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For non-negative integers n, let int(n)=(sum_(k=0)^(n)((k+1)/(n+2)pi)sin((k+2)/(n+2)pi))/(sum_(k=0)^(n)sin^2((k+1)/(n+2)pi)) Assuming cos^(-1)x takes values in [0.pi], which of the following options is/are correct? |
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Answer» If `mu=tan (cos^(-1)//(6),then" "ALPHA^2+2alpha-1=0` `int (n)=(sum _(k=0)^(n)sin((k+1)/(n+2))sin((k+2)/(n+2)pi))/(sum_(k=0)^(n)sin^2((k+1)/(n+2)pi))` `int (n)=(sum _(k=0)^(n)(cos.(pi)/(n+2)-cos((2k+3)/(n+2)pi)))/(sum_(k=0)^(n)(1-cos((2k+2)/(n+2)pi)))` `[because 2 sin B=cos(A-B)-cos (A+B)and 2sin^2A=1-cos 2A]` = ((cos(pi/(n+2)))sum_(k=0)^(n)1-{{:(cos'(3pi)/(n+2)+,cos'(5pi)/(n+2)+cos'(7pi)/(n+2)),(,+"........"+cos'((2n+3)/(n+2)pi)):}})/(sum_(k=0)^(n)1-{{:(cos'(2pi)/(n+2)+,cos'(4pi)/(n+2)+cos'(6pi)/(n+2)),(,+"........"+cos'((2n+2)/(n+2)pi)):}})` `[because cos (alpha)+cos(alpha+beta)+ cos (alpha+2beta)+.......` `+cos (alpha)+cos (alpha+beta)+cos(alpha+beta)+...` `+cosalpha+(n-1)(beta)=(sin((nbeta)/2))/(sin((beta)/2))cos((2alpha+(n-1)beta)/(2))]` `=((n+1)cos((pi)/(n+2))-(sin(pi-(pi)/(n+2)))/(sin((pi)/(n+2)))cos(pi+(pi)/(n+2)))/((n+1)-(sin((pi)/(n+2)))/(sin((pi)/(n+2)))cos(pi))` `=((n+1)cos((pi)/(n+2))+(sin((pi)/(n+2)))/(sin((pi)/(n+2)))cos((pi)/(n+2)))/((n+1)+(sin((pi)/(n+2)))/(sin((pi)/(n+2))))` `=((n+2)cos((pi)/(n+2)))/((n+2))=cos ((pi)/(n+2))` `rArr f(n)=cos ((pi)/(n+2))` Now, `f(6)=cos(pi)/(8)` `because alpha =tan cos^-1f(((6))=tan(pi)/(8)""{{:(because cos^-1cosx =X),(if x in (0,(pi)/(2))):}}` `=sqrt(2)-1` `rArr (alpha+1)=sqrt(2)rArr(alpha+1)^2=2rArr alpha^2+2alpha+1=2` `rArr alpha^2+2alpha-1=0` Now, `f(4)=cos ((pi)/(4+2))=cos (pi)/(6)=(sqrt(3))/(2)`, Now, `sin (7cos ^-1f(5))=sin(7cos^-1((pi)/(5+2)))` `sin(7(pi)/(7))=sinpi=0` and Now, `LIM(NTO oo)f(x)=lim(n to oo) cos (pi)/(n+2)=cos=1` Hence, options (a),(b) and (c) are correct. |
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