Question

For the expression where x belongs to the set of real numbers find the maximum and minimum value of the following given expression stated as :-

Answer 1

\[\huge \frac{ 1 }{ \sin^6 x + \cos^6 x }\]

Answer 2

it simplifies to :-

\[\huge \frac{ 1 }{ 1-3\sin^2xcos^2x }\]

Answer 3

start with sin^6x+cos^6x

Answer 4

(sin^2x)^3 + (cos^2x)^3 i did that

Answer 5

\[\sin^6x+\cos^6x=(\sin^2x)^3+(\cos^2x)^3\\= (\sin^2 x+ \cos^2x) [ (\sin^2 x)^2 + \sin^2 x \cos^2x +(\cos^2x)^2 ]\]

Answer 6

yes?

Answer 7

(a+b)^3 -3ab(a+b)

Answer 8

\[a^3+b^3=(a+b)(a^2+ab+b^2)\]

Answer 9

(sin^4x + cos^4 x +(sinxcosx)^2)

Answer 10

ahh that ab will be -ab

Answer 11

=

-sin^2 x cos ^2x

Answer 12

yeash

Answer 13

now
simplify it more\[\sin^2x=\frac{ 1-\cos2x }{ 2 }\\cos^2x=\frac{ 1+\cos2x }{ 2 }\]
put them in the equation

Answer 14

(cos^2 2x -1)/4

Answer 15

so more simplification will give
\[\sin^6x+\cos^6x=[(\frac{ 1-\cos2x }{ 2 })^2+(\frac{ 1+\cos2x }{ 2 })^2-\frac{ \sin2x }{ 2 }]\]

Answer 16

ok mota bai?

Answer 17

yes

Answer 18

ok let me simplify it more, wait

Answer 19

finee :)

Answer 20

yeah now we have
\[\sin^6x+\cos^6x=\frac{ 1-(\sin^22x+\sin2x) }{ 2 }\]

Answer 21

simplified a little

Answer 22

now it will be min when this term
\[\sin^2x+\sin2x~will~be~\max\]

Answer 23

i will show you what i did
(a+b)^3 = a^3 +b^3 + 3ab(a+b)

a^3 + b^3 = (a+b)^3 -3ab(a+b)

so writing it as

(sin^2 x ) ^3 + (cos^2 x ) ^3
we get

1- 3(sinxcosx)^2 max of this value

how to proceed further

Answer 24

\[1-3\sin^2x.\cos^2x\]
this will be min when \[\sin^2x.\cos^2x~will~be~\max.\]

Answer 25

yeahh

Answer 26

but what i did is correct right?

Answer 27

yes it seems ok

Answer 28

how to proceed

Answer 29

so that term will be max when x=45 deg.

Answer 30

uhh yes

Answer 31

got it got it , thank you

Answer 32

so 1/4,right?

Answer 33

yeah so reciprocol of that 4 is answer

Answer 34

and how to find min

Answer 35

same procedure

Answer 36

min of
sin^2x cos^2 x

Answer 37

yes

Answer 38

yeah what would it be

Answer 39

aah connection lost again

Answer 40

diffrentiate it and see for what value 2nd derivative is positive