Question

The intercepts on $x$-axis made by tangents to curve, $y=\underset{0}{\overset{x}{\int }}|t|dt,x\in R$, which are parallel to the line $y=2x$, are equal to:

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

Answer 1

The correct option is A $\pm 1$

$y=2x...\left(1\right)$
$\Rightarrow m_1=\frac{dy}{dx}=2$
Now, $y=\underset{0}{\overset{x}{\int }}|t|dt,x\in R$
Case$-1$: if $t>0$
$\Rightarrow y=\underset{0}{\overset{x}{\int }}tdt$
$\Rightarrow y=\frac{x^2}{2}...\left(2\right)$
$\Rightarrow m_2=\frac{dy}{dx}=\frac{2x}{2}=x$
As, $m_1=m_2\Rightarrow x=2$
From equation $\left(2\right),y=2$
The equation of line pasing through $(2,2)$ and have slope $2$ is
$y-2=2(x-2)$
It has intercept on $x-$ axis, so, $y=0$
$\Rightarrow 0-2=2(x-2)\Rightarrow x=1$
$\therefore$ intercept on $x$-axis $=1$
Case$-2$: If $t<0$
$y=\underset{0}{\overset{x}{\int }}-tdt$
$\Rightarrow y=-\frac{x^2}{2}...\left(3\right)$
$\Rightarrow m_3=\frac{dy}{dx}=-\frac{2x}{2}=-x$
As, $m_3=m_1\Rightarrow x=-2$
From equation $\left(3\right),y=-2$
The equation of line pasing through $(-2,-2)$ and have slope $2$ is
$y+2=2(x+2)$
It has intercept on $x-$ axis, so put, $y=0$
$\Rightarrow 0+2=2(x+2)\Rightarrow x=-1$
$\therefore$ intercept on $x$-axis $=-1$
Hence, from case$-1$ and case$-2$, intercept on $x$-axis $=\pm 1$