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

$ \left(0, \frac{9}{2}\right) $

$ \left(\frac{9}{2}, \infty\right) $

$ \left(-\frac{9}{2}, \frac{9}{2}\right) $

JEE MATHEMATICS MCQ: MAXIMA & MINIMA

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Analyzing +4 | -1

Let the function

f(x)=2 x^{3}+(2 p-7) x^{2}+3(2 p-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

View answer and explanation

Correct option: D

\left(-\infty, \frac{9}{2}\right)

Option D (

\left(-\infty, \frac{9}{2}\right)

) is correct because the condition for maxima at x < 0 and minima at x > 0 implies that

f'(0) < 0

, leading to

p < \frac{9}{2}

Practice All Chapter MCQs →

f(x)=\left(1+b^2\right) x^2+2 b x+1

m(b)

f(x)

b

m(b)

f: R \rightarrow R

f(x)=(x-3)^{n_1}(x-5)^{n_2}, n_1, n_2 \in N

|2 a x-3|+|a x+1|+|5-a x|=\frac{1}{2}

f\left( x \right)

f\left( x \right) = {\tan ^{ - 1}}x - \frac{1}{2}{\log _e}x

\left[ {1/\sqrt 3 ,\;\sqrt 3 } \right]

f(x)=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^{2} x}, x \in\left(0, \frac{\pi}{2}\right),

\frac{k}{e}

\left(\frac{k}{e}\right)^{8}+\frac{k^{8}}{e^{5}}+k^{8}