If `lambda!=0,a,b,c,lambda` are real such that `a^(2)+b^(2)+c^(2)-2 lambda a-4b lambda-8 lambda c+21 lambda^(2)<=0` then (A) `a,b,c` are in A.P (B) `a,b,c` are in G.P (C) `a,b,c` are in H.P (D) `a,b,c` are in A.G.P

lambda ^ (3) -6 lambda ^ (2) -15 lambda-8 = 0

|[a^(2)+lambda^(2),ab+c lambda,ca-b lambdaab-c lambda,b^(2)+lambda^(2),bc+aca+b lambda,bc-a lambda,c^(2)+lambda^(2) then he value of lambda is 8 b.27, c.1 d.-1=(1+a^(2)+b^(2)+c^(2))^(3)

If |(a^2,b^2,c^2),((a+lambda)^2,(b+lambda)^2,(c+lambda)^2),((a-lambda)^2,(b-lamda)^2,(c-lambda)^2)|=klambda|(a^2,b^2,c^2),(a,b,c),(1,1,1)|lambda!=0 then k is equal to : (A) 4 lambda abc (B) -4 lambda abc (C) 4 lambda^2 (D) -4 lambda^2

If p lambda^(4) + q lambda^(3) + r lambda^(2) + s lambda + t = |{:(b^(2)+c^(2), a^(2) + lambda , a^(2) + lambda),(b^(2)+lambda,c^(2)+a^(2),b^(2)+lambda),(c^(2)+lambda,c^(2)+lambda,a^(2)+b^(2)):}| is an identity in lambda wherep, q, r,s , t are constants, then the value of t is

If a lambda^(4)+b lambda^(3)+c lambda^(2)+d lambda+e=|[lambda^(2)-3,lambda-1,lambda+sqrt(3)],[lambda^(2)+lambda,2 lambda-4,3 lambda+1],[lambda^(2)-3 lambda,3 lambda+5,lambda-3]| then |a| equals

p(lambda)^4+q(lambda)^3+r(lambda)^2+s(lambda)+t = |((lambda^2 + 3lambda) , lambda -1 , lambda+3) , (lambda +1 , -2lambda , lambda-4 ) ,(lambda-3 , lambda+4 , 3lambda)| find t=?

If a lambda^(4)+b lambda^(3)+c lambda^(2)+d lambda+e = |[lambda^(2)-3,-1,lambda+sqrt(3)],[lambda^(2)+lambda,2 lambda-4,3 lambda+1],[lambda^(2)-3 lambda,3 lambda+5,lambda-3]| then |a| equals

If the two roots of the equation (lambda-2)(x^2+x+1)^2-(lambda+2)(x^4+x^2+1) are real and equal for lambda=lambda_1,lambda_2 then lambda_1+lambda_2=0 (b) |lambda_1-lambda_2|=6 (c) lambda_1+lambda_2=32 (d) all of these

Two particle are moving perpendicular to each with de-Broglie wave length lambda_(1) and lambda_(2) . If they collide and stick then the de-Broglie wave length of system after collision is : (A) lambda = (lambda_(1) lambda_(2))/(sqrt(lambda_(1)^(2) + lambda_(2)^(2))) (B) lambda = (lambda_(1))/(sqrt(lambda_(1)^(2) + lambda_(2)^(2))) (C) lambda = (sqrt(lambda_(1)^(2) + lambda_(2)^(2)))/(lambda_(2)) (D) lambda = (lambda_(1) lambda_(2))/(sqrt(lambda_(1) + lambda_(2)))

det [[lambda ^ (4) +2 lambda ^, 2 lambda ^ (3) -3,3 lambda ^ (2) + lambda2,3,15,0,2]] = a lambda ^ (4) + b lambda ^ (3) + c lambda ^ (2) + d lambda + e