Question

The co-ordinates of the point on the curve \(y = xlogx\) at which the normal is parallel to the line \(2x-2y = 3\) are...

• A. \((0,0)\)
• B. \((e,e)\)
• C. \((e^2,2e^2)\)
• D. \((e^{-2},-2e^{-2})\)

Hint:

The normal is parallel to slope 1, so the tangent has slope -1.

Answer 1

The Correct Option is D

Step 1: Slope of normal:
Normal slope $=-\dfrac1{y'}$. Set equal to $1$: $y'=-1$.

Step 2: Solve:
$\ln x+1=-1$, so $x=e^{-2}$.

Step 3: Verify:
Only option (D) has $x=e^{-2}$, and $y=-2e^{-2}$ matches $x\ln x$.

Final Answer:
Tangent slope must be -1. \[ \boxed{D} \]