Question

A polynomial function f(x) satisfies the condition $f(x+1)=f\left(x\right)+2x+1$. Find $f\left(x\right)$ if $f\left(0\right)=1$. Find also the equations of the pair of tangents from the origin on the curve $y=f\left(x\right)$ and compute the area enclosed by the curve and the pair of tangents.

• A. $f\left(x\right)=x^2+1;$, $y=\pm 2x;$ , $A=\frac{2}{3}$ sq.units
• B. $f\left(x\right)=x^2-1;$, $y=\pm 2x;$ , $A=\frac{2}{3}$ sq.units
• C. $f\left(x\right)=x^2+1;$, $y=\pm 2x;$ , $A=\frac{3}{2}$ sq.units
• D. $f\left(x\right)=x^2-1;$, $y=\pm 2x;$ , $A=\frac{3}{2}$ sq.units

Answer 1

The correct option is A $f\left(x\right)=x^2+1;$, $y=\pm 2x;$ , $A=\frac{2}{3}$ sq.units

Let $f\left(x\right)=x^2+1$
Hence $f\left(0\right)=1$ and
$f(x+1)$
$=(x+1{)}^2+1$
$=x^2+2x+2$
$=(x^2+1)+2x+1$
$=f\left(x\right)+2x+1$.
Therefore $f\left(x\right)=x^2+1$ satisfies all the conditions required.
Let the tangents be $y=mx$
Hence
$f^{\prime }(x{)}_{x=h}=m$
Hence
$2x=m$
$x=\frac{m}{2}$ ...(i)
Now
$mx=x^2+1$
$x^2-mx+1=0$
$x=\frac{m\pm \sqrt{m^2-4}}{2}$
$=\frac{m}{2}$ ... from i.
Hence
$\sqrt{m^2-4}=0$
Or
$m=\pm 2$.
Hence the equations of the tangents are $y=\pm 2x$
The required area will be
$=|2{\int }_0^1{x}^2+1-2x.dx|$ or $|2{\int }_0^{-1}{x}^2+1+2x.dx|$
Hence
$=|2{\int }_0^1{x}^2+1-2x.dx|$
$=2{\int }_0^1(x-1{)}^2.dx$
$=2.|\left[\frac{(x-1{)}^3}{3}\right]{|}_0^1$
$=\frac{2}{3}$sq units.