Then | AX | is equal to : |
AB |
Let C(O, r) is a circle and let M is a point within it where O is the centre and r is the radius of the circle.
Let CD is another chord passes through point M.
We have to prove that AB < CD.
Now join OM and draw OL perpendicular to CD.
In right angle triangle OLM,
OM is the hypotenuse.
So OM > OL
? chord CD is nearer to O in comparison to AB.
? CD > AB
? AB < CD
So all chords of a circle of a circle at a given point within it, the smallest is one which is bisected at that point.
Hence required answer will be option D .
Required answer = [64 log10 2] + 1
= [ 64 x 0.3010 ] + 1
= 19.264 + 1
= 19 + 1
= 20
? loga x = ( logabx) / (logaba)
? The given expression = [[(logabx) / (logaba)] / [( logabx )] - ( logab)
= (1/logaba) - logab = logaab - logab = loga(ab/b)
= logaa = 1
Given Exp.= log23 x log 32 x log34 x log43
= (log3 / log2) x ( log2 / log3) x (log4 / log3) x (log3 / log4) = 1
log2 10 = log 10 / log 2
= 1 / log 2
= 1.0000 / 0.3010
= 1000 / 301
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