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mastermind890

_{[x ]}

_{[ 1]}(2-x)^{tan (X/2)}

titun

Let _{[x ]}

_{[ 1]}(2-x)^{tan (X/2)}= P

Therefore, ln P = _{[x ]}

_{[ 1]}ln ( 2 - x ) / cot (

x / 2 )

This is in 0/0 form. So, applying, L.Hospitals rule,

ln P = _{[x ]}

_{[ 1]}2 /

(2-x) (cosec²(

x / 2)) = 2 /

So, P = e ^{2 /}

Cheers !!

amar.gupta

good work titun

I.e.irodov

Let [x ][ 1] (2-x)tan (X/2) = P

Therefore, ln P = [x ][ 1] ln ( 2 - x ) / cot ( x / 2 )

This is in 0/0 form. So, applying, L.Hospitals rule,

ln P = [x ][ 1] 2 / (2-x) (cosec2( x / 2)) = 2 /

So, P = e 2 /

Cheers !!

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