Q. The minimum value of `3^(sin^6x)` + `3^(cos^6x)` is: (Note: It is 3 raised to the power of `sin^6x` ) A) ```(2.3)^(1/8)` B) `(3.2)^(1/8)` C) `(2.3)^(7/8)` D) `(3.2)^(1/8)`



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llltkl's profile pic

Posted on (Answer #1)

Let `y =3^(sin^6x) + 3^(cos^6x)`

In order to determine the minimum value of y, we have to find dy/dx first.

Let `p=3^(sin^6x)` and `q= 3^(cos^6x)`  So that y = p+q,

and, `(dy)/dx = (dp)/dx+(dq)/dx`


`(dp)/dx= [3^ (sin^6x)]'`

Applying chain rule of differentiation:

`(dp)/dx = 3^ (sin^6x)*ln 3*(sin^6x)'`

`=3^ (sin^6x)*ln 3*6sin^5x(sinx)'`

`=3^ (sin^6x)*ln 3*6sin^5x*cosx`


Similarly, `q= 3^(cos^6x)`

`(dq)/dx=3^ (cos^6x)*ln 3*(cos^6x)'`

`=3^ (cos^6x)*ln 3*6cos^5x(cosx)'`

`=3^ (cos^6x)*ln 3*6cos^5x*(-sinx)`


So, `(dy)/dx =18*ln3*sinxcosx(sin^10x-cos^10x)`

At extrema (maxima or minima) of y, dy/dx = 0

`rArr 18ln3sinxcosx(sin^10x-cos^10x)=0`

This leads to two solutions, one for the maximum and the other representing the minimum:

First, sinxcosx=0,

`rArr` either [sinx=0, cosx=1] or, [sinx=1, cosx=0] --- (i)

and then, `sin^10x-cos^10x=0`

`rArr sinx=cosx`

This leads us to the solution, `x=pi/4` ------- (ii)

Putting solution set (i) in the given function, we get

`y=3^0 + 3^1=1+3=4`

Putting solution set (ii) in the given function, we get `y=3^((sinpi/4)^6) + 3^((cospi/4)^6)`

`= 3^((1/sqrt2)^6)+3^((1/sqrt2)^6)`

`=3^((1/2)^3) (1+1)`


Comparing y obtained from x in solution sets (i) and (ii), it is clear that solution set (i) reprtesents maximum, whereas set (ii) represents the minimum of y.

Hence the minimum value of y is `2*3^(1/8)` 

It does not match with any of the options given here. Options B) and D) are same. So, there must be something wrong in it. Recheck for your right option.


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user8235304's profile pic

Posted on (Reply #1)

It can also be done using A.M. and G.M. method.

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