JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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The minimum value of

f(x)=\frac{(x-2)^{10}+(x-2)^{11}+3(x-2)^{12}+(x-2)^{15}+(x-2)^{23}+(x-2)^{25}}{(x-2)^{15}}

\forall \, x > 2

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Correct option: D

Option D (8) is correct because applying AM >= GM to the terms results in a minimum value of 8.

More Questions from SEQUENCE OF SERIES

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A=\{1,2,3\}, B=\{3,4\}, C=\{4,5,6\}

{A} \cup({B} \cap {C})

A-(B \cap C)

(B \cup C)-A

a,b,c

a,x,b

b,y,c

{x^2},{b^2},{y^2}

P(x)=1-x+x^2-x^3+\ldots+x^{2012}

y

Q(y)=a_0+a_1 y+\ldots+a_{2012} y^{2012}

y=x-2

\sum\limits_{{i=0}}^{2012} a_i

\log \left(\frac{5 c}{a}\right), \log \left(\frac{3 b}{5 c}\right)

\log \left(\frac{a}{3 b}\right)

A P

a, b

G P

a, b, c

\frac{3}{4}