Maximum value of f-x-sin x-cos-x is121-2-2-

The correct option is D √(2)y=f(x)=sin⁡x+cos⁡x nbsp;dydx=cos⁡ x sin⁡x For dydx=0 ,sin⁡ x=cos⁡ x , tan⁡ x=1 ∴ nbsp; x=45^∘ y=sin⁡ 45^∘+cos45^∘ =1√(2)+1√(2)=2√ ...

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Standard XII

Physics

Basic Differentiation Rule

Maximum value...

Question

Maximum value of $f (x) = sin x + cos x$ is

1

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2

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\frac{1}{\sqrt{2}}

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\sqrt{2}

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Solution

The correct option is D $\sqrt{2}$
$y = f (x) = sin x + cos x$
$\frac{d y}{d x} = cos x - s i n x$
For $\frac{d y}{d x} = 0, sin x = cos x, t a n x = 1$
$∴ x = 45^{\circ}$
$y = sin 45^{\circ} + c o s 45^{\circ}$
$= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$

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Maximum value of f(x)=sinx+cosx is

The correct option is D √2y=f(x)=sinx+cosx dydx=cosx−sinx For dydx=0,sinx=cosx, tanx=1 ∴x=45∘ y=sin45∘+cos45∘ =1√2+1√2=2√2=√2

Maximum value of $f (x) = sin x + cos x$ is

The correct option is D $\sqrt{2}$
$y = f (x) = sin x + cos x$
$\frac{d y}{d x} = cos x - s i n x$
For $\frac{d y}{d x} = 0, sin x = cos x, t a n x = 1$
$∴ x = 45^{\circ}$
$y = sin 45^{\circ} + c o s 45^{\circ}$
$= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$