The minimum value of a tan 2 x - b 2 x equals the maximum value of a sin 2- b cos 2- where a - b -0- thenA- a-2 bB- a-3 bC- a-4 bD- a-b

The correct option is C a = 4bMinimum value of a tan^2 x + b cot^2 x is 2√(ab) and maximum value of a sin^2 θ + b cos^2 θis a. ∵ a sin^2 θ + b cos^2 θ = ( ...

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Standard XII

Mathematics

Global Maxima

The minimum v...

Question

The minimum value of $a t a n^{2} x + b c o t^{2} x$ equals the maximum value of a $s i n^{2} θ + b c o s^{2} θ w h e r e a > b > 0,$ then

a = b

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a = 2b

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a = 3b

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a = 4b

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Solution

The correct option is D
a = 4b

Minimum value of a $t a n^{2} x + b c o t^{2} x i s 2 \sqrt{a b}$ and maximum value of a $s i n^{2} θ + b c o s^{2} θ$ is a. $∵ a s i n^{2} θ + b c o s^{2} θ = (a - b s i n^{2} θ + b)$
Given, $a = 2 \sqrt{a b}$ Hence (d) is the correct answer.

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Similar questions

Q. The minimum value of

a t a n^{2} x + b c o t^{2} x

equals the maximum value of a

s i n^{2} θ + b c o s^{2} θ w h e r e a > b > 0,

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The minimum value of a tan2 x+b cot2 x equals the maximum value of a sin2θ+b cos2 θ where a>b>0, then

The correct option is D a = 4bMinimum value of a tan2x+b cot2 x is 2√ab and maximum value of a sin2θ+b cos2θis a. ∵a sin2 θ+b cos2θ=(a−b sin2θ+b) Given,a=2√ab Hence (d) is the correct answer.

The minimum value of $a t a n^{2} x + b c o t^{2} x$ equals the maximum value of a $s i n^{2} θ + b c o s^{2} θ w h e r e a > b > 0,$ then

The correct option is D
a = 4b

Minimum value of a $t a n^{2} x + b c o t^{2} x i s 2 \sqrt{a b}$ and maximum value of a $s i n^{2} θ + b c o s^{2} θ$ is a. $∵ a s i n^{2} θ + b c o s^{2} θ = (a - b s i n^{2} θ + b)$
Given, $a = 2 \sqrt{a b}$ Hence (d) is the correct answer.