Vector `vecP` angle `alpha,beta` and `gamma` with the x,y and z axes respectively.
Then `sin^(2)alpha+sin^(2)beta+sin^(2)gamma=`
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A nucleus with Z =92 emits the following in a sequence: `alpha,beta^(-),beta^(-),alpha,alpha,alpha,alpha,alpha,beta^(-),beta^(-),alpha,beta^(+),beta^(
Correct Answer - B `8 alpha` have been emitted `4 beta^(-)` have been emitted `2 beta^(+)` have emitted `alpha` reduces atomic number by 2 `beta^(-)` increases atomic number by 1 `beta^(+)`...
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Using peoperties of determinants in questions 11 to 15, prove that : `|{:(alpha,alpha^(2),beta+gamma),(beta,beta^(2),gamma+alpha),(gamma,gamma^(2),alp
L.H.S.=`|{:(alpha,alpha^(2),beta+gamma),(beta,beta^(2),gamma+alpha),(gamma,gamma^(2),alpha+beta):}|=|{:(alpha,alpha^(2),alpha+beta+gamma),(beta,beta^(2),alpha+beta+gamma),(gamma,gamma^(2),alpha+beta+gamma):}|` `(C_(3)toC_(3)+C_(1))` `=(alpha+beta+gamma)|{:(alpha,alpha^(2),1),(beta,beta^(2),1),(gamma,gamma^(2),1):}|` `(alpha+beta+gamma)|{:(alpha,alpha^(2),1),(beta-alpha,beta^(2)-alpha^(2),0),(gamma-alpha,gamma^(2)-alpha^(2),0):}|` `(R_(2)toR_(2)-R_(1),R_(3)toR_(3)-R_(1))` `(alpha+beta+gamma)(beta-alpha)(gamma-alpha)|{:(alpha,alpha^(2),1),(1,beta-alpha,0),(1,gamma-alpha,0):}|` `=-(alpha-beta)(gamma-alpha)(alpha+beta+gamma)1.|{:(1,beta+alpha),(1,gamma+alpha)` `=-(alpha-beta)(gamma-alpha)(alpha+beta+gamma)(gamma+alpha-beta-alpha)` `=(alpha-beta)(beta-gamma)(gamma-alpha)(alpha+beta+gamma)` =R.H.S.
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If `"sin"(alpha + beta) "sin" (alpha-beta) = "sin" gamma(2"sin" beta + "sin"gamma) " where " 0 lt alpha, beta, lt pi,` then the straight line whose eq
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...
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If `A(alpha, beta)=[("cos" alpha,sin alpha,0),(-sin alpha,cos alpha,0),(0,0,e^(beta))]`, then `A(alpha, beta)^(-1)` is equal to
Correct Answer - A We have, `A(alpha, beta)^(-1)=1/e^(beta) [(e^(beta) cos alpha,-e^(beta) sin alpha,0),(e^(beta) sin alpha,e^(beta) cos alpha,0),(0,0,1)]` `=A(-alpha, -beta)`
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If `alpha, beta, gamma` are three real numbers and `A=[(1,cos (alpha-beta),cos(alpha-gamma)),(cos (beta-alpha),1,cos (beta-gamma)),(cos (gamma-alpha),
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)...
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The value of `(alpha^3)/2cos e c^2(1/2tan^(-1)alpha/beta)+(beta^3)/2sec^2(1/2tan^(-1)(beta/alpha))i se q u a lto` `(alpha+beta)(alpha^2+beta^2)` (b) `
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)...
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The minimum value of the expression `sin alpha + sin beta+ sin gamma`, where `alpha,beta,gamma` are real numbers satisfying `alpha+beta+gamma=pi` is
Correct Answer - C For `alpha=-pi//2,beta=pi//2andgamma=2pi` `sinalpha+sinbeta=singamma=-2` Hence, the minimum value of the expression is negative.
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Let `alpha,beta` be two real numbers satisfying the following relations `alpha^2+beta^2=5, 3(alpha^5+beta^5)=11(alpha^3+beta^3)1.` Possible value of `
Correct Answer - A `alpha^(2)+beta^(2)=5` `3(alpha^(5)+beta^(5))=11(alpha^(3)+beta^(3))` `(alpha^(5)+beta^(5))/(alpha^(3)+beta^(3))=(11)/(3)` ` :. ((alpha^(3)+beta^(3))(alpha^(2)+beta^(2))-(alpha^(2)beta^(2)(alpha+beta)))/(alpha^(3)+beta^(3))=(11)/(3)` `:. alpha^(2)+beta^(2)-(alpha^(2)beta^(2)(alpha+beta))/((alpha+beta)(alpha^(2)+beta^(2)-alphabeta))=(11)/(3)` ` :. 5-(alpha^(2)beta^(2))/(5-alphabeta)=(11)/(3)` ltbrlt `:. (25-5alphabeta-alpha^(2)beta^(2))/(5-alphabeta)=(11)/(3)` Let `alphabeta=t` `(25-5t-t^(2))/(5-t)=(11)/(3)` `75-15t-3t^(2)=55-11 t` `75-15t-3t^(2)-55+11t=0` `-3t^(2)-4t+20=0` `(t-2)(3t+10)=0` ` :. t=2` or `(-10)/(3)` So `alpha...
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If `cos alpha+cos beta+cos gamma=0=sin alpha+sin beta+sin gamma`, then `(sin3alpha+sin3beta+sin3gamma)/(sin(alpha+beta+gamma))` is equal to
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)`
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If `alpha, beta. gamma` are the roots of `x^3 + px^2 + q = 0,` where `q=0,` ther `Delta=[(1/alpha,1/beta,1/gamma),(1/beta,1/gamma,1/alpha),(1/gamma,1/
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].
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