If $a{x^2} + \frac{b}{x} \ge c$ for all positive $x$, where , $a,b, > 0$ then

JEE MATHEMATICS MCQ: MAXIMA & MINIMA

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a{x^2} + \frac{b}{x} \ge c

for all positive

x

, where ,

a,b, > 0

then

View answer and explanation

Correct option: A

27a{b^2} \ge 4{c^3}

Option A (

27a{b^2} \ge 4{c^3}

) is correct because it represents the condition for the inequality to hold for all positive x, derived using calculus and optimization.

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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(x)=2 x^{3}+(2 p-7) x^{2}+3(2 p-9) x

x<0

x>0

p

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}