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AP EAMCET · Maths · Statistics

The mean of numbers \(a, b, 8,5,10\) is 6 and their variance is 6.80. Then \(\operatorname{Tan}^{-1} \frac{1}{a}+\operatorname{Tan}^{-1} \frac{1}{b}=\)

  1. A \(\operatorname{Tan}^{-1} \frac{7}{12}\)
  2. B \(\operatorname{Tan}^{-1}\left(-\frac{7}{11}\right)\)
  3. C \(\operatorname{Tan}^{-1} \frac{11}{7}\)
  4. D \(\operatorname{Tan}^{-1} \frac{7}{11}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\operatorname{Tan}^{-1} \frac{7}{11}\)

Step-by-step Solution

Detailed explanation

\(\frac{a+b+8+5+10}{5}=6\) \(a+b=30-23=7\) \(\frac{a^2+b^2+8^2+5^2+10^2}{5}-6^2=6.80\) \(\frac{a^2+b^2+64+25+100}{5}=36+6.80=42.80\) \(a^2+b^2+189=214\) \(a^2+b^2=25\) \((a+b)^2=a^2+b^2+2ab\) \(7^2=25+2ab\) \(49=25+2ab\) \(2ab=24\) \(ab=12\) From \(a+b=7\) and \(ab=12\),…