यदि घनाभ की भुजाएँ V1 = logabc a3, V2 = logabc b3 और V3 = logabc cहैं, तो V12+V22+V32+2(V1V2+V2V3+V1V3) का मान कितना है?

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MP Excise Constable Official Paper (Held On: 21 Feb 2023 Shift 2)
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  1. 16
  2. 9
  3. 8
  4. 27

Answer (Detailed Solution Below)

Option 2 : 9
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दिया गया है:

V1 = logabc a3, V2 = logabc b3 और V3 = logabc c3  

V12+V22+V32+2(V1V2+V2V3+V1V3) = ?

प्रयुक्त अवधारणा:

Logab = logcblogca

log10abc = log10a + log10b + log10c

(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

गणना:

V1 = logabc alog10a3log10abc = 3log10alog10abc 

V2 = logabc b3 log10b3log10abc = 3log10blog10abc

V3 = logabc c3 = log10c3log10abc = 3log10clog10abc

V1 + V2 + V3 = 3log10alog10abc + 3log10blog10abc3log10clog10abc

⇒ 3 × log10a+log10b+log10clog10a+log10b+log10c = 3

अब, V12+V22+V32+2(V1V2+V2V3+V1V3) 

⇒ (V1 + V2 + V3)2 = 32 = 9

∴ V12+V22+V32+2(V1V2+V2V3+V1V3) = 9

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