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Need clarity, kindly explain! - Trigonometry - JEE Main-2

What is the maximum value of the expression ? 

 f(x) = 5 \sin ( x+ \pi /4 )+ 3 \cos( x+ \pi/4)

  • Option 1)

    4

  • Option 2)

    4 \sqrt2

  • Option 3)

    \sqrt {34}

  • Option 4)

    none of these

 
Answers (1)
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As we have learned

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

 

 f(x )= 5 \left ( \frac{\sin x }{\sqrt 2}+ \frac{\cos }{\sqrt2} \right )+ 3 \left ( \frac{\cos x }{\sqrt2}- \frac{\sin x }{\sqrt 2 } \right )

\sqrt 2 \sin x + 4\sqrt 2 \cos x

max value = \sqrt{\sqrt {2}^2+ 4\sqrt{2}^2}= \sqrt{2+16 \times 2}= \sqrt {34}

 

 

 

 


Option 1)

4

Option 2)

4 \sqrt2

Option 3)

\sqrt {34}

Option 4)

none of these

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