A) \[\pm 1\]
B) \[\pm 2\]
C) \[\pm 3\]
D) \[\pm 4\]
Correct Answer: A
Solution :
\[y=\int\limits_{0}^{x}{|t|}\,dt\] Case − I: If \[t>0\] \[y=\left[ \frac{{{t}^{2}}}{2} \right]_{0}^{x}=\frac{{{x}^{2}}}{2}=\frac{dy}{dx}=x=2\] \[\Rightarrow \]\[x=2\] and y = 2 \[(y-2)=2(x-2)\]\[\Rightarrow \]\[y-2x+2=0.\]Hence\[x\]intercept = 1. Case − II: \[t<0\] Similarly, x intercept = −1.
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