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Q. The minimum value of `3^(sin^6x)` + `3^(cos^6x)` is: (Note: It is 3 raised to the...
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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
This leads to two solutions, one for the maximum and the other representing the minimum:
`rArr` either [sinx=0, cosx=1] or, [sinx=1, cosx=0] --- (i)
and then, `sin^10x-cos^10x=0`
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)`
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.
Posted by llltkl on July 1, 2013 at 1:42 PM (Answer #1)
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