Home
Class 12
MATHS
The least value of the product xyz for w...

The least value of the product xyz for which the determinant `|(x,1,1),(1,y,1),(1,1,z)|` is non-negative, is: (A) `-16sqrt2` (B) `-2sqrt2` (C) `-1` (D) `-8`

Promotional Banner

Similar Questions

Explore conceptually related problems

If x , y in R , then the determinant = |(cos x , -sin x,1),(sin x, cos x,1),(cos(x+y), -sin(x+y), 0)| lies in the interval (a) [-sqrt(2),sqrt(2)] (b) [-1,1] (c) [-sqrt(2),1] (d) [-1,-sqrt(2)]

Given that xyz = -1 , the value of the determinant |(x,x^(2),1 +x^(3)),(y,y^(2),1 + y^(3)),(z,z^(2),1 +z^(3))| is

If x , y in R , then the determinant =|[cosx,-sinx,1],[sinx,cosx,1],[cos(x+y),-sin(x+y),0]| lies in the interval (a) [-sqrt(2),sqrt(2)] (b) [-1,1] (c) [-sqrt(2),1] (d) [-1,-sqrt(2)]

If |z-4/z|=2 then the greatest value of |z| is (A) sqrt(5)-1 (B) sqrt(5)+1 (C) sqrt(5) (D) 2

d/(dx)(sin^(-1)x+cos^(-1)x) is equal to : (A) (1)/(sqrt(1-x^(2))), (B) (2)/(sqrt(1-x^(2))), (C) 0 (D) sqrt(1-x^(2))

The equation of the straight line making angles 60^(@),60^(@)and45^(@) with positive direction of the coordinate axes and passing through the point (2,1,-1) is a) sqrt(2)(x-2)=sqrt(2)(y-1)=(z+1) b) (x-2)=sqrt(2)(y-1)=(z+1) c) sqrt(2)(x-2)=(y-1)=(z+1) d) (x-2)=sqrt(2)(y-1)=sqrt(2)(z+1)

If x , y in R , then the determinant =|cosx-sinx1sinxcosx1cos(x+y)-sin(x+y)0| lies in the interval [-sqrt(2),sqrt(2)] (b) [-1,1] [-sqrt(2),1] (d) [-1,-sqrt(2)]

For the curve sqrt(x)+sqrt(y)=1 , (dy)/(dx) at (1//4,\ 1//4) is (a) 1//2 (b) 1 (c) -1 (d) 2

The value of sin(1/4sin^(-1)(sqrt(63))/8) is 1/(sqrt(2)) (b) 1/(sqrt(3)) (c) 1/(2sqrt(2)) (d) 1/(3sqrt(3))

The value of sin(1/4sin^(-1)((sqrt(63))/8)) is (a) 1/(sqrt(2)) (b) 1/(sqrt(3)) (c) 1/(2sqrt(2)) (d) 1/(3sqrt(3))