JEE MATHEMATICS MCQ: MAXIMA & MINIMA

1 Hard

1 of 1

Analyzing +4 | -1

The function

f\left( x \right) = ax + \frac{b}{x},\;b,\;x > 0

takes the least value at

x

equal to

View answer and explanation

Correct option: D

\sqrt {\frac{b}{a}}

Option D is correct because the function has a minimum at x =

\sqrt{{b/a}}

, which is found by setting the derivative equal to zero and confirming a positive second derivative.

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

m(b)

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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}

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x<0

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p

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\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}

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