Question

A tangent to the curve $y={\int }_0^x\left|t\right|dt$, which is parallel to the line y=x, cuts off an intercept from the y-axis equal to

• A. $1$
• B. $-\frac{1}{2}$
• C. $\frac{1}{2}$
• D. $-1$

Answer 1

Answer: (B), (C)

The correct options are
B $\frac{1}{2}$
C $-\frac{1}{2}$

Since the integral is in the first quadrant $|t|=t$
Thus
${\int }_{x=0}^x|t|.dt$ $={\int }_{x=0}^{x}t.dt$
$=\frac{x^2}{2}$
Or
$2y=x^2$ is the equation of the curve.
Now
$2y^{\prime }=2x$
Or
$y^{\prime }=x$ ...(i)
Now the tangent is parallel to $y=x$
Hence slope$=y^{\prime }=1$
Thus the x-coordinate of point of contact is 1.
Hence
$y=\frac{1}{2}$.
Hence equation of the tangent will be
$y-\frac{1}{2}=(x-1).$
Or
$x-y=\frac{1}{2}$
Or
$y=x-\frac{1}{2}$.
Hence intercept cut on the y axis is $\frac{-1}{2}$.