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Sagot :
Sure, let's solve each of the given questions step-by-step:
### Question 16:
Given [tex]\( y = x^4 e^{3x} \tan x \)[/tex], we need to find [tex]\(\frac{dy}{dx}\)[/tex].
The derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[ \frac{dy}{dx} = x^4 ( \tan x)^2 e^{3x} + 3x^4 e^{3x} \tan x + 4x^3 e^{3x} \tan x \][/tex]
Comparing this with the given options:
1. [tex]\(\text{Option (a):} \ \frac{4}{x} + \frac{\sec x}{\ln x}\)[/tex]
Clearly, [tex]\(\frac{dy}{dx}\)[/tex] contains terms involving [tex]\(e^{3x}\)[/tex] and [tex]\( (\tan x)^2 \)[/tex], which are not present in this option.
2. [tex]\(\text{Option (b):} \ \frac{4}{x} + 3 + \frac{\sec^2 x}{\tan x}\)[/tex]
This option has terms that resemble parts of the derivative involving [tex]\(\frac{4}{x}\)[/tex], [tex]\(3\)[/tex], and [tex]\(\frac{\sec^2 x}{\tan x}\)[/tex], which does correspond to the structure of the derived expression:
[tex]\[ 4/x + 3 + \sec^2(x)/\tan(x) \][/tex]
which matches the form of the solution.
3. [tex]\(\text{Option (c):} \ \frac{x}{4} + 3\)[/tex]
This does not match any parts of the derived expression.
4. [tex]\(\text{Option (d):} \ \frac{\sec x}{x}\)[/tex]
Again, this does not correspond to the derived expression.
Thus, the correct answer is:
(b) [tex]\(\frac{4}{x} + 3 + \frac{\sec^2 x}{\tan x}\)[/tex]
### Question 17:
We need to use the distance formula to show that the points [tex]\((2, -1)\)[/tex], [tex]\((5, 5)\)[/tex], and [tex]\((6, -5)\)[/tex] are vertices of a right triangle.
The distance formula between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
1. Calculate the distance between [tex]\((2, -1)\)[/tex] and [tex]\((5, 5)\)[/tex]:
[tex]\[ d_{AB} = \sqrt{(5 - 2)^2 + (5 + 1)^2} = \sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \][/tex]
2. Calculate the distance between [tex]\((5, 5)\)[/tex] and [tex]\((6, -5)\)[/tex]:
[tex]\[ d_{BC} = \sqrt{(6 - 5)^2 + (-5 - 5)^2} = \sqrt{1^2 + (-10)^2} = \sqrt{1 + 100} = \sqrt{101} \][/tex]
3. Calculate the distance between [tex]\((2, -1)\)[/tex] and [tex]\((6, -5)\)[/tex]:
[tex]\[ d_{AC} = \sqrt{(6 - 2)^2 + (-5 + 1)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \][/tex]
We need to check if these distances form a right triangle. For a right triangle, the square of one side should be equal to the sum of the squares of the other two sides ([tex]\(c^2 = a^2 + b^2\)[/tex]).
Checking:
[tex]\[ d_{AC}^2 = 32 \][/tex]
[tex]\[ d_{AB}^2 + d_{BC}^2 = (3\sqrt{5})^2 + (\sqrt{101})^2 = 45 + 101 = 146 \][/tex]
However, we can see
[tex]\[ d_{BC}^2 = 101 \][/tex]
[tex]\[ d_{AB}^2 + d_{AC}^2 = 45 + 32 = 77 \][/tex]
None of these satisfy [tex]\( c^2 = a^2 + b^2 \)[/tex] with the calculated sides. Therefore, the calculated distances should be rechecked for correctness. There might be a mistake in the original setup. It turns out they do not seem to form vertices of a right triangle with the current analysis.
Please double-check the problem details and reevaluate any potential arithmetic mistakes.
### Question 16:
Given [tex]\( y = x^4 e^{3x} \tan x \)[/tex], we need to find [tex]\(\frac{dy}{dx}\)[/tex].
The derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[ \frac{dy}{dx} = x^4 ( \tan x)^2 e^{3x} + 3x^4 e^{3x} \tan x + 4x^3 e^{3x} \tan x \][/tex]
Comparing this with the given options:
1. [tex]\(\text{Option (a):} \ \frac{4}{x} + \frac{\sec x}{\ln x}\)[/tex]
Clearly, [tex]\(\frac{dy}{dx}\)[/tex] contains terms involving [tex]\(e^{3x}\)[/tex] and [tex]\( (\tan x)^2 \)[/tex], which are not present in this option.
2. [tex]\(\text{Option (b):} \ \frac{4}{x} + 3 + \frac{\sec^2 x}{\tan x}\)[/tex]
This option has terms that resemble parts of the derivative involving [tex]\(\frac{4}{x}\)[/tex], [tex]\(3\)[/tex], and [tex]\(\frac{\sec^2 x}{\tan x}\)[/tex], which does correspond to the structure of the derived expression:
[tex]\[ 4/x + 3 + \sec^2(x)/\tan(x) \][/tex]
which matches the form of the solution.
3. [tex]\(\text{Option (c):} \ \frac{x}{4} + 3\)[/tex]
This does not match any parts of the derived expression.
4. [tex]\(\text{Option (d):} \ \frac{\sec x}{x}\)[/tex]
Again, this does not correspond to the derived expression.
Thus, the correct answer is:
(b) [tex]\(\frac{4}{x} + 3 + \frac{\sec^2 x}{\tan x}\)[/tex]
### Question 17:
We need to use the distance formula to show that the points [tex]\((2, -1)\)[/tex], [tex]\((5, 5)\)[/tex], and [tex]\((6, -5)\)[/tex] are vertices of a right triangle.
The distance formula between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by
[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
1. Calculate the distance between [tex]\((2, -1)\)[/tex] and [tex]\((5, 5)\)[/tex]:
[tex]\[ d_{AB} = \sqrt{(5 - 2)^2 + (5 + 1)^2} = \sqrt{3^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \][/tex]
2. Calculate the distance between [tex]\((5, 5)\)[/tex] and [tex]\((6, -5)\)[/tex]:
[tex]\[ d_{BC} = \sqrt{(6 - 5)^2 + (-5 - 5)^2} = \sqrt{1^2 + (-10)^2} = \sqrt{1 + 100} = \sqrt{101} \][/tex]
3. Calculate the distance between [tex]\((2, -1)\)[/tex] and [tex]\((6, -5)\)[/tex]:
[tex]\[ d_{AC} = \sqrt{(6 - 2)^2 + (-5 + 1)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \][/tex]
We need to check if these distances form a right triangle. For a right triangle, the square of one side should be equal to the sum of the squares of the other two sides ([tex]\(c^2 = a^2 + b^2\)[/tex]).
Checking:
[tex]\[ d_{AC}^2 = 32 \][/tex]
[tex]\[ d_{AB}^2 + d_{BC}^2 = (3\sqrt{5})^2 + (\sqrt{101})^2 = 45 + 101 = 146 \][/tex]
However, we can see
[tex]\[ d_{BC}^2 = 101 \][/tex]
[tex]\[ d_{AB}^2 + d_{AC}^2 = 45 + 32 = 77 \][/tex]
None of these satisfy [tex]\( c^2 = a^2 + b^2 \)[/tex] with the calculated sides. Therefore, the calculated distances should be rechecked for correctness. There might be a mistake in the original setup. It turns out they do not seem to form vertices of a right triangle with the current analysis.
Please double-check the problem details and reevaluate any potential arithmetic mistakes.
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