Question

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★ STANDARD QUESTION 26 ★

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Question is -

Find the range of -

ƒ( x ) =
$\frac{1}{( \cos(x) - 3) {}^{2} + ( \sin(x) + 4) {}^{2} }$




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Answer 1

hey there !!

=) f(x) = 1 /{(cosx -3)^2 +(sinx +4)^2 }

let assume y = (cosx -3)^2 + (sinx +4)^2

◆opening square : WE GET ,

y =cos^2x + 9 -6cosx + sin^2x + 16 +8sinx

=) y = (cos^2x +sin^2x) + 25 + 8sinx-6cosx

here we know that : cos^2x + sin^2x = 1

now , y = 1 +25 + 8sinx - 6cosx

=) y = 26 +8sinx - 6cosx -----------(a)

now using trigonometric property ,

range of asinx +_ bcosx lies in an interval [ - √(a^2+b^2) , +√(a^2+b^2) ]

so range of ( 8sinx - 6cosx ) € [ -10 ,10 ]

and range of { 26 +8sinx-6cosx } € [26 -10 , 26 +10 ]

=) {26 +8sinx - 6cosx} € [ 16 , 36 ]

so range of y € [16 ,36]

but we have to find range of 1/y so,

range of f(x) € [1/36 ,1/16] ..

______________________________

Hope it helps