JEE MainMathematicsToughDefinite Integration

If $f(x) = \int_{0}^{x}\mspace{2mu} 2|t|dt$ , then the tangent to
f(x) which are parallel to the angle bisector of the axes, in the first
quadrant, are

Show answer and solution

Line parallel to the bisector of the angle in the first
quadrant is y = x + c

f(x) = \int_{0}^{x}\mspace{2mu} 2|t|dt

\Rightarrow

f^{'}(x) = 2|x| = 1

\Rightarrow

x = \pm \frac{1}{2}

When

x = \frac{1}{2}f\left( \frac{1}{2} \right) = \int_{0}^{1/2}\mspace{2mu}|t|dt = 2\int_{0}^{1/2}\mspace{2mu} tdt = \frac{1}{4}

Hence,

\frac{1}{4} = \frac{1}{2} + C

\Rightarrow C = - \frac{1}{4}

When

x = - \frac{1}{2}f\left( - \frac{1}{2} \right) = 2\int_{0}^{- 1/2}\mspace{2mu}|t|dt = - 2\int_{0}^{- 1/2}\mspace{2mu} tdt = - \frac{1}{4}

Hence,

- \frac{1}{4} = - \frac{1}{2} + C

\Rightarrow C = \frac{1}{4}

Thus the tangent lines are y = x +

\frac{1}{4}

and y = x -

\frac{1}{4}

Source: Grass-G / M-13 Mathematics Paper 4 Nov.docx

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