Question

A polynomial function f(x) satisfies the condition $f(x+1)=f\left(x\right)+2x+1$ Find f(x) 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. $f\left(x\right)=x^2+1;y=\pm x;A=\frac{2}{3}sq.units$
• B. $f\left(x\right)=x^2+1;y=\pm x;A=\frac{4}{3}sq.units$
• C. $f\left(x\right)=x^2+1;y=\pm 2x;A=\frac{2}{3}sq.units$
• D. $f\left(x\right)=x^2+1;y=\pm 2x;A=\frac{4}{3}sq.units$

Answer 1

Answer: (C)

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

$f(x+1)=f\left(x\right)+2x+1$
$\Rightarrow f^{\prime \prime }(x+1)=f^{\prime \prime }\left(x\right)\forall \times \in R$ Let $f"\left(x\right)=a\Rightarrow$ $f^{\prime }\left(x\right)=ax+b\Rightarrow f\left(x\right)=\frac{a{x}^2}{2}+bx+c\Rightarrow c=1$ $[\because f(0)=1]$
Now $f(x+1)-f\left(x\right)=2x+1\Rightarrow [\frac{a}{2}{(x+1)}^2+b(x+1)+c]-[\frac{a{x}^2}{2}+bx+c]=2x+1$ $\Rightarrow ax+\frac{a}{2}+b=2x+1$ on comparing we get a = 2
or $\frac{a}{2}+b=1\Rightarrow b=0$
$\therefore f\left(x\right)=x^2+\mathrm{1.........}\left(i\right)$
Now let equation of tangent be $y=mx.......\left(ii\right)$
From (i) and (ii), we get
$x^2-mx+1=0\Rightarrow m=\pm 2$
$\therefore tangentsarey=2xory=-2x$ $A=2{\int }_0^1(x^2+1-2x)dx=\frac{2}{3}$