Question

Let $y=f\left(x\right)$ be a parabola, having its axis parallel to y-axis, which is touched by the line $y=x$ at $x=1$, then

• A. $2f\left(0\right)=1-f^{\prime }\left(0\right)$
• B. $f\left(0\right)+f^{\prime }\left(0\right)+f"\left(0\right)=1$
• C. $f^{\prime }\left(1\right)=1$
• D. $f^{\prime }\left(0\right)=f^{\prime }\left(1\right)$

Answer 1

Answer: (A), (C)

The correct options are
A $2f\left(0\right)=1-f^{\prime }\left(0\right)$
C $f^{\prime }\left(1\right)=1$

The general equation of a parabola having its axis parallel to y-axis is
$y=a{x}^2+bx+c$ ........ (i)
The line $y=x$ touches the required parabola. $y=a{x}^2+bx+c$ at $(1,1)$.
Hence, the slope of the parabola at $x=1$ is $1$.
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(1,1)}=1$
$\Rightarrow 2a+b=1$ .....(ii)
Also, (1, 1) lies on the parabola.
$\Rightarrow a+b+c=1$ ...(iii)
From the equations (ii) and (iii)
$a-c=0$
$\Rightarrow a=c$
Using $a=c$ in (iii), we get $2c+b=1$.
$\Rightarrow 2f\left(0\right)+f^{\prime }\left(0\right)=1$ $[\because f(0)=cand{f}^{\prime }(0)=b]$
or, $2f\left(0\right)=1-f^{\prime }\left(0\right)$
The other options are not satisfied.
Hence, option A and C are correct.