IIT JEE , AIEEE Forum - Mathematics , Algebra - a problem: find minimum value of "(2^sin¤)+(2^cos¤)"

towards.destination

a problem: find minimum value of "(2^sin¤)+(2^cos¤)"

Decoder

method 1...

take this as y = 2^sinx + 2^cosx...
just differentiate..

dy/dx = cosx.2^sinxlog2 - sinx.2^cosx.log2..

this method wasn't good ..

2^sinx,cosx both r positive...

lets use inequalities..

a-m - gm..

min. value is 2 root { 2^ (sinx + cosx)...}

sinx+ cosx at max. gives u. root2...

allamraju

The ans is 2^1-1/rt2.The same question was asked by somebody previously and was answered.It can be solved using AM-GM.

Decoder

yes.. it can be done in 10 secs. by just observation.
.put cosx = sinx = 1/root2..

budokai_tenkaichi_returns

generally when d question is symetric about sinx and cosx

the maximum or minimum value is holded by the function when sinx = cosx

this is quite useful in eliminating options (when they r given)

so better try this next time a similaer question comes.

computer001

$({2}^{sinx}+{2}^{cosx})/2\; \ge\; {2}^{(sinx+cosx)/2}$ $\ge {2}^{-\sqrt{2}/2}$

$so\; {2}^{sinx}+{2}^{cosx}\ge{2}^{1-1/\sqrt{2}}$

computer001

computer001

budokai_tenkaichi_returns

happy noew

computer001

budokai_tenkaichi_returns

well it was becoming too heavy for me too

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