Find The Value Of Sinh Left Log Left 2+sqrt5 Right Class 11 Maths CBSE

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AnswerQuestion Answers for Class 12Class 12 BiologyClass 12 ChemistryClass 12 EnglishClass 12 MathsClass 12 PhysicsClass 12 Social ScienceClass 12 Business StudiesClass 12 EconomicsQuestion Answers for Class 11Class 11 EconomicsClass 11 Computer ScienceClass 11 BiologyClass 11 ChemistryClass 11 EnglishClass 11 MathsClass 11 PhysicsClass 11 Social ScienceClass 11 AccountancyClass 11 Business StudiesQuestion Answers for Class 10Class 10 ScienceClass 10 EnglishClass 10 MathsClass 10 Social ScienceClass 10 General KnowledgeQuestion Answers for Class 9Class 9 General KnowledgeClass 9 ScienceClass 9 EnglishClass 9 MathsClass 9 Social ScienceQuestion Answers for Class 8Class 8 ScienceClass 8 EnglishClass 8 MathsClass 8 Social ScienceQuestion Answers for Class 7Class 7 ScienceClass 7 EnglishClass 7 MathsClass 7 Social ScienceQuestion Answers for Class 6Class 6 ScienceClass 6 EnglishClass 6 MathsClass 6 Social ScienceQuestion Answers for Class 5Class 5 ScienceClass 5 EnglishClass 5 MathsClass 5 Social ScienceQuestion Answers for Class 4Class 4 ScienceClass 4 EnglishClass 4 MathsFind the value of $\sinh \left( \log \left( 2+\sqrt{5} \right) \right)$, here logarithm is to the natural base e.(a) 2(b) 3(c) $2-\sqrt{5}$(d) $2+\sqrt{5}$AnswerVerified574.8k+ viewsHint: We start solving the problem by recalling the definition of sine hyperbolic function as $\sinh \left( y \right)=\dfrac{{{e}^{y}}-{{e}^{-y}}}{2}$. We then substitute $\log \left( 2+\sqrt{5} \right)$ in place of ‘y’ and then make use of the results $-\log a=\log \dfrac{1}{a}$ and ${{e}^{\log a}}=a$ to proceed through the problems. We then make use of the result ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and then perform the necessary calculations to get the required result.Complete step-by-step answer:According to the problem, we are asked to find the value of $\sinh \left( \log \left( 2+\sqrt{5} \right) \right)$, where logarithm is to the natural base e.We know that the sine hyperbolic function is defined as $\sinh \left( y \right)=\dfrac{{{e}^{y}}-{{e}^{-y}}}{2}$. Let us use this definition to find the value of the given $\sinh \left( \log \left( 2+\sqrt{5} \right) \right)$.So, we have \[\sinh \left( \log \left( 2+\sqrt{5} \right) \right)=\dfrac{{{e}^{\log \left( 2+\sqrt{5} \right)}}-{{e}^{-\log \left( 2+\sqrt{5} \right)}}}{2}\] ---(1).We know that $-\log a=\log \dfrac{1}{a}$. Let us use this result in equation (1).\[\Rightarrow \sinh \left( \log \left( 2+\sqrt{5} \right) \right)=\dfrac{{{e}^{\log \left( 2+\sqrt{5} \right)}}-{{e}^{\log \left( \dfrac{1}{2+\sqrt{5}} \right)}}}{2}\] ---(2).We know that ${{e}^{\log a}}=a$. Let us use this result in equation (2).\[\Rightarrow \sinh \left( \log \left( 2+\sqrt{5} \right) \right)=\dfrac{2+\sqrt{5}-\left( \dfrac{1}{2+\sqrt{5}} \right)}{2}\].\[\Rightarrow \sinh \left( \log \left( 2+\sqrt{5} \right) \right)=\dfrac{\left( \dfrac{{{\left( 2+\sqrt{5} \right)}^{2}}-1}{2+\sqrt{5}} \right)}{2}\] ---(3).We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. Let us use this result in equation (3).\[\Rightarrow \sinh \left( \log \left( 2+\sqrt{5} \right) \right)=\dfrac{\left( \dfrac{{{2}^{2}}+{{\left( \sqrt{5} \right)}^{2}}+2\left( 2 \right)\left( \sqrt{5} \right)-1}{2+\sqrt{5}} \right)}{2}\].\[\Rightarrow \sinh \left( \log \left( 2+\sqrt{5} \right) \right)=\dfrac{\left( 4+5+4\sqrt{5} \right)-1}{2\left( 2+\sqrt{5} \right)}\].\[\Rightarrow \sinh \left( \log \left( 2+\sqrt{5} \right) \right)=\dfrac{8+4\sqrt{5}}{4+2\sqrt{5}}\].\[\Rightarrow \sinh \left( \log \left( 2+\sqrt{5} \right) \right)=2\].So, we have found the value of $\sinh \left( \log \left( 2+\sqrt{5} \right) \right)$ as 2.∴ The correct option for the given problem is (a).So, the correct answer is “Option (a)”.Note: If the given logarithm is not to the natural base, then the result will not be the same as we get in this problem. We should always take logarithms to the natural base if nothing is mentioned in the problem about the base while solving problems related to hyperbolic functions. We can also solve this problem as shown below:Let us assume $\sinh \left( \log \left( 2+\sqrt{5} \right) \right)=x$.$\Rightarrow \log \left( 2+\sqrt{5} \right)={{\sinh }^{-1}}\left( x \right)$.We know that ${{\sinh }^{-1}}\left( x \right)=\log \left( x+\sqrt{{{x}^{2}}+1} \right)$.$\Rightarrow \log \left( 2+\sqrt{5} \right)=\log \left( x+\sqrt{{{x}^{2}}+1} \right)$.$\Rightarrow 2+\sqrt{5}=x+\sqrt{{{x}^{2}}+1}$.Comparing rational and irrational parts on both sides we get,$\Rightarrow x=2$.Recently Updated PagesWrite a brief account of abscisic acid class 11 biology CBSEPhotolysis of water involves a Excitement of water class 11 biology CBSEBoth wind and water pollinated flowers are not very class 11 biology CBSEWhich among the following has specialized tissue for class 11 biology CBSEGive one point of difference between the notochord class 11 biology CBSEWhat are the factors that are essential for photos class 11 biology CBSEWrite a brief account of abscisic acid class 11 biology CBSEPhotolysis of water involves a Excitement of water class 11 biology CBSEBoth wind and water pollinated flowers are not very class 11 biology CBSEWhich among the following has specialized tissue for class 11 biology CBSEGive one point of difference between the notochord class 11 biology CBSEWhat are the factors that are essential for photos class 11 biology CBSE

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