If A = 1 + 2P and B = 1 + 2-P, then what is the value of B?
A. $$\frac{{A + 1}}{{A - 1}}$$
B. $$\frac{{A + 2}}{{A + 1}}$$
C. $$\frac{A}{{A - 1}}$$
D. $$\frac{{A - 2}}{{A + 1}}$$
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & A = 1 + {2^P} \cr & B = 1 + {2^{ - P}} \cr & {\text{Put }}P = 1 \cr & A = 3,\,B = \frac{3}{2} \cr & {\text{Option C is correct}} \cr & \cr & {\bf{Alternate:}} \cr & A = 1 + {2^P}\,......\,\left( {\text{i}} \right) \cr & B = 1 + {2^{ - P}} \cr & = 1 + \frac{1}{{{2^P}}} \cr & = \frac{{{2^P} + 1}}{{{2^P}}} \cr & = \frac{A}{{A - 1}}\,\,\,\,\,\left( {{\text{from equation }}\left( {\text{i}} \right)} \right) \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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