"Let the function $f(x) = 2{x^3} + (2p - 7){x^2} + 3(2p - 9)x - 6$ have a maxima for some value of $x 0$. Then, the set of all values of p is"

"$f^{\\prime}(x)=6 x^{2}+2 x(2 p-7)+3(2 p-9)$\n \n$x_{1} 0$\n \n$\\Rightarrow f^{\\prime}(0) \n$\\Rightarrow p \n$p \\in \\left( { - \\infty ,{9 \\over 2}} \\right)$"

Medium MCQ +4 / -1 PYQ · JEE Mains 2023

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

A $\left( { - {9 \over 2},{9 \over 2}} \right)$
B $\left( {{9 \over 2},\infty } \right)$
C $\left( {0,{9 \over 2}} \right)$
D $\left( { - \infty ,{9 \over 2}} \right)$ Correct answer

Solution

$f^{\prime}(x)=6 x^{2}+2 x(2 p-7)+3(2 p-9)$ $x_{1}<0, x_{2}>0$ $\Rightarrow f^{\prime}(0)<0$ $\Rightarrow p<\frac{9}{2}$ $p \in \left( { - \infty ,{9 \over 2}} \right)$

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Subject: Mathematics · Chapter: Application of Derivatives · Topic: Tangents and Normals

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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

Solution

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