Question

If the tangent to the curve, $y={{x}^{3}}+ax-b$ at the point $\left( 1,-5 \right)$ is perpendicular to the line, $-x+y+4=0$ , then which one of the following points lies on the curve?
A. \[\left( -2,2 \right)\]
B. \[\left( 2,-2 \right)\]
C. \[\left( 2,-1 \right)\]
D. \[\left( -2,1 \right)\]

Answer 1

Hint: We will start by finding the relation between b and a , this we will do by putting the point $\left( 1,-5 \right)$ in the equation of the curve given in the question. Then we will find the slopes of the tangent at this point and of the line given in the question, we will then apply the condition of perpendicularity on both these slopes and find the value of a and eventually b. Finally we get the equation of the curve and then we will check putting points from the options given and get the answer.

Complete step by step answer:
We are given the equation of curve: $y={{x}^{3}}+ax-b\text{ }.......\text{ Equation 1}\text{.}$ Now given that point $\left( 1,-5 \right)$ lies on the curve as tangent is at that point, therefore we will put this point in the equation no. 1, we have:
$\begin{align}
  & \Rightarrow y={{x}^{3}}+ax-b \\
 & \Rightarrow \left( -5 \right)={{\left( 1 \right)}^{3}}+a\left( 1 \right)-b\Rightarrow -5=1+a-b \\
 & \Rightarrow b-a=6\text{ }........\text{Equation 2}\text{.} \\
\end{align}$
Now we will find the slope ${{m}_{1}}$ of tangent to the curve$y={{x}^{3}}+ax-b$:
${{m}_{1}}=\dfrac{dy}{dx}=~\dfrac{d\left( {{x}^{3}}+ax-b \right)}{dx}$
We will apply the power rule for differentiation that is : $f\left( x \right)={{x}^{n}}\Rightarrow f'\left( x \right)=n{{x}^{n-1}}$
$\therefore {{m}_{1}}=\dfrac{dy}{dx}=~\dfrac{d\left( {{x}^{3}}+ax-b \right)}{dx}\Rightarrow \dfrac{dy}{dx}=~3{{x}^{2}}+a$

We will find the slope at point $\left( 1,-5 \right)$
$\begin{align}
  & \Rightarrow {{\left[ {{m}_{1}} \right]}_{\left( 1,-5 \right)}}={{\left[ 3{{x}^{2}}+a \right]}_{\left( 1,-5 \right)}}\Rightarrow {{\left[ {{m}_{1}} \right]}_{\left( 1,-5 \right)}}=3\times {{\left( 1 \right)}^{2}}+a \\
 & \Rightarrow {{\left[ {{m}_{1}} \right]}_{\left( 1,-5 \right)}}=3+a \\
\end{align}$
Now we will find the slope ${{m}_{2}}$ of the line given in the question \[-x+y+4=0\]
$\begin{align}
  & -x+y+4=0\Rightarrow y=x-4 \\
 & {{m}_{2}}=\dfrac{dy}{dx}=1\Rightarrow {{m}_{2}}=1 \\
 & \\
\end{align}$
Now we are given that tangent having slope ${{m}_{1}}=a+3$ at the point $\left( 1,-5 \right)$ is perpendicular to the line \[-x+y+4=0\] whose slope is ${{m}_{2}}=1$.
Now we know that according to the condition of perpendicularity, the product of the slopes of the lines is -1 which are perpendicular to each other.
Therefore: ${{m}_{1}}{{m}_{2}}=-1$ now we will be putting values of ${{m}_{1}}$ and ${{m}_{2}}$ in this condition that is ${{m}_{1}}=a+3$ and ${{m}_{2}}=1$:
$\begin{align}
  & {{m}_{1}}{{m}_{2}}=-1 \\
 & \Rightarrow \left( a+3 \right)\left( 1 \right)=-1\Rightarrow a+3=-1\Rightarrow a=-4 \\
\end{align}$
Now we will put this value of $a$ in equation 2 that is $b-a=6$ ,
$b=6+a\Rightarrow b=6+\left( -4 \right)\Rightarrow b=2$
We got the value of $a=-4$ and $b=2$ , we will now put these values in equation 1 that is $y={{x}^{3}}+ax-b\Rightarrow y={{x}^{3}}+\left( -4 \right)x-\left( 2 \right)\Rightarrow y={{x}^{3}}-4x-2$
So we have the equation of curve: $y={{x}^{3}}-4x-2$

Now, from options we will check and we will find that: $\left( 2,-2 \right)$ lies on it.

So, the correct answer is “Option B”.

Note: When you check the points by putting them in the equation of curve, take care of the sign students can make mistakes there. Note that when it is given that point lies on a given curve that means it will satisfy the given equation.