If x = \;12/5, then 3x + 3x = [TG EAPCET 10 May 2024 Shift 2]

Correct Answer is : 125\;125125

Correct Answer is : 125\;125125

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TS EAMCET 10 May 2024 Shift 2 Solved Paper

Section: Mathematics

Question : 27 of 160

Marks: +1, -0

\sinh x = \;\frac{12}{5}

\sinh 3x + \cosh 3x =

Solution:

Given that

\sinh x = \;\frac{12}{5}

, we denote

\cosh x = t

. We utilize the identity for hyperbolic functions:

\cosh^2 x - \sinh^2 x = 1

Substituting the known value, we have:

\;t^2 - \left(\;\frac{12}{5}\right)^2 = 1

\;t^2 - \;\frac{144}{25} = 1

Solving for

t^2

, we get:

t^2 = \;\frac{169}{25}

Thus,

t = \;\frac{13}{5}

Now, we need to find

\sinh 3x + \cosh 3x

. We use the identities for triple angles:

\;\sinh 3x = 3 \sinh x + 4 \sinh^3 x

\;\cosh 3x = 4 \cosh^3 x - 3 \cosh x

So,

\sinh 3x + \cosh 3x = 3 \sinh x + 4 \sinh^3 x + 4 \cosh^3 x - 3 \cosh x

We substitute the values:

\;= 3 \left(\;\frac{12}{5} - \;\frac{13}{5}\right) + 4 \left(\;\frac{12}{5} + \;\frac{13}{5}\right) \left(\;\frac{144}{25} + \;\frac{169}{25} - \;\frac{156}{25}\right)

\;= 3 \left(\;\frac{12}{5} - \;\frac{13}{5}\right) + 4 \left(\;\frac{25}{5}\right) \left(\;\frac{157}{25}\right)

\;= \;\frac{-3}{5} + 4 \cdot \;\frac{157}{25}

\;= \;\frac{-3}{5} + \;\frac{628}{25}

\;= \;\frac{-3}{5} + \;\frac{628}{25} = \;\frac{-3}{5} + \;\frac{2512}{125}

\;= \;\frac{-75+2512}{125} = \;\frac{2437}{125}

Thus, the calculated value is:

= 125

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