JEE Main 24 Feb 2021 Shift 2 Solved Paper

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
⇒‌‌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.