The curve $y=f\left(x\right)$ Passes through the origin O and the tangent to the curve at the origin is the x-axis . The tangent at any point $P(x,y)$ on the curve makes an angle $\psi$ with the x-axis and the arc length OP is s. Then the relation between $\psi$ and s will be called the intrinsic equation of the curve.

 

 

 

 

 

 

Also, the length $l$ of the arc AB on the curve $y=f\left(x\right)$, where $A=(x_1,y_1)$ and  $B=(x_2,y_2)$ is given by $l=\underset{x=x_1}{\overset{x_2}{\int }}ds$ where ${\left(ds\right)}^2={\left(dx\right)}^2+{\left(dy\right)}^2$ and so $\frac{ds}{dx}=\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}$ based on above information which of the following statements is/are correct?

Both points are on the curve. So, the required distance is the arc length of the curve $y=x^2+1$  from the point $(0,1)$ and $(1,2)$ .
$\therefore$ the requires distance 
$$\begin{gathered} =\underset{x=0}{\overset{1}{\int }}ds=\underset{0}{\overset{1}{\int }}\frac{ds}{dx}dx=\underset{0}{\overset{1}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx \\ =\underset{0}{\overset{1}{\int }}\sqrt{1+{\left(2x\right)}^2}dx \\ ={\left[\frac{1}{2}\{\frac{2x}{2}\sqrt{4x^2+1}+\frac{1}{2}\log (2x+\sqrt{4x^2+1})\}\right]}_0^1 \\ =\frac{\sqrt{5}}{2}+\frac{1}{4}\log (2+\sqrt{5}) \end{gathered}$$

So, the correct answer is A

2. Here $y^2=4x \Rightarrow \frac{dy}{dx}=\frac{2}{y}.$

So arc $POQ=2\times arcOP=2\underset{0}{\overset{1}{\int }}ds$

$$\begin{gathered} \therefore arc POQ=2\underset{0}{\overset{1}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx=2\underset{0}{\overset{1}{\int }}\sqrt{1+\frac{4}{y^2}}dx =2\underset{0}{\overset{1}{\int }}\sqrt{1+\frac{4}{4x}}dx=2\underset{0}{\overset{1}{\int }}\sqrt{\frac{x+1}{x}}dx \\ =2\underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\int }}\sqrt{\frac{{\sec }^2\theta }{{\tan }^2\theta }}.2\tan \theta .{\sec }^2\theta d\theta \end{gathered}$$

(putting $x={\tan }^2\theta$)

$=4\underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\int }}{\sec }^3\theta d\theta$

Now, $I=\underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\int }}{\sec }^3\theta d\theta ={(\sec \theta .\tan \theta )}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}-\underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\int }}\tan \theta .\sec \theta .\tan \theta .d\theta$

$\begin{array}{l}=\sqrt{2}-\underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\int }}({\sec }^3\theta -\sec \theta )d\theta \\ =\sqrt{2}-I+{\left[\log (\sec \theta +\tan \theta )\right]}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\\ \therefore 2I=\sqrt{2}+\log (\sqrt{2}+1).\end{array}$