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" Dumes "1quad |3x-2|<=(1)/(2)...

" Dumes "1quad |3x-2|<=(1)/(2)

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The equation |{:((1+x)^(2),(1-x)^(2),-(2+x^(2))),(2x+1,3x,1-5x),(x+1,2x,2-3x):}|+|{:((1+x)^(2),2x+1,x+1),((1-x)^(2),3x,2x),(1-2x,3x-2,2x-3):}|=0

If a,b,c are in A.P .and f(x)=|[x^2+x+a+1, x^2+1, 1] , [2x^2+x+b-1, 2x^2-1, 1] , [3x^2+x+c-2, 3x^2-2, 1]| then f'(x) is

Let tan^(-1)y=tan^(-1)x+tan^(-1)((2x)/(1-x^2)) , where |x|<1/(sqrt(3)) . Then a value of y is : (1) (3x-x^3)/(1-3x^2) (2) (3x+x^3)/(1-3x^2) (3) (3x-x^3)/(1+3x^2) (4) (3x+x^3)/(1+3x^2)

Let tan^(-1)y=tan^(-1)x+tan^(-1)((2x)/(1-x^2)) , where |x|<1/(sqrt(3)) . Then a value of y is : (1) (3x-x^3)/(1-3x^2) (2) (3x+x^3)/(1-3x^2) (3) (3x-x^3)/(1+3x^2) (4) (3x+x^3)/(1+3x^2)

Let tan^(-1)y=tan^(-1)x+tan^(-1)((2x)/(1-x^2)) , where |x|<1/(sqrt(3)) . Then a value of y is : (1) (3x-x^3)/(1-3x^2) (2) (3x+x^3)/(1-3x^2) (3) (3x-x^3)/(1+3x^2) (4) (3x+x^3)/(1+3x^2)

The constant term in the expansion of |(3x +1,2x-1,x+2),( 5x-1, 3x+2,x+1),(7x-1,3x+1,4x-1)| is

If |(x^2+x,x+1,x-2),(2x^2+3x-1,3x,3x-3),(x^2+2x+3,2x-1,2x-1)|=ax-12 then 'a' is equal to (1) 12 (2) 24 (3) -12 (4) -24

If |(x^2+x,x+1,x-2),(2x^2+3x-1,3x,3x-3),(x^2+2x+3,2x-1,2x-1)|=ax-12 then 'a' is equal to (1) 12 (2) 24 (3) -12 (4) -24

If |(x^2+x,x+1,x-2),(2x^2+3x-1,3x,3x-3),(x^2+2x+3,2x-1,2x-1)|=ax-12 then 'a' is equal to (1) 12 (2) 24 (3) -12 (4) -24