JEE MAIN - Mathematics (2023 - 25th January Evening Shift - No. 9)
Let the function $$f(x) = 2{x^3} + (2p - 7){x^2} + 3(2p - 9)x - 6$$ have a maxima for some value of $$x < 0$$ and a minima for some value of $$x > 0$$. Then, the set of all values of p is
$$\left( { - {9 \over 2},{9 \over 2}} \right)$$
$$\left( {{9 \over 2},\infty } \right)$$
$$\left( {0,{9 \over 2}} \right)$$
$$\left( { - \infty ,{9 \over 2}} \right)$$
Explanation
$f^{\prime}(x)=6 x^{2}+2 x(2 p-7)+3(2 p-9)$
$x_{1}<0, x_{2}>0$
$\Rightarrow f^{\prime}(0)<0$
$\Rightarrow p<\frac{9}{2}$
$$p \in \left( { - \infty ,{9 \over 2}} \right)$$
$x_{1}<0, x_{2}>0$
$\Rightarrow f^{\prime}(0)<0$
$\Rightarrow p<\frac{9}{2}$
$$p \in \left( { - \infty ,{9 \over 2}} \right)$$
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