CTET 2 Math and Science Jan 2016 Paper(HM)

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Question : 54 of 150

Marks: +1, -0

यदि

2 m-\;{1}/{2 m}=2

है, जहाँ

m ≠ 0,

तब

m^{2}+\;{1}/{16 m^{2}}

का मान है

$2 \;{1}/{2}$
2
$1 \;{1}/{2}$
4

Solution:

∵ \;\; 2 m-\;{1}/{2 m}=2

दोनों तरफ वर्ग करने पर,

4 m^{2}+\;{1}/{4 m^{2}}-2 \;\; =4

⇒ \;\; 4 m^{2}+\;{1}/{4 m^{2}} \;\; =6

⇒ \;\; m^{2}+\;{1}/{16 m^{2}} \;\; =\;{6}/{4}=\;{3}/{2}

\;\; =1 \;{1}/{2}

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