Question

After completion of the topic Maxima
and Minima, a function is given f(x)=
(sin 2x) - X, (-T/2) = x < Tt/2 and the
following questions are asked:
The local maximum value of f(x) 1 point
is *
O (1/12) - (1/4)
O (13/2) - (11/6)
O (13/2) + (1/6)
O (1/12) + (11/4)​

Answer 1

EXPLANATION.

A function is given by

f(x) = sin2x - x, -π/2 ≤ x ≤ π/2.

Differentiate the function w.r.t to x.

F'(x) = 2cos2x - 1 , -π/2 ≤ x ≤ π/2.

F'(x) = 0

F'(x) = 2cos2x - 1 = 0 , -π/2 ≤ x ≤ π/2.

2cos2x - 1 = 0.

2cos2x = 1.

cos2x = 1/2.

2x = π/3 , -π/3.

x = π/6 , -π/6.

When x = π/6.

F(π/6) = Sin2(π/6) - π/6.

F(π/6) = Sin(π/3) - π/6.

F(π/6) = √3/2 - π/6.

When x = -π/6.

F(-π/6) = Sin2(-π/6) - ( -π/6).

F(-π/6) = Sin(-π/3) + π/6.

F(-π/6) = -√3/2 + π/6.

F(-π/6) = π/6 - √3/2.

Answer 2

$\huge{\boxed{\rm{\red{Question}}}}$

After completion of the topic Maxima

and Minima, a function is given f(x)=

(sin 2x) - X, (-T/2) = x < Tt/2 and the

following questions are asked:

The local maximum value of f(x).

• (1/12) - (1/4)
• (13/2) - (11/6)
• (13/2) + (1/6)
• (1/12) + (11/4)

$\huge{\boxed{\rm{\red{Answer}}}}$

${\bigstar}\large{\boxed{\sf{\pink{Given \: that}}}}$

• A function is given f(x)= (sin 2x) - X, (-T/2) = x < Tt/2

${\bigstar}\large{\boxed{\sf{\pink{To \: find}}}}$

• The local maximum value of f(x)

${\bigstar}\large{\boxed{\sf{\pink{Full \: Solution}}}}$

$\large\purple{\texttt{Function is given below ↓}}$

$\implies$ f(x) sin2x - x - π/-2 < x < π/2

$\mapsto$ $\large{\boxed{\sf{π \: is \: pronounced \: as \: pi}}}$

$\mapsto$ $\large{\boxed{\sf{value \: of \: π \: 22/7}}}$

$\large\gray{\texttt{Let's carry on}}$

$\large\purple{\texttt{Differentiate the functions}}$

$\implies$ F¹ (x) = 2cos2x - 1 , -π/2 < x < π/2

$\implies$ F¹ (x) = 0

$\implies$ F¹ (x) = 2cos2x - 1 = 0 , -π/2 < x < n/2

$\implies$ 2cos2x - 1 = 0

$\implies$ 2cos2x = 1

$\implies$ cos2x = 1/2

{In this situation 2cos2x = 1 , 2 is as multiply but in next situation cos2x = 1/2 we send 2 in right side of equal to sign so it become as divide from multiply}

$\implies$ 2x = π/3 , -π/3

$\implies$ x = π/6 , -π/6

{ 2 × 3 = 6 that's why from here 2x = π/3 , -π/3 .... then the result x = π/6 , -π/6 } { It's bcz we multiply 2 by 3 = 6 }

$\large{\boxed{\rm{\orange{When \: x \: = \: π/6}}}}$

$\implies$ F (π/6) = Sin2(π/6) - π/6

$\implies$ F (π/6) = Sin (π/3) -π/6

{It's come this value Sin (π/3) insted of Sin2(π/6) because we divide 6 by 2 hence, the result is 3}

$\implies$ F (π/6) = √3/2 -π/6

$\mapsto$ $\large{\boxed{\sf{√ \: means \: square \: root}}}$

$\large{\boxed{\rm{\orange{When \: x \: = \: -π/6}}}}$

$\implies$ F = (-π/6) = Sin2 (-π/6) - (-π/6)

$\implies$ F = (-π/6) = Sin (-π/3) +π/6

{ 6 ÷ 2 = 3 Hence we get new equation from F = (-π/6) = Sin2 (-π/6) - (-π/6) to F = (-π/6) = Sin (-π/3) +π/6 }

{ As we know - - = + }

$\implies$ F = (-π/6) = √3/2 + π/6

$\implies$ F = π/6 - √3/2

@Itzbeautyqueen23

Hope it's helpful

Thank you :)