let f(x)=(ax+5)/(x^2+2) then find the value of a for which the function is invertible - Maths -

Question

let f(x)=(ax+5)/(x^2+2) then find the value of a for which the
function is invertible

Aarushi Mishra · Accepted Answer

fx=ax+5x2+2For fx to be invertible, it should be one-one and ontof'x=x2+2d dxax+5-ax+5d dxx2+2x2+22f'x=ax2+2-2xax+5x2+22f'x=ax2+2a-2ax2-10xx2+22f'x=-ax2-10x+2ax2+22for fx to be one-one f'x≥0 or f'x≤0Note: x2+22>0 for all x, and for f'x≥0 or f'x≤0 discriminant D of -ax2-10x+2a, D≤0D=-102-4×2a×-a D=100+8a2>0But D≤0
Hence there is no value of a