If ${\log _{10}}\left\{ {98 + \sqrt {{x^2} - 12x + 36} } \right\} = 2,$ then $x = $

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

1 of 1

Application +4 | -1

{\log _{10}}\left\{ {98 + \sqrt {{x^2} - 12x + 36} } \right\} = 2,

then

x =

View answer and explanation

Correct option: D

4,\;8

Option D (4, 8) is correct because both values satisfy the given logarithmic equation after considering the absolute value.

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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

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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)

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A P

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G P

a, b, c

\frac{3}{4}