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

The general equation of a parabola having its axis parallel to y-axis is
$y=a{x}^2+bx+c$
$\because$ parabola passes through $(1,1)$
$\therefore a+b+c=1\cdots \left(1\right)$
Equation of tangent to parabola at $(1,1)$ is $T=0$
$\Rightarrow \frac{1}{2}(y+1)=ax\left(1\right)+\frac{b}{2}(x+1)+c$
$\Rightarrow y+1=2ax+bx+b+2c$
$\Rightarrow y=(2a+b)x+(b+2c-1)$
But, the given equation is $y=x$
$\Rightarrow 2a+b=1,$ and
$b+2c-1=0$
$\Rightarrow 1-2a+2c-1=0$
$\Rightarrow a=c\cdots \left(2\right)$
$\therefore$ The equation of parabola is $y=a{x}^2+(1-2a)x+a$
$\Rightarrow f\left(0\right)=a$
$f^{\prime }\left(x\right)=2ax+(1-2a)$
$\Rightarrow f^{\prime }\left(0\right)=1-2a,f^{\prime }\left(1\right)=1$