If `sqrt(9x^2+6x+1)<2-x` then

Number of real values of x satisfying the equation sqrt(x^2 - 6x+9) + sqrt(x^2 - 6x +6) = 1 is (i)0 (ii)1 (iii)2

Using properties of proportion, solve for x : (i) (sqrt(x + 5) + sqrt(x - 16))/ (sqrt(x + 5) - sqrt(x - 16)) = (7)/(3) (ii) (sqrt(x + 1) + sqrt(x - 1))/ (sqrt(x + 1) - sqrt(x - 1)) = (4x -1)/(2) . (iii) (3x + sqrt(9x^(2) -5))/(3x - sqrt(9x^(2) -5)) = 5 .

If int (x +(cos^-1 3x)^2)/sqrt(1 - 9x^2) dx = A sqrt(1 - 9x^2) + B (cos^-1 3x)^3 + c then A, B are respectively

lim_(x rarr3)((x^(2)-9-sqrt(x^(2)-6x+9)))/(|x-1|-2)

(1)/(log_(3)(x+1))<(1)/(2log_(9)sqrt(x^(2)+6x+9))

Find the range of f(x)=sqrt(1-sqrt(x^2-6x+9))

Find the range of f(x)=sqrt(1-sqrt(x^(2)-6x+9))

Find the range of f(x)=sqrt(1-sqrt(x^2-6x+9))

Find the range of f(x)=sqrt(1-sqrt(x^2-6x+9))

lim_(x rarr oo)((sqrt(1+25x^(2))+sqrt(9x^(2)-1))/(sqrt(1+25x^(2))-sqrt(9x^(2)-1)))=