Question

If \( \text{sech}^{-1}x = \log 2 \) and \( \text{cosech}^{-1}y = -\log 3 \), then \( (x+y) = \)

• A. \( \frac{1}{6} \)
• B. \( \frac{1}{20} \)
• C. \( 6 \)
• D. \( 20 \) Correct Answer

Hint:

Know the definitions or logarithmic forms of inverse hyperbolic functions: \( \text{sech}^{-1}x = \cosh^{-1}(1/x) = \log(1/x + \sqrt{(1/x)^2-1}) = \log(\frac{1+\sqrt{1-x^2}}{x}) \) for \( 0<x \le 1 \). \( \text{cosech}^{-1}y = \sinh^{-1}(1/y) = \log(1/y + \sqrt{(1/y)^2+1}) \). Alternatively, if \( \text{sech}^{-1}x = A \), then \( x = \text{sech} A = \frac{1}{\cosh A} \). If \( A = \log 2 \), then \( \cosh A = \frac{e^A+e^{-A}}{2} = \frac{2+1/2}{2} = \frac{5/2}{2} = 5/4 \). So \( x = 4/5 \). If \( \text{cosech}^{-1}y = B \), then \( y = \text{cosech} B = \frac{1}{\sinh B} \). If \( B = -\log 3 = \log(1/3) \), then \( \sinh B = \frac{e^B-e^{-B}}{2} = \frac{1/3-3}{2} = \frac{(1-9)/3}{2} = \frac{-8/3}{2} = -4/3 \). So \( y = 1/(-4/3) = -3/4 \).

Answer 1

The Correct Option is B

Step 1: Use the logarithmic forms of inverse hyperbolic functions.
\( \text{sech}^{-1}x = \log\left(\frac{1+\sqrt{1-x^2}}{x}\right) \) for \( 0<x \le 1 \).
Given \( \text{sech}^{-1}x = \log 2 \).
So, \( \log\left(\frac{1+\sqrt{1-x^2}}{x}\right) = \log 2 \).
\[ \frac{1+\sqrt{1-x^2}}{x} = 2 \] \[ 1+\sqrt{1-x^2} = 2x \] \[ \sqrt{1-x^2} = 2x-1 \] Square both sides: \( 1-x^2 = (2x-1)^2 = 4x^2-4x+1 \).
\[ 0 = 5x^2-4x \] \[ x(5x-4) = 0 \] Since \( 0<x \le 1 \), we have \( x \ne 0 \).
So \( 5x-4=0 \implies x = \frac{4}{5} \).
Check condition for squaring: \( 2x-1 \ge 0 \implies 2(\frac{4}{5})-1 = \frac{8}{5}-1 = \frac{3}{5} \ge 0 \).
Valid.
So, \( x = \frac{4}{5} \).

Step 2: Use the logarithmic forms for \( \text{cosech}^{-1}y \).
\( \text{cosech}^{-1}y = \log\left(\frac{1}{y} + \frac{\sqrt{1+y^2}}{|y|}\right) \).
Or, \( \text{cosech}^{-1}y = \sinh^{-1}(\frac{1}{y}) = \log\left(\frac{1}{y} + \sqrt{\left(\frac{1}{y}\right)^2+1}\right) \).
Given \( \text{cosech}^{-1}y = -\log 3 = \log(3^{-1}) = \log(\frac{1}{3}) \).
So, \( \log\left(\frac{1}{y} + \sqrt{\frac{1}{y^2}+1}\right) = \log(\frac{1}{3}) \).
\[ \frac{1}{y} + \sqrt{\frac{1+y^2}{y^2}} = \frac{1}{3} \] \[ \frac{1}{y} + \frac{\sqrt{1+y^2}}{|y|} = \frac{1}{3} \] Since \( \text{cosech}^{-1}y = -\log 3<0 \), \( y \) must be negative.
(Because \( \text{cosech}^{-1}y \) has the same sign as \(y\)).
So \( |y| = -y \).
\[ \frac{1}{y} + \frac{\sqrt{1+y^2}}{-y} = \frac{1}{3} \] \[ \frac{1-\sqrt{1+y^2}}{y} = \frac{1}{3} \] \[ 3(1-\sqrt{1+y^2}) = y \] \[ 3-y = 3\sqrt{1+y^2} \] Square both sides: \( (3-y)^2 = 9(1+y^2) \).
\[ 9-6y+y^2 = 9+9y^2 \] \[ 0 = 8y^2+6y \] \[ 2y(4y+3) = 0 \] Since \( y \ne 0 \), we have \( 4y+3=0 \implies y = -\frac{3}{4} \).
Check condition for squaring: \( 3-y \ge 0 \).
For \( y = -3/4 \), \( 3 - (-3/4) = 3+3/4 = 15/4 \ge 0 \).
Valid.
So, \( y = -\frac{3}{4} \).

Step 3: Calculate \( (x+y) \).
\[ x+y = \frac{4}{5} + \left(-\frac{3}{4}\right) = \frac{4}{5} - \frac{3}{4} \] \[ = \frac{4 \cdot 4 - 3 \cdot 5}{5 \cdot 4} = \frac{16-15}{20} = \frac{1}{20} \] This matches option (2).