Question

The intercept on the x-axis made by the tangent to the curve, $ y=\int\limits_{0}^{x}{\left| t \right|dt} $ , $ x\in R $ , which are parallel to the line y = 2x, are equal to
a). $ \pm 2 $
b). $ \pm 3 $
c). $ \pm 4 $
d). $ \pm 1 $

Answer 1

Hint: To solve the above question we will first differentiate the curve equation and get the slope of the general tangent and since the required tangent is parallel to y = 2x, so we will equate the slope of tangent obtained after differentiation to the slope of line y = 2x and then find the value of y from $ y=\int\limits_{0}^{x}{\left| t \right|dt} $ and get the tangent equation.

Complete step by step answer:
We know that the given curve is $ y=\int\limits_{0}^{x}{\left| t \right|dt} $ and we also know that the differentiation of any curve gives the general tangent equation.
So, after differentiating the curve $ y=\int\limits_{0}^{x}{\left| t \right|dt} $ , we will get:
 $ \Rightarrow \dfrac{dy}{dx}=\left| x \right| $
So, the slope of the tangent at any point (x, y) to the given curve is equal to $ \left| x \right| $ .
Also, we know that the required tangent is parallel to line y = 2x. and since the line is in the form of y = mx where ‘m’ is the slope of the line.
So, we can say that slope of the line y = 2x is equal to 2.
Then, we say that $ \dfrac{dy}{dx}=\left| x \right|=2 $
So, x = $ \pm 2 $
Now, we know that $ y=\int\limits_{0}^{x}{\left| t \right|dt} $ , so when x = 2:
 $ \Rightarrow y=\int\limits_{0}^{2}{\left| t \right|dt} $
 $ \Rightarrow y=\int\limits_{0}^{2}{tdt} $
 $ \Rightarrow y=\left| \dfrac{{{t}^{2}}}{2} \right|_{0}^{2}=2 $
Also, when x = -2, then:
 $ \Rightarrow y=\int\limits_{0}^{-2}{\left| t \right|dt} $
 $ \Rightarrow y=\int\limits_{0}^{-2}{-tdt} $
 $ \Rightarrow y=-\left| \dfrac{{{t}^{2}}}{2} \right|_{0}^{-2}=-2 $
So, equation of line with slope m = 2 and passing through point (2, 2) is given by:
\[\begin{align}
  & \Rightarrow \left( y-2 \right)=2\left( x-2 \right) \\
 & \Rightarrow y=2x-2 \\
\end{align}\]
And, the equation of the line with slope m = 2 and passing through the point (-2, -2) is given by:
\[\begin{align}
  & \Rightarrow \left( y+2 \right)=2\left( x+2 \right) \\
 & \Rightarrow y=2x+2 \\
\end{align}\]
So, the intercept made by the tangent y = 2x + 2 on the x- axis is equal to:
x = -1
And, the intercept made by the tangent y = 2x – 2 on the x-axis is equal to:
x = 1
So, x = $ \pm 1 $ .
Hence, option (d) is the correct answer.

Note:
Students are required to note that when we differentiate any given curve then we always get the slope of the tangent at any general point (x, y) on the curve, and the equation obtained after differentiating the curve is a locus of the slope of tangents to the curve.