Question

The arrangement of the slopes of the normal to the curve y = ${{e}^{\log \left( \cos x \right)}}$ in the ascending order at the points given below.
A. $x=\dfrac{\pi }{6}$ B. $x=\dfrac{7\pi }{4}$ C. $x=\dfrac{11\pi }{6}$ D. $x=\dfrac{\pi }{3}$
(a) C, B, D, A
(b) B, C, A, D
(c) A, D, C, B
(d) D, A, C, B

Answer 1

Hint: We will use the concept of derivative here with the help of which we will be able to find the slope. Here the slope is taken out by the substitution of x. This is because the slope of any tangent to the given curve is $\dfrac{dy}{dx}$. We will use the concept of relationship between tangent and normal to the curve. This relationship is given by the product of slope of tangent and slope of normal equal to – 1. Numerically, this is given as $\text{slope of tangent }\times \text{ slope of normal = - 1}$

Complete step-by-step answer:
We will consider the curve given to us as y = ${{e}^{\log \left( \cos x \right)}}$ ...(i). Now, by using the formula ${{e}^{\log x}}=x$ the equation (i) changes into y = $\cos x$.
Now, we will differentiate y with respect to x. Thus, we will get \[\dfrac{d}{dx}\left( y \right)=\dfrac{d}{dx}\left( \cos x \right)\]. By using the formula $\dfrac{d}{dx}\left( \cos x \right)=-\sin x+c$ we get \[\dfrac{dy}{dx}=-\sin x+c\]. Now, this is the slope of the curve to the tangent \[\dfrac{dy}{dx}=-\sin x+c\].
Now, we will use the concept of relationship between tangent and normal to the curve. This relationship is given by the product of slope of tangent and slope of normal equal to – 1. Numerically, this is given as $\text{slope of tangent }\times \text{ slope of normal = - 1}$.
As the slope of the tangent is \[\dfrac{dy}{dx}=-\sin x+c\]. Therefore, we get $\text{- sinx }\times \text{ slope of normal = - 1}$. After further solving we will get
$\begin{align}
  & \text{- sinx }\times \text{ slope of normal = - 1} \\
 & \Rightarrow \text{ slope of normal =}\dfrac{\text{ - 1}}{\text{- sinx}} \\
 & \Rightarrow \text{ slope of normal =}\dfrac{1}{\sin x}...\text{(i)} \\
\end{align}$
Now, we will substitute the values of x one by one in this equation (i).

A. First we will substitute $x=\dfrac{\pi }{6}$ in equation (i). Thus, we get
$\begin{align}
  & \text{slope of normal =}\dfrac{1}{\sin x} \\
 & \Rightarrow \text{slope of normal =}\dfrac{1}{\sin \left( \dfrac{\pi }{6} \right)} \\
\end{align}$
As we know that $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$ therefore we get $\text{slope of normal =}\dfrac{1}{\dfrac{1}{2}}$ and by the property $\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{ad}{bc}$ we get $\text{slope of normal =2}$.

B. Now, we will substitute $x=\dfrac{7\pi }{4}$ in equation (i). Thus, we get
$\begin{align}
  & \text{slope of normal =}\dfrac{1}{\sin x} \\
 & \Rightarrow \text{slope of normal =}\dfrac{1}{\sin \left( \dfrac{7\pi }{4} \right)} \\
\end{align}$
As $\sin \left( \dfrac{7\pi }{4} \right)=\sin \left( 2\pi -\dfrac{\pi }{4} \right)$ therefore we get $\text{slope of normal =}\dfrac{1}{\sin \left( 2\pi -\dfrac{\pi }{4} \right)}$.
Since, $\sin \left( 2\pi -\dfrac{\pi }{4} \right)=-\sin \left( \dfrac{\pi }{4} \right)$. Thus we now have $\text{slope of normal =}\dfrac{1}{-\sin \left( \dfrac{\pi }{4} \right)}$.
Since, $\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ so, we get $\text{slope of normal =}\dfrac{1}{-\dfrac{1}{\sqrt{2}}}$ and by the property $\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{ad}{bc}$ we get $\text{slope of normal = - }\sqrt{2}$.

C. Now we will put $x=\dfrac{11\pi }{6}$ in equation (i). Thus, we get
$\begin{align}
  & \text{slope of normal =}\dfrac{1}{\sin x} \\
 & \Rightarrow \text{slope of normal =}\dfrac{1}{\sin \left( \dfrac{11\pi }{6} \right)} \\
\end{align}$
As $\sin \left( \dfrac{11\pi }{6} \right)=\sin \left( 2\pi -\dfrac{\pi }{6} \right)$ therefore we get $\text{slope of normal =}\dfrac{1}{\sin \left( 2\pi -\dfrac{\pi }{6} \right)}$.
Since, $\sin \left( 2\pi -\dfrac{\pi }{6} \right)=-\sin \left( \dfrac{\pi }{6} \right)$. Thus we now have $\text{slope of normal =}\dfrac{1}{-\sin \left( \dfrac{\pi }{6} \right)}$.
Since, $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$ so, we get $\text{slope of normal =}\dfrac{1}{-\dfrac{1}{2}}$ and by the property $\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{ad}{bc}$ we get $\text{slope of normal = -2}$.

D. Now we will write $x=\dfrac{\pi }{3}$ in equation (i). Thus, we get
$\begin{align}
  & \text{slope of normal =}\dfrac{1}{\sin x} \\
 & \Rightarrow \text{slope of normal =}\dfrac{1}{\sin \left( \dfrac{\pi }{3} \right)} \\
\end{align}$
As $\sin \left( \dfrac{\pi }{3} \right)=\sin \left( 2\pi -\dfrac{\pi }{3} \right)$ therefore we get $\text{slope of normal =}\dfrac{1}{\sin \left( 2\pi -\dfrac{\pi }{3} \right)}$.
Since, $\sin \left( 2\pi -\dfrac{\pi }{3} \right)=-\sin \left( \dfrac{\pi }{3} \right)$. Thus we now have $\text{slope of normal =}\dfrac{1}{-\sin \left( \dfrac{\pi }{3} \right)}$.
Since, $\sin \left( \dfrac{\pi }{3} \right)=\dfrac{\sqrt{3}}{2}$ so, we get $\text{slope of normal =}\dfrac{1}{-\dfrac{\sqrt{3}}{2}}$ and by the property $\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{ad}{bc}$ we get $\text{slope of normal = }\dfrac{-2}{\sqrt{3}}$. By substituting the value of $\sqrt{3}=1.73$ we will get $\text{slope of normal = -1}\text{.15}$.
Thus, as we clearly know that $-2<-1.15<-\sqrt{2}<2$ Therefore we get that $\text{CBDA}$.
Hence the correct option is (a).

Note: We have used the concept of derivative here with the help of which we will be able to find the slope. Here the slope is taken out by the substitution of x. This is because the slope of any tangent to the given curve is $\dfrac{dy}{dx}$. There is a lot of difference between the equation of tangent and equation of normal. This is because the slope of the tangent is $\dfrac{dy}{dx}$ while the slope of normal is $-\dfrac{dx}{dy}$. We can alternatively use the slope of normal as $-\dfrac{dx}{dy}$ directly to solve the question and get the desired answer. Here while using the angle in terms of $2\pi $ one can make mistakes of negative and positive signs. As we know that sine is positive only in the first and second quadrant so, accordingly we have used the signs in this solution.