An element `_(z)M^(A)` undergoes an `alpha -` emission followed by two successive `beta- emission.` The element formed is ……
A. Statement I is true , Statement II is true, Statement II is the correct explanation of Statement I
B. Statement I is true, Statement II is true, Statement II is not the correct explanation of Statement I
C. Statement I is true, Statement II is false
D. Statement I is false, Statement II is true
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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 a variable takes the discrete values `alpha-4`, `alpha -(7)/(2), alpha-(5)/(2), alpha-2,alpha+(1)/(2), alpha-(1)/(2), alpha+5(alpha gt 0)`, then th
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)`
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If alpha + beta is equal to minus 3 and alpha beta is equal to minus 5 by 2 then find the cube roots are alpha and beta
α + β = - 3 .......(1) and αβ = \(\frac{-5}{2}\) .........(2) (α - β)2 = (α + β)2 - 4αβ = (-3)2 - 4 x \(\frac{-5}{2}\) = 9 + 10 = 19 α - β = ±√19 ........(3) By adding equations (1) and...
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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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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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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))` Poss
Correct Answer - B `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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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))` Quad
Correct Answer - D `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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