Question

Extreme values of
$\rm \: 4 \cos( {x}^{2} ) . \cos \bigg( \dfrac{ \pi}{3} + {x}^{2} \bigg). \cos \bigg( \dfrac{ \pi}{3} - {x}^{2} \bigg)$
over R , are

a] -1 , 1

b] -2 , 2

c] -3 , 3

d] -4 , 4​

Answer 1

EXPLANATION.

Extreme value of,$\sf \implies 4cos(x^{2} ). cos\bigg(\dfrac{\pi}{3} + x^{2}\bigg).cos\bigg(\dfrac{\pi}{3} - x^{2} \bigg)$

As we know that,Formula of : cos(A ± B).

⇒ Cos(A + B) = Cos(A).Cos(B) - Sin(A).Sin(B).

⇒ Cos(A - B) = Cos(A).Cos(B) + Sin(A).Sin(B).

Using the formula in equation, we get.

$\sf \implies 4cos( x^{2} ) \bigg(cos\dfrac{\pi}{3} .cos(x^{2} ) - sin\dfrac{\pi}{3} .sin(x^{2} ) \bigg). \bigg(cos\dfrac{\pi}{3}.cos(x^{2} ) + sin\dfrac{\pi}{3}.sin(x^{2} ) \bigg)$

As we know that,

Formula of :⇒ cos(π/3) = cos(180/3) = cos(60°) = 1/2.

⇒ sin(π/3) = sin(180/3) = sin(60°) = √3/2.

Using the formula in equation, we get.

$\sf \implies 4cos(x^{2} ) \bigg(\dfrac{1}{2}cos(x^{2} ) -\dfrac{\sqrt{3} }{2} sin(x^{2} ) \bigg). \bigg(\dfrac{1}{2}cos(x^{2} ) + \dfrac{\sqrt{3} }{2}sin(x^{2} ) \bigg)$

As we know that,

Formula of :⇒ (a² - b²) = (a + b)(a - b).

Using the formula in equation, we get.

$\sf \implies 4cos(x^{2} ) \bigg(\dfrac{1}{2}cos(x^{2} ) \bigg)^{2} .- \bigg(\dfrac{\sqrt{3} }{2} sin(x^{2}) \bigg)^{2}$

$\sf \implies 4cos(x^{2} ) \bigg(\dfrac{1}{4} cos^{2}x^{2} - \dfrac{3}{4}sin^{2} x^{2} \bigg)$

$\sf \implies \dfrac{4cos(x^{2} )}{4} \bigg(cos^{2} x^{2} - 3sin^{2} x^{2} \bigg)$

As we know that,

Formula of :

⇒ sin²∅ + cos²∅ = 1.⇒ sin²∅ = 1 - cos²∅.

Using this formula in equation, we get.

$\sf \implies cos(x^{2} ) \bigg(cos^{2}x^{2} - 3(1 - cos^{2} x^{2} ) \bigg)$

$\sf \implies cos^{} (x^{2} )\bigg(cos^{2} x^{2} -3+3cos^{2} x^{2} \bigg)$

$\sf \implies cos(x^{2} ) \bigg(4cos^{2}x^{2} - 3\bigg)$

$\sf \implies 4cos^{3} (x^{2}) - 3cos(x^{2})$

As we know that,

Formula of :

⇒ cos3∅ = 4cos³∅ - 3cos∅.

Using this formula in equation, we get.

⇒ cos(3x²).

As we know that,

Range of cos∅.⇒ cos∅ = -1 < cos∅ < 1.

⇒ range = [-1,1].

So,⇒ cos3x² = -1 < cos3x² < 1.

Extreme value of cos(3x²) = [ -1, 1].

Option [A] is correct answer.

                                                                                                                                      MORE INFORMATION.

Domain & Range of inverse trigonometric functions.

(1) = sin⁻¹x

Domain = [-1, 1].

Range = [ -π/2, π/2].

(2) = cos⁻¹x

Domain = [-1, 1].

Range = [0, π].

(3) = tan⁻¹x

Domain = (-∞. ∞).

Range = (-π/2, π/2).

(4) = cot⁻¹x

Domain = (-∞, ∞).

Range = (-π/2, π/2).

(5) = sec⁻¹x

Domain = (-∞, -1] ∪ [1,∞).

Range = [0,π/2) ∪ (π/2, π].

(6) = cosec⁻¹x

Domain = (-∞, -1] ∪ [1,∞).

Range = [-π/2, 0 ) ∪ (0, π/2].

Answer 2

Answer:

As we know that,

Formula of :

⇒ cos3θ = 4cos³θ - 3cosθ

Using this formula in equation, we get.

⇒ cos(3x²).

As we know that,

Range of cosθ⇒ cosθ = -1 & π; cosθ & π; 1.

⇒ range = [-1,1].

So,⇒ cos3x² = -1 & π; cos3x² &π; 1.

Extreme value of cos(3x²) = [ -1, 1].

Option [A] is correct answer.