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Any tangent to the curve \[y = 2{x^7} + 3x + 5\],
A. Is parallel to x-axis
B. Is parallel to y-axis
C. Makes an acute angle with x-axis
D. Makes an obtuse angle with x-axis

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Last updated date: 14th Jun 2024
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Answer
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Hint: This problem can be solved by using derivatives. Slope of the Tangent line of the curve will be obtained. Then based on the value of slope, angle of the tangent with the x-axis can be easily computed.

Complete step-by-step answer:
Equation of the curve given in the question is,
\[y = 2{x^7} + 3x + 5\]
Now, its first derivative will be done with respect to x, to get the value of $\dfrac{{dy}}{{dx}}$.
The given expression has three terms each with a simple exponential value of variable x or some constant.
So, $
  \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(2{x^7} + 3x + 5) \\
   \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(2{x^7}) + \dfrac{d}{{dx}}(3x) + \dfrac{d}{{dx}}(5)...(1) \\
    \\
 $
We know that,
$\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$
Thus by putting value of n as 7 we get,
$\dfrac{d}{{dx}}({x^7}) = 7{x^6}$
And similarly by putting value of n as 1 we get,
$\dfrac{d}{{dx}}({x^1}) = {x^0} = 1$
Also the derivative of the constant term is always 0.
Now, putting all above values in equation (1), we get
$
  \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(2{x^7}) + \dfrac{d}{{dx}}(3x) + \dfrac{d}{{dx}}(5) \\
   \Rightarrow \dfrac{{dy}}{{dx}} = 2 \times 7{x^6} + 3 \times 1 + 0 \\
   \Rightarrow \dfrac{{dy}}{{dx}} = 14{x^6} + 3 \\
    \\
 $
The clear value of $\dfrac{{dy}}{{dx}}$ will be positive as, ${x^6}$ is always a positive for any real value of x.
Thus $\tan \theta $ will also be positive. So, $\theta $ means the angle of the tangent line with x-axis will lie in the first quadrant.

Thus, the tangent to the curve will definitely make an acute angle with the x-axis.

So, the correct answer is “Option C”.

Note: All trigonometric ratios are having specific properties of their values based on their quadrant. This property helps a lot to find the possible range of values of the angle by knowing the positive or negative signs of the trigonometric ratios. Also the first derivative of any curve will give the slope of the tangent line of the curve. So, in the above problem we have used these properties to get the correct solution.