सिद्ध कीजिए - `|(x+lambda,2x,2x),(2x,x+lambda,2x),(2x,2x,x+lambda)|=(5x+lambda)(lambda-x)^2`.

If |(x+lambda,x,x),(x,x+lambda,x),(x,x,x+lambda)| = 0, (lambda != 0) then a) x = - lambda//3 b) x = 3 lambda c)x = 0 d)None of these

Let f be a real valued function such that for any real x, f(lambda+x)=f(lambda-x) and f(2lambda+x)=f(2lambda-x) for some lambda>0 . Then

the values of lambda for which (lambda^(2)-3 lambda+2)x^(2)+(lambda^(2)-5 lambda+6)+lambda^(2)-4=0 is identity in x is

If the equation (lambda^(2)-5 lambda+6)x^(2)+(lambda^(2)-3 lambda+2)x+(lambda^(2)-4)=0 is satisfied by more then two values of x then lambda=

The period of the function f(x) = lambda| sin x| + lambda^(2) | cos x| + g (lambda) is (lambdai)/(2) , if lambda is

If x=(4 lambda)/(1+lambda^(2)) and y=(2-2 lambda^(2))/(1+lambda^(2)) where lambda is a real parameter and Z=x^(2)+y^(2)-xy then possible values of Z can be

The equation lambda^(2)x^(2)+(lambda^(2)-5 lambda+4)xy+(3lambda-2)y^(2)-8x+12y-4=0 will represent a circle, if lambda=

The equation lambda^(2)x^(2)+(lambda^(2)-5 lambda+4)xy+(3lambda-2)y^(2)-8x+12y-4=0 will represent a circle, if lambda=

The values of lambda for which the function f(x)=lambda x^(3)-2 lambda x^(2)+(lambda+1)x+3 lambda is increasing through out number line (a) lambda in(-3,3)( b) lambda in(-3,0)( c )lambda in(0,3) (d) lambda in(1,3)