
The total number of parallel tangents of ${{f}_{1}}(x)={{x}^{2}}-x+1$ and ${{f}_{2}}(x)={{x}^{3}}-{{x}^{2}}-2x+1$ is
A. 2
B. 0
C. 1
D. infinite
Answer
163.5k+ views
Hint: Here we have to differentiate the given expressions first. The derivative of both functions gives us the slope of each function or expression. Since the tangents are parallel, their slope will also be equal. After that, we have to equate them to find the number of parallel tangents.
Formula used:
$\frac{\text{d}x^{n}}{\text{d}x}=nx^{n-1}$
Complete Step-by-step solution:
Given ${{f}_{1}}(x)={{x}^{2}}-x+1........(1)$
${{f}_{2}}(x)={{x}^{3}}-{{x}^{2}}-2x+1.............(2)$
Apart from integration, differentiation is one of the two key principles. A technique for determining a function's derivative is differentiation. Mathematicians use a procedure called differentiation to determine a function's instantaneous rate of change based on one of its variables.
We need to differentiate $(1)$ with respect to $x$.
Differentiation of ${{x}^{2}}$ is 2x and the differentiation of a constant is 0.
so,
$f_{1}^{\prime}(x)=2 x-1$ (By using the formula $\frac{\text{d}x^{n}}{\text{d}x}=nx^{n-1}$)
Now, we need to differentiate $(2)$ with respect to $x$.
Differentiation of ${{x}^{3}}$ is $3{x}^2$, ${{x}^{2}}$ is 2x and the differentiation of 2x is 2.
Therefore,
$f_{2}^{\prime}(x)=3 x^{2}-2 x-2$ (By using the formula $\frac{\text{d}x^{n}}{\text{d}x}=nx^{n-1}$)
$f_{1}^{\prime}(x)=f_{2}^{\prime}(x)$
Let the tangents given at $\left(x_{1}, f_{1}\left(x_{1}\right)\right)$and $\left(x_{2}, f_{2}\left(x_{2}\right)\right)$to the curves $y=f_{1}(x)$and $y=f_{2}(x)$be parallel.
Therefore both the slopes i,e, both the derivative will be equal to each other as the tangents are parallel.
$\Rightarrow~2 x-1=3 x^{2}-2 x-2$
Then,
$2 x_{1}-1=3 x_{2}^{2}-2 x_{2}-2$
$2 x_{1}=3 x_{2}^{2}-2 x_{2}-1$
Here, there are no restrictions on the value of $x_{1}$ and $x_{2}$. We can give any value to both $x_{1}$ and $x_{2}$. This implies that there are infinite solutions to the above equation. If there have been any restrictions, then the result would have been finite.
Also, If we use a formula or method for infinite solutions to simplify the equation, we will get both sides equal, hence, it is an infinite solution.
Hence there are an infinite number of parallel tangents.
So the correct answer is D.
Note: Note that If an equation meets certain requirements for infinite solutions, it will result in an infinite solution. The lines must coincide and have the same y-intercept in order to obtain an endless solution. The two lines that have the same slope and y-intercept are the same line. To put it another way, the system would produce an endless answer if the two lines shared a single line. Therefore, if an equational system has an infinite number of solutions, the system will be consistent.
Formula used:
$\frac{\text{d}x^{n}}{\text{d}x}=nx^{n-1}$
Complete Step-by-step solution:
Given ${{f}_{1}}(x)={{x}^{2}}-x+1........(1)$
${{f}_{2}}(x)={{x}^{3}}-{{x}^{2}}-2x+1.............(2)$
Apart from integration, differentiation is one of the two key principles. A technique for determining a function's derivative is differentiation. Mathematicians use a procedure called differentiation to determine a function's instantaneous rate of change based on one of its variables.
We need to differentiate $(1)$ with respect to $x$.
Differentiation of ${{x}^{2}}$ is 2x and the differentiation of a constant is 0.
so,
$f_{1}^{\prime}(x)=2 x-1$ (By using the formula $\frac{\text{d}x^{n}}{\text{d}x}=nx^{n-1}$)
Now, we need to differentiate $(2)$ with respect to $x$.
Differentiation of ${{x}^{3}}$ is $3{x}^2$, ${{x}^{2}}$ is 2x and the differentiation of 2x is 2.
Therefore,
$f_{2}^{\prime}(x)=3 x^{2}-2 x-2$ (By using the formula $\frac{\text{d}x^{n}}{\text{d}x}=nx^{n-1}$)
$f_{1}^{\prime}(x)=f_{2}^{\prime}(x)$
Let the tangents given at $\left(x_{1}, f_{1}\left(x_{1}\right)\right)$and $\left(x_{2}, f_{2}\left(x_{2}\right)\right)$to the curves $y=f_{1}(x)$and $y=f_{2}(x)$be parallel.
Therefore both the slopes i,e, both the derivative will be equal to each other as the tangents are parallel.
$\Rightarrow~2 x-1=3 x^{2}-2 x-2$
Then,
$2 x_{1}-1=3 x_{2}^{2}-2 x_{2}-2$
$2 x_{1}=3 x_{2}^{2}-2 x_{2}-1$
Here, there are no restrictions on the value of $x_{1}$ and $x_{2}$. We can give any value to both $x_{1}$ and $x_{2}$. This implies that there are infinite solutions to the above equation. If there have been any restrictions, then the result would have been finite.
Also, If we use a formula or method for infinite solutions to simplify the equation, we will get both sides equal, hence, it is an infinite solution.
Hence there are an infinite number of parallel tangents.
So the correct answer is D.
Note: Note that If an equation meets certain requirements for infinite solutions, it will result in an infinite solution. The lines must coincide and have the same y-intercept in order to obtain an endless solution. The two lines that have the same slope and y-intercept are the same line. To put it another way, the system would produce an endless answer if the two lines shared a single line. Therefore, if an equational system has an infinite number of solutions, the system will be consistent.
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