Question

Coordinates of the point on the curve $y = x\log x$ at which the normal is parallel to the line $2x - 2y = 3$ are:
(A) $\left( {0,0} \right)$
(B) $\left( {e,e} \right)$
(C) $\left( {{e^2},2{e^2}} \right)$
(D) $\left( {{e^{ - 2}}, - 2{e^{ - 2}}} \right)$

Answer 1

Hint: In the given question, we are provided with the equation of a curve and we have to find the coordinates of a point in the curve such that the normal to the curve is parallel to the straight line $2x - 2y = 3$. So, we first differentiate the equation of the curve to find the slope of the normal at any point and then equate it to the slope of the straight line.

Complete step-by-step answer:
So, the equation of the curve is $y = x\log x$.
Now, differentiating both sides of the equation with respect to x, we get,
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {x\log x} \right)$
Now, we use the product rule of differentiation \[\dfrac{{d\left( {f\left( x \right) \times g\left( x \right)} \right)}}{{dx}} = f\left( x \right) \times \dfrac{d}{{dx}}\left[ {g\left( x \right)} \right] + g\left( x \right) \times \dfrac{d}{{dx}}\left[ {f\left( x \right)} \right]\]. So, we get,
$ \Rightarrow \dfrac{{dy}}{{dx}} = x\dfrac{d}{{dx}}\left( {\log x} \right) + \log x\dfrac{d}{{dx}}\left( x \right)$
We know that the derivative of x with respect to x is one. So, we get,
$ \Rightarrow \dfrac{{dy}}{{dx}} = x \times \dfrac{1}{x} + \log x$
Also, we know that the derivative of $\log x$ with respect to x is $\dfrac{1}{x}$. So, we get,
$ \Rightarrow \dfrac{{dy}}{{dx}} = 1 + \log x$
Now, we know that the expression $ - \dfrac{{dx}}{{dy}}$ represents the slope of a normal. So, we find the value of $ - \dfrac{{dx}}{{dy}}$ by taking the reciprocal of the equation and multiplying it with a negative sign.
\[ \Rightarrow - \dfrac{{dx}}{{dy}} = \dfrac{{ - 1}}{{1 + \log x}}\]
Now, the slope of normal at any point on the curve $y = x\log x$ is \[\dfrac{{ - 1}}{{1 + \log x}}\].
Now, we have to find the coordinates of the point on the curve such that the normal is parallel to the line $2x - 2y = 3$. So, we find the slope of this line.
We represent this in the slope intercept form. So, isolating the y term, we get,
$ \Rightarrow - 2y = 3 - 2x$
Dividing both the sides of equation by $ - 2$, we get,
$ \Rightarrow y = \dfrac{{3 - 2x}}{{ - 2}}$
Simplifying further, we get,
$ \Rightarrow y = x - \dfrac{3}{2}$
So, the slope of the line $2x - 2y = 3$ is $1$. Now, we equate the expression for slope of the normal of the curve to one as we have to find the point where the normal is parallel to the line $2x - 2y = 3$.
So, we get,
\[\dfrac{{ - 1}}{{1 + \log x}} = 1\]
Cross multiplying the terms, we get,
\[ \Rightarrow 1 + \log x = - 1\]
Shifting the terms in the equation and taking exponential function on both sides, we get,
\[ \Rightarrow \log x = - 1 - 1 = - 2\]
\[ \Rightarrow x = {e^{ - 2}}\]
So, we get the value of x coordinate as \[{e^{ - 2}}\]. Putting this in the equation of the curve, we get the value of y coordinate as,
$y = x\log x$
$ \Rightarrow y = {e^{ - 2}}\log \left( {{e^{ - 2}}} \right)$
We know the logarithmic property $\log {x^n} = n\log x$. So, we get,
$ \Rightarrow y = - 2{e^{ - 2}}$
Hence, the coordinates of point are: $\left( {{e^{ - 2}}, - 2{e^{ - 2}}} \right)$. Therefore, the option (D) is the correct answer.
So, the correct answer is “Option D”.

Note: We must know the core concepts and concepts of applications of derivatives in order to solve the given question. We should take care while calculating the slope of normal using differentiation and use the correct formula for the same. We also must know methods of solving equations in order to reach the final answer with the help of simplification rules and transposition.