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

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^(+)`...

`L.H.S.=|{:(1,x+alpha,y+z-alpha),(1,t+beta,+x-beta),(1,z+gamma,x+y-gamma):}|` `=|{:(1,x+alpha,y+z-alpha),(1,t+beta,+x-beta),(1,z+gamma,x+y-gamma):}|(C_(3)toC_(3)+C_(2))` `=(x+y+z)|{:(1,x+alpha,1),(1,t+beta,1),(1,z+gamma,1):}|` `=0 " "(therefore C_(1)" and "C_(3)" are same")` `=R.H.S." "Hence Proved`.

Correct Answer - A Arrange the data as follows : `alpha-(7)/(2), alpha-3, alpha-(5)/(2),alpha-2, alpha-(1)/(2),alpha+(1)/(2),alpha+4,alpha+5` Median `=(1)/(2)` [value of 4th item+value of 5th item] `therefore " Median"=(alpha-2+alpha-(1)/(2))/(2)=(2alpha-(5)/(2))/(2)=alpha-(5)/(4)`

Correct Answer - C `cos^(2)alpha+cos^(2)beta+1-sin_(gamma)^(2)=1` `1-sin^(2)alpha+sin^(2)beta+1-sin^(2)gamma=1` `3-(sin^(2)alpha+sin^(2)beta+sin^(2)gamma)=1` `sin^(2)alpha+sin^(2)beta+sin^(2)gamma=3-1=2`

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...

We have, tr (A)=tr (B), `:. alpha^(2)+beta^(2)+gamma^(2)=2alpha+2beta+2 gamma-3` `implies (alpha^(2)-2alpha+1)+(beta^(2)-2beta+1)+(gamma^(2)-2gamma+1)=0` `implies (alpha-1)^(2)+(beta-1)^(2)+(gamma-1)^(2)=0` `implies alpha=1, beta=1, gamma=1` `:. (alpha^(-1)+beta^(-1)+gamma^(-1))=(1+1+1)=3`

Given equation is `[(x,y),(z,t)]^(2)=[(0,0),(0,0)]` `implies [(x,y),(z,t)][(x,y),(z,t)]=[(x^(2)+yz,xy+yt),(zx+tz,zy+t^(2))]=[(0,0),(0,0)]` `implies x^(2)+yz=0` (1) `y(x+t)=0` (2) `z(x+t)=0` (3) `yz+t^(2)=0` (4) From (1) and (4), we have `x^(2)=t^(2)` or `x= pm t` Case I : If...

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)`

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)...