JEE MATHEMATICS MCQ: MAXIMA & MINIMA

1 Easy

1 of 1

Analyzing +4 | -1

Let

a_1, a_2, a_3, \ldots

be an A. P. with

a_6=2

. Then the common difference of this A.P., which maximises the product

a_1 a_4

a_5

, is :

View answer and explanation

Correct option: B

\frac{8}{5}

Option B (

\frac{8}{5}

) is correct. By expressing the product in terms of the common difference and maximizing it using calculus, the common difference is found to be

\frac{8}{5}

More Questions from MAXIMA & MINIMA

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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}

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x<0

x>0

p

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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}