IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

ashish_banga

the maximum value of the function sinx( 1 + cosx)

djdylan2000

Is it 1 ?

ashish_banga

Mr.IITIAN007

(3 x root 3) / 4

ashish_banga

perfect answer please explain

allamraju

Let it be f(x) then f'(x)=cosx+cos2x=2cos²x+cosx-1=0iff cosx=-1 or 1/2.We get f_max at cosx=1/2 and equal to 3rt3/4

Mr.IITIAN007

Its easy ...in any kind of maximum or minima question ....differentiate the given function f(x) with respect to its parameter ( which is x here ) .
so u get f'(x) .. Now equate f'(x) to zero ..u will get a quadratic equation in cos x .

f(x) = sinx( 1 + cosx)

f'(x) = 2 cos (sq) x + cos x - 1

Now ...f'(x) =1

=> (cos x +1 ) ( 2 cos x - 1 ) =0

so x= 2n pie + 0 ans x = 2n pie + pie / 3 where n belongs to the set of integers (using general solution of cos x = cos alpha )

putting some of the value of n .........u will still find pie / 3 as being the source of greatest value of f(x) ...so just pie / 3 in f(x) and get max value of f(x) .

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Let it be f(x) then f'(x)=cosx+cos2x=2cos2x+cosx-1=0iff cosx=-1 or 1/2.We get fmax at cosx=1/2 and equal to 3rt3/4

Let it be f(x) then f'(x)=cosx+cos2x=2cos²x+cosx-1=0iff cosx=-1 or 1/2.We get f_max at cosx=1/2 and equal to 3rt3/4