If `alpha` and `beta`, `alpha` and `gamma`, `alpha` and `delta` are the roots of the equations `ax^(2)+2bx+c=0`, `2bx^(2)+cx+a=0` and `cx^(2)+ax+2b=0`

Correct Answer - C `"sin"(alpha+beta)"sin"(alpha-beta)="sin" gamma(2"sin"beta + "sin"gamma)` `"or ""sin"^(2)alpha-("sin"beta + "sin" gamma)^(2)=0` `"or "("sin"alpha+"sin" beta+"sin"gamma)("sin"alpha-"sin"beta-"sin"gamma)=0` `"Since " 0 lt alpha, beta, gamma lt pi," we have"` `"sin"alpha+"sin"beta+"sin"gamma ne 0` `therefore...

Correct Answer - A::B::D `A=[(cos alpha,sin alpha,0),(cos beta,sin beta,0),(cos gamma,sin gamma,0)][(cos alpha,cos beta,cos gamma),(sin alpha,sin beta,sin gamma),(0,0,0)]` Clearly, A is symmetric and `|A|=0`, hence, singular and not invertiable. Also, `A A^(T)...

Correct Answer - C `(alpha^(3))/(2) cosec^(2) ((1)/(2) tan^(-1). (alpha)/(beta)) + (beta^(3))/(2) sec^(2) ((1)/(2) tan^(-1).(beta)/(alpha))` `= alpha^(3) (1)/(1 - cos(tan^(-1) ((alpha)/(beta)))) + beta^(3) (1)/(1 + cos (tan^(-1).(beta)/(alpha)))` `= alpha^(3) (1)/(1 -cos (cos^(-1)...

Correct Answer - C For `alpha=-pi//2,beta=pi//2andgamma=2pi` `sinalpha+sinbeta=singamma=-2` Hence, the minimum value of the expression is negative.

Correct Answer - C `(c )` Because `b^(2)=ac`, the roots of both the equations are equal. `:.alpha=beta` and `gamma=delta`. But `gamma=(1)/(alpha)` and `delta=(1)/(beta)` as the given equations are reciprocal to each...

Correct Answer - B `(b)` `P(x)=x^(6)-x^(5)-x^(3)-x^(2)-x` and `Q(x)=x^(4)-x^(3)-x^(2)-1` Here` P(x)=Q(x)(x^(2)+1)+x^(2)-x+1` `:. sumP(alpha)=sumalpha^(2)-sumalpha+4` `=(sumalpha)^(2)-2sum(alphabeta)-sumalpha+4` `=1-2(-1)-(1)+4` `=6`

Correct Answer - A `(a)` `a=e^(ialpha)`, `b=e^(ibeta)`, `c=e^(igamma)` Clearly `a+b+c=0` `impliesa+b+c=0=(1)/(a)+(1)/(b)+(1)/(c )` `impliesa^(3)+b^(3)+x^(3)=3abc` `impliessin3alpha+sin3beta+sin3gamma=sin(alpha+beta+gamma)`

Correct Answer - B::C `(b,c)` `alpha,beta,gamma,delta` are in `H.P.` `implies(1)/(alpha)`, `(1)/(beta)`, `(1)/(gamma)`, `(1)/(delta)` are in `A.P.` Let `d` be the common difference of the `A.P.` Since `alpha`, `gamma` are roots of...

Correct Answer - C `(c )` The given equation is `px^(3)+qx^(2)-rx-r=0` `alpha+beta+gamma=(-q)/(p)`, `alphabeta+betagamma+gammaalpha=-(r )/(p)`, `alphabetagamma=(r )/(p)` `D=(1+alpha)(1+beta)(1+gamma)+1+1-(1+alpha)-(1+beta)-(1+gamma)` `=alphabetagamma+alphabeta+alphagamma+betagamma` `=(r )/(p)-(r )/(p)=0`

Correct Answer - D `(d)` We have `betaalpha+gammaalpha+alphabeta=0` `Delta=(1)/(alpha^(3)beta^(3)gamma^(3))|{:(betagamma,gammaalpha,alphabeta),(gammaalpha,alphabeta,betagamma),(alphabeta,betagamma,gammaalpha):}|` `=(1)/(alpha^(3)beta^(3)gamma^(3))|{:(betagamma+gammaalpha+alphabeta,gammaalpha,alphabeta),(gammaalpha+alphabeta+betagamma,alphabeta,betagamma),(alphabeta+betagamma+gammaalpha,betagamma,gammaalpha):}|` [using `C_(1)toC_(1)+C_(2)+C_(3)`] `=(1)/(alpha^(3)beta^(3)gamma^(3))|{:(0,gammaalpha,alphabeta),(0,alphabeta,betagamma),(0,betagamma,gammaalpha):}|=0` [all zero property].