Given, curve ⇒y=ax2+bx+c,x∈R and point (1,2) ∵ The given curve passes through (1,2). ∴2=a+b+c Also, slope of tangent of y=ax2+bx+c is
dy
dx
=2ax+b ∵ Tangent passes through origin (0,0) . ∴
dy
dx
|(0,0)=2a×0+b=b⋅⋅⋅⋅⋅⋅⋅(i) According to the question, tangent at origin is y=x ∴ Its slope is 1.⋅⋅⋅⋅⋅⋅⋅(ii) From Eqs. (i) and (ii), b=1 Also, a+b+c=2 cc⇒a+c+1=2 ⇒a+c=1 From the option look for b=1 and a+c=1 The only correct order triplet is a=1, b=1,c=0.