Question

Q - Find the maximum value of
$\sqrt{3} \sin(x)+ \cos(x)$ and x for which a maximum value occurs.​

Answer 1

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Let √3 Sin x + Cos x = F(x)

F(x) = √3 Sin x + Cos x

Differentiate both sides w.r.t x we have

F'(x) = √3 Cos x - Sin x

F'(x) = 0 { For Critical points }

√3 Cos x - Sin x = 0

√3 Cos x = Sin x

Tan x = √3

Tan x = Tan 60 { Becoz Tan 60 = √3 }

x = 60°

Now, This point x = 60° May, be either Max OR Mina of Given Function F(x). To check we Differentiate The Function Second Time, i,e we take it's Second Derivative. IF second Derivative comes > 0 Then This point x = 60° is Point of Minima, And IF Second Derivative comes < 0 Then this point x = 60° is pont of Maxima.

Differentiate F'(x) w.r.t x we have

F''(x) = -√3Sin x - Cos x

At x = 60°

F"(x) = -√3 Sin 60° - Cos 60°

F"(x) = -√3 (√3/2) - (1/2)

F"(x) = -(√3)²/2 - (1/2)

F"(x) = -(3/2) - (1/2)

F"(x) = (-3 - 1)/2

F"(x) = -4/2

F"(x) = -2

Here, F"(x) < 0 So, x = 60° is point of Maxima!!)

To Find Maximum Value of Function!) put x = 60° In F(x)

F(x) = √3 Sin 60 + Cos 60

F(x) = √3 (√3)/2 + (1/2)

F(x) = (3 + 1)/2

F(x) = 2

So, The Maximum value of F(x) is +2

Therefore, Maximum Value of F(x) is +2 at x = 60°

Answer 2

Answer:

Maximum value = + 2

x = 60°

Step-by-step explanation:

Refer the attached picture.

Formulae used :

• sin ( A + B ) = sinA cos B + cosA sin B
• $\mathsf{\dfrac{\sqrt{3}}{2}\:=\:cos\:30°\:and \:{\dfrac{1}{2}\:=\:sin\:30°}}$