If sinx+cosx=y+y1,xϵ[0,π] and y>0, then-

We know that for two positive numbers A.M≥G.My+y1≥2 ...(i)1+sin2x=y+y1 ...(ii)From (i) and (ii), we get sin2x≥1⇒sin2x=1⇒2x=2π⇒x=4πHence, option 'A' is correct.

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If { x } + x= {y+ {1}{y}}, x [0, ] and y>0, then-

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If $sin x + cos x = y + \frac{1}{y}, x ϵ [0, π]$ and $y > 0$ , then-

x = π /4

x = \frac{π}{2}

x = \frac{π}{6}

x = 3 π /4

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Oliver Discussed

sin x + cos x = y + \frac{1}{y}, x ϵ [0, π]

and

y > 0

, then-

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9 mins ago

Text solutionVerified

We know that for two positive numbers

A . M \geq G . M

y + \frac{1}{y} \geq 2

...(i)

1 + s in 2 x = y + \frac{1}{y}

...(ii)
From (i) and (ii), we get

s in 2 x \geq 1

\Rightarrow s in 2 x = 1

\Rightarrow 2 x = \frac{π}{2}

\Rightarrow x = \frac{π}{4}

Hence, option 'A' is correct.

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Question Text	If $sin x + cos x = y + \frac{1}{y}, x ϵ [0, π]$ and $y > 0$ , then-
Answer Type	Text solution:1
Upvotes	150

If sinx+cosx=y+y1​​,xϵ[0,π] and y>0, then-

Text solutionVerified

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If $sin x + cos x = y + \frac{1}{y}, x ϵ [0, π]$ and $y > 0$ , then-