3}{4} + \frac{( - a)}{4} - \frac{1}{2} - 3 = 0 \\ \implies \frac{ - 17}{4} + \frac{( - a)}{4} = 0 \\ \implies \frac{4 \times \frac{ - 17}{4} }{4} + \frac{( - a)}{4} = 0 \\ \implies - \frac{ - 17 - a}{4} = 0 \\ \implies \frac{4( - a - 17)}{4} = 0 \\ \implies - a - 17 = 0 \\ \implies - a - 17 = 0 \\ = a = -17$

$\text{a=-17}$

$\text{hope\:it\: helps\:you}$