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TG EAMCET 2 May 2025 Shift 2 Paper

Question : 27 of 160

Marks: +1, -0

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

a+b+c=

Solution:

We know that

\; \sinh^{-1} x = \log (x + \sqrt{x^2+1})

\; \therefore \sinh^{-1} 2 = \log (2 + \sqrt{4+1}) = \log (2 + \sqrt{5})

\; \cosh^{-1} x = \log (x + \sqrt{x^2-1})

\; \therefore \cosh^{-1} \sqrt{6} = \log (\sqrt{6} + \sqrt{6-1}) = \log (\sqrt{6} + \sqrt{5})

\; \therefore \sinh^{-1} 2 + \cosh^{-1} \sqrt{6}

\; = \log (2 + \sqrt{5}) + \log (\sqrt{6} + \sqrt{5})

\; = \log [(2 + \sqrt{5})(\sqrt{6} + \sqrt{5})]

\; e^{(\sinh^{-1} \sqrt{2} + \cosh^{-1} \sqrt{6})} = e^{\log ((2+\sqrt{5})(\sqrt{6}+\sqrt{5})]}

\; = (2+\sqrt{5})(\sqrt{6}+\sqrt{5}) = 2\sqrt{6} + 2\sqrt{5} + \sqrt{30} + 5

\; = 5 + 2\sqrt{5} + 2\sqrt{6} + \sqrt{30} = 5 + 2\sqrt{5} + 2\sqrt{6} + \sqrt{5}\sqrt{6}

\; = 5 + (2+\sqrt{6})\sqrt{5} + 2\sqrt{6} = a + (b+\sqrt{c})\sqrt{a} + b\sqrt{c}

\; \therefore a=5, b=2, c=6

\; \therefore \;\; a+b+c=5+2+6=13

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