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  • Directions : Question are Assertion- Reason type questions. Each of these questions contains two statements. Statement-1 (Assertion) and Statement-2 (Reason). Each of these questions also has four

Directions : Question are Assertion- Reason type questions. Each of these questions contains two statements. Statement-1 (Assertion) and Statement-2 (Reason). Each of these questions also has four alternative choices,only one of which is the correct answer. You have to select the correct choice:

Question : let    f:R\rightarrow R   be a  continuous function defined by f(x)=\frac{1}{e^{x}+2e^{-x}}.

Statement-1 :  f(c)=1/3,for\; some\; c\in R.

Statement-2 :  0< f(x)\leq \frac{1}{2\sqrt{2}},for\; all\; x\in R.

 

 

 

Option: 1

Statement -1 is true, statement -2 is true ; statement -2 is a correct explanation of statement-1.


Option: 2

Statement -1 is true, statement -2 is true ; statement -2 is not a correct explanation of statement-1.


Option: 3

Statement -1 is true, statement -2 is false


Option: 4

Statement -1 is false, statement -2 is true.


Answers (1)

best_answer

f (x)= \frac{1}{e^x+2e^{-x}}=\frac{1}{y}

Let  y = e^x + 2 e^{-x}

 \because (A.M\geqslant GM)

\\\therefore \frac{e^{x}+2e^{-x}}{2}\geqslant \sqrt{e^{x}\times 2e^{-x}}

\\\therefore {e^{x}+2e^{-x}}\geqslant {2}\sqrt{2e^{x-x}} \\\therefore y\geqslant {2}\sqrt{2}

\therefore \frac{1}{y}\leqslant \frac{1}{2\sqrt{2}}

\therefore 0< f(x) \leq \frac{1}{2\sqrt{2}}

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