WBJEE · Maths · Basic of Mathematics
If \(x=\log _a b c, y=\log _b c a, z=\log _c a b\), then the value of \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\) will be
- A \(x+y+z\)
- B 1
- C \(a b+b c+c a\)
- D \(\mathrm{abc}\)
Answer & Solution
Correct Answer
(B) 1
Step-by-step Solution
Detailed explanation
Hints: \(1+x=\log _a a+\log _a b c=\log _a a b c\) \(\begin{aligned} & \frac{1}{1+x}=\log _{a b c} a, \text { Similarly } \frac{1}{1+y}=\log _{a b c} b \\ & \frac{1}{1+z}=\log _{a b c} c \text {, Ans. }=\log _{(a b c)} a b c=1 \end{aligned}\)
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