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The number of solutions of \cos x+ \sqrt{3}\sin x=5, 0\leq x\leq 5\pi is

 

  • Option 1)

    4

  • Option 2)

    0

  • Option 3)

    5

  • Option 4)

    None of these

 

Answers (1)

best_answer

As we learnt

Maximum and minimum values -

Max. value = \sqrt{a^{2}+b^{2}}

Min. value = -\sqrt{a^{2}+b^{2}}

- wherein

The maximum and minimum value of a\cos \Theta + b\sin \Theta

 

 \cos x + \sqrt{3 \sin x}=5

min value of 

\cos x + \sqrt{3 \sin x} is - \sqrt{1+3}=-\sqrt{4}=-2

& max value is 2

So  -2 \leq \sqrt{3} \sin x+ \cos x \leq 2

 


Option 1)

4

Incorrect option 

Option 2)

0

correct option 

Option 3)

5

Incorrect option 

Option 4)

None of these

Incorrect option 

Posted by

Plabita

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