Question

If \( e^{(\sinh^{-1} 2 + \cosh^{-1} \sqrt{6})} = a + (b+\sqrt{c})\sqrt{a} + b\sqrt{c} \), then \( a+b+c = \)

• A. 13
• B. 15
• C. 17
• D. 11

Hint:

Remember \( \sinh^{-1} x = \ln(x+\sqrt{x^2+1}) \) and \( \cosh^{-1} x = \ln(x+\sqrt{x^2-1}) \). These definitions convert inverse hyperbolic problems into simple algebraic ones.

Answer 1

The Correct Option is A

Step 1: Simplify LHS:

Using logarithmic definitions: \( \sinh^{-1} x = \ln(x + \sqrt{x^2+1}) \implies \sinh^{-1} 2 = \ln(2 + \sqrt{5}) \). \( \cosh^{-1} x = \ln(x + \sqrt{x^2-1}) \implies \cosh^{-1} \sqrt{6} = \ln(\sqrt{6} + \sqrt{5}) \). Sum \( S = \ln((2+\sqrt{5})(\sqrt{6}+\sqrt{5})) \). LHS \( = e^S = (2+\sqrt{5})(\sqrt{6}+\sqrt{5}) \). Expand: \[ \text{LHS} = 2\sqrt{6} + 2\sqrt{5} + \sqrt{30} + 5 \] \[ = 5 + 2\sqrt{5} + 2\sqrt{6} + \sqrt{30} \]
Step 2: Analyze RHS Structure:

RHS \( = a + (b+\sqrt{c})\sqrt{a} + b\sqrt{c} \) Expand: \[ \text{RHS} = a + b\sqrt{a} + \sqrt{ac} + b\sqrt{c} \]
Step 3: Compare Coefficients:

Equate LHS and RHS: \[ 5 + 2\sqrt{5} + 2\sqrt{6} + \sqrt{30} = a + b\sqrt{a} + b\sqrt{c} + \sqrt{ac} \] By inspection: Integer part: \( a = 5 \). Term with \( \sqrt{a} = \sqrt{5} \): \( b\sqrt{5} = 2\sqrt{5} \implies b = 2 \). Term with \( b\sqrt{c} = 2\sqrt{c} \): Matches \( 2\sqrt{6} \implies c = 6 \). Check the last term \( \sqrt{ac} \): \( \sqrt{5 \times 6} = \sqrt{30} \). Matches perfectly. Values are \( a=5, b=2, c=6 \). We need \( a+b+c = 5 + 2 + 6 = 13 \).
Step 4: Final Answer:

The sum is 13.