"If Rolle's theorem holds for the function $f(x) = {x^3} - a{x^2} + bx - 4$, $x \in [1,2]$ with $f'\left( {{4 \over 3}} \right) = 0$, then ordered pair (a, b) is equal to :"

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Accepted Answer

"$f(1) = f(2)$

$\Rightarrow 1 - a + b - 4 = 8 - 4a + 2b - 4$

$3a - b = 7$ ..... (1)

$f'(x) = 3{x^2} - 2ax + b$

$$ \Rightarrow f'\left( {{4 \over 3}} \right) = 0 \Rightarrow 3 \times {{16} \over 9} - {8 \over 3}a + b = 0$$

$\Rightarrow - 8a + 3b = - 16$ ..... (2)

$\therefore$ $a = 5,b = 8$"

If Rolle's theorem holds for the function $f(x) = {x^3} - a{x^2} + bx - 4$, $x \in [1,2]$ with $f'\left( {{4 \over 3}} \right) = 0$, then ordered pair (a, b) is equal to :

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If Rolle's theorem holds for the function $f(x) = {x^3} - a{x^2} + bx - 4$, $x \in [1,2]$ with $f'\left( {{4 \over 3}} \right) = 0$, then ordered pair (a, b) is equal to :

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