Question

Let $y=f\left(x\right)$ be the equation of a parabola which is touches by the line $y=x$ at the point where $x=1$. Then

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

Answer 1

Answer: (A), (B), (C), (D)

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

Solving $y=f\left(x\right)$ and $y=x$ we get
$f\left(x\right)=x$
Differentiating this w.r.t x we get
$f^{{}^{\prime }}\left(x\right)=1$ ...(1)
At $x=1$ we get $f^{{}^{\prime }}\left(1\right)=1$
Differentiating (1) w.r.t x we get
$f^{\prime \prime }\left(x\right)=0$
For $x=0$ $f\left(0\right)=0$, $f^{{}^{\prime }}\left(0\right)=1$ and $f^{\prime \prime }\left(0\right)=0$
Thus, $f\left(0\right)+f^{{}^{\prime }}\left(0\right)+f^{\prime \prime }\left(0\right)=0+1+0=1$
$f^{\prime }\left(0\right)=f^{\prime }\left(1\right)=1$ and $2f^{\prime }\left(0\right)=1-f^{\prime }\left(0\right)=0$
Hence, options 'A', 'B', 'C' and 'D' are correct.