If the functions f ( x ) = x 3 3 + 2 b x + a x 2 2 and g ( x ) = x 3 3 + a x + b x 2 , a 2 b , have a common extreme point, then a + 2 b + 7 is equal to [2023]

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Solution

Q.
If the functions $f (x) = \frac{x^{3}}{3} + 2 b x + \frac{a x^{2}}{2}$ and $g (x) = \frac{x^{3}}{3} + a x + b x^{2}, a \neq 2 b,$ have a common extreme point, then $a + 2 b + 7$ is equal to [2023]

1 $\frac{3}{2}$

2 3

3 6

4 4

Ans.
(3)

We have, $f (x) = \frac{x^{3}}{3} + 2 b x + \frac{a x^{2}}{2}$

and $g (x) = \frac{x^{3}}{3} + a x + b x^{2}, a \neq 2 b$

For critical points,

$f' (x) = x^{2} + 2 b + a x = 0 ...(i)$

$g' (x) = x^{2} + 2 b x + a = 0 ...(ii)$

Since $f (x)$ and $g (x)$ have a common extreme point,

$∴$ condition for a common root is

$α = \frac{a_{2} c_{1} - a_{1} c_{2}}{a_{1} b_{2} - a_{2} b_{1}} = \frac{b_{1} c_{2} - b_{2} c_{1}}{a_{2} c_{1} - a_{1} c_{2}}, α \neq 0$ $= \frac{2 b - a}{2 b - a} = \frac{a^{2} - 4 b^{2}}{2 b - a}$

$\Rightarrow \frac{(a + 2 b) (a - 2 b)}{- (a - 2 b)} = 1 \Rightarrow a + 2 b = - 1$

$∴$ $a + 2 b + 7 = - 1 + 7 = 6$

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