MHT CET · Maths · Application of Derivatives
The co-ordinates of a point on the curve \(y=x \log x\) at which the normal is parallel to the line \(2 x-2 y=3\) are
- A \(\left(-\mathrm{e}^{-2}, 2 \mathrm{e}^{-2}\right)\)
- B \(\left(-\mathrm{e}^{-2},-2 \mathrm{e}^{-2}\right)\)
- C \(\quad\left(\mathrm{e}^{-2}, 2 \mathrm{e}^{-2}\right)\)
- D \(\left(\mathrm{e}^{-2},-2 \mathrm{e}^{-2}\right)\)
Answer & Solution
Correct Answer
(D) \(\left(\mathrm{e}^{-2},-2 \mathrm{e}^{-2}\right)\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& y=x \log x \\
\therefore \quad & \frac{\mathrm{~d} y}{\mathrm{~d} x}=1+\log x
...(i)\end{aligned}\)
Slope of the normal \(=-\frac{1}{\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)}=\frac{-1}{1+\log x}\)
Slope of the given line is 1 .
Since the normal is parallel to the given line.
\(\begin{aligned}
\therefore \quad & \frac{-1}{1+\log x}=1 \\
& \Rightarrow \log x=-2 \\
& \Rightarrow x=\mathrm{e}^{-2}
\end{aligned}\)
From (i), \(y=-2 \mathrm{e}^{-2}\)
\(\therefore \quad\) Co-ordinates of the point are \(\left(\mathrm{e}^{-2},-2 \mathrm{e}^{-2}\right)\).
& y=x \log x \\
\therefore \quad & \frac{\mathrm{~d} y}{\mathrm{~d} x}=1+\log x
...(i)\end{aligned}\)
Slope of the normal \(=-\frac{1}{\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)}=\frac{-1}{1+\log x}\)
Slope of the given line is 1 .
Since the normal is parallel to the given line.
\(\begin{aligned}
\therefore \quad & \frac{-1}{1+\log x}=1 \\
& \Rightarrow \log x=-2 \\
& \Rightarrow x=\mathrm{e}^{-2}
\end{aligned}\)
From (i), \(y=-2 \mathrm{e}^{-2}\)
\(\therefore \quad\) Co-ordinates of the point are \(\left(\mathrm{e}^{-2},-2 \mathrm{e}^{-2}\right)\).
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