Question

The intercepts on $x-$axis, made by tangents to the curve, $y={\int }_0^x\left|t\right|dt,x\in \mathrm{ℝ}$ which are parallel to the line $y=2x$, are

• A. $\pm 1$
• B. $\pm 2$
• C. $\pm 3$
• D. $\pm 4$

Answer 1

The correct option is A

$\pm 1$

Explanation for the correct answer:

Step 1: Find the slope of the given line

The equation of the curve: $y={\int }_0^x\left|t\right|dt,x\in \mathrm{ℝ}$.

The equation of the straight line: $y=2x$.

Differentiate both sides of the equation with respect to $x$.

$\frac{dy}{dx}=2$.

As the tangents of the curve are parallel to the given straight line.

Thus, the slope of the tangents can be given by, $\frac{dy}{dx}=2$.

Case-1: For $x\ge 0$.

Step 2:Find the slope of the tangent for Case 1

The equation of the curve: $y={\int }_0^x\left|t\right|dt$.

$$\begin{gathered} \Rightarrow y={\int }_0^{x}tdt \\ \Rightarrow y={\left[\frac{t^2}{2}\right]}_0^x \\ \Rightarrow y=\frac{x^2}{2} \end{gathered}$$

Differentiate both sides of the equation with respect to $x$.

$\frac{dy}{dx}=\frac{d}{dx}\left(\frac{x^2}{2}\right)=\frac{2x}{2}=x$.

Thus, the slope of the tangent $x=2$.

So, $y=\frac{x^2}{2}=\frac{2^2}{2}=2$.

Step 3:Find the equation of the tangent for Case 1

Hence, the equation of the tangent with slope $m=2$ and passing through the point $\left(2,2\right)$ can be given by,

$$\begin{gathered} \mathbf{(}\mathbf{y}\mathbf{-}{\mathbf{y}}_{\mathbf{1}}\mathbf{)}\mathbf{=}\mathit{m}\mathbf{(}\mathbf{x}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}\mathbf{)} \\ \Rightarrow \left(y-2\right)=2\left(x-2\right) \\ \Rightarrow 2x-y=2 \\ \Rightarrow \frac{x}{1}+\frac{y}{-2}=1 \end{gathered}$$

Compare the equation with the intercept form of a straight line $\frac{x}{a}+\frac{y}{b}=1$.

Thus, the $x-$intercept of the tangent, $a=1$.

Case-2: For $x<0$.

Step 4:Find the slope of the tangent for Case 2

The equation of the curve: $y={\int }_0^x\left|t\right|dt$.

$$\begin{gathered} \Rightarrow y=-{\int }_0^{x}tdt \\ \Rightarrow y=-{\left[\frac{t^2}{2}\right]}_0^x \\ \Rightarrow y=-\frac{x^2}{2} \end{gathered}$$

Differentiate both sides of the equation with respect to $x$.

$\frac{dy}{dx}=\frac{d}{dx}\left(-\frac{x^2}{2}\right)=-\frac{2x}{2}=-x$.

Thus, the slope of the tangent $\left(-x\right)=2$.

$\Rightarrow x=-2$

So, $y=-\frac{x^2}{2}=-\frac{{\left(-2\right)}^2}{2}=-2$.

Step 5:Find the equation of the tangent for Case 2

Hence, the equation of the tangent with slope $m=2$ and passing through the point $\left(-2,-2\right)$ can be given by,

$$\begin{gathered} \mathbf{(}\mathbf{y}\mathbf{-}{\mathbf{y}}_{\mathbf{1}}\mathbf{)}\mathbf{=}\mathit{m}\mathbf{(}\mathbf{x}\mathbf{-}{\mathbf{x}}_{\mathbf{1}}\mathbf{)} \\ \Rightarrow \left(y+2\right)=2\left(x+2\right) \\ \Rightarrow 2x-y=-2 \\ \Rightarrow \frac{x}{-1}+\frac{y}{2}=1 \end{gathered}$$

Compare the equation with the intercept form of a straight line $\frac{x}{a}+\frac{y}{b}=1$.

Thus, the $x-$intercept of the tangent, $a=-1$.

Therefore. the $x-$intercepts of the tangent are $\pm 1$.