The number of real solution of \sin e^{x}.\cos e^{x}=2^{x-2}+2^{-x-2} is 

  • Option 1)

    Zero

  • Option 2)

    One

  • Option 3)

    Two 

  • Option 4)

    Infinite

 

Answers (1)

As learnt in

Trigonometric Ratios of Functions -

\sin \Theta = \frac{Opp}{Hyp}

\cos \Theta = \frac{Base}{Hyp}

\tan \Theta = \frac{Opp}{Base}

- wherein

Trigonometric Ratios of Functions

 

 \sin e^{x} \cos e^{x}=2^{x-2}+2^{-x-2}

                              =\frac {2x}{2}+ \frac {2^{-x}}{2}

2 \sin e^{x} \cos e^{x}=2 ^{x}+2^{-x}

\sin e^{x}=2 ^{x}+2^{-x}\geqslant 2

which is impossible because

-1 \leq \sin 2 e^{x}\leq 1 and (2^{x}+2^{-x})=2 minvalue AM \geqslant G.M

 


Option 1)

Zero

This option is correct

Option 2)

One

This option is incorrect

Option 3)

Two 

This option is incorrect

Option 4)

Infinite

This option is incorrect

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