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Chapter 10: Problem 844

Equation of the tangent of the curvey \(\mathrm{y}=1-\mathrm{e}^{(\mathrm{X} / 2)}\) when intersect to \(\mathrm{y}\) -axis than \(=\) (a) \(x+y=0\) (b) \(x+2 y=0\) (c) \(2 \mathrm{x}+\mathrm{y}=0\) (d) \(x-y=0\)

Short Answer

Expert verified
The equation of the tangent to the curve \(y = 1 - e^{(X/2)}\) when it intersects the y-axis is \(x + 2y = 0\), which is option \((b)\).

Step by step solution

01

Find the y-coordinate of the intersection

When the curve intersects the y-axis, the x-coordinate is 0. So, we can plug in 0 for X in the equation and find the corresponding y-coordinate: y = 1 - e^(0/2) y = 1 - e^0 y = 1 - 1 y = 0 The point of intersection is (0, 0).
02

Differentiate the curve and find the slope of the tangent

We now need to find the slope of the tangent to the curve. To do this, we need to differentiate the given equation with respect to X: y = 1 - e^(X/2) Differentiating both sides with respect to X, we get: dy/dX = 0 - 1/2 * e^(X/2) Now, we need to find the slope of the tangent at the point of intersection (0, 0). We can do this by plugging in the x-coordinate of the point into the derivative: dy/dX = -1/2 * e^(0/2) dy/dX = -1/2 * e^0 = -1/2 The slope of the tangent is -1/2.
03

Use the point-slope formula to find the equation of the tangent

With the point of intersection (0, 0) and the slope of the tangent at this point (-1/2), we can use the point-slope formula to find the equation of the tangent: y - y1 = m(x - x1) Using the point (0, 0) and the slope m = -1/2: y - 0 = (-1/2)(x - 0) Simplify the equation: y = -1/2x Multiply both sides by 2 to remove the fraction: 2y = -x Rearrange the equation to match the given options: x + 2y = 0 From the given choices, the equation of the tangent to the curve when it intersects the y-axis is x + 2y = 0, which is option (b).

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Most popular questions from this chapter

If \(\mathrm{y}^{2}=\mathrm{p}(\mathrm{x})\) is polynomial function more than 3 degree then \(2(\mathrm{~d} / \mathrm{dx})\left|\mathrm{y}^{3} \mathrm{y}_{2}\right|=\) (a) \(\mathrm{p}^{\prime}(\mathrm{x}) \cdot \mathrm{p}^{\prime \prime}(\mathrm{x})\) (b) \(\mathrm{p}(\mathrm{x}) \cdot \mathrm{p}^{\prime \prime}(\mathrm{x})\) (c) \(\mathrm{p}^{\prime}(\mathrm{x}) \cdot \mathrm{p}^{\prime \prime \prime}(\mathrm{x})\) (d) \(p(x) \cdot p^{\prime \prime \prime}(x)\)

If \(\mathrm{y}=3 \mathrm{x}^{(3 / 2)}(\mathrm{x}-1)\) then \(|[\mathrm{dy} / \mathrm{d} \mathrm{x}]|_{(\mathrm{x}=1)}=\) (a) 6 (b) 3 (c) 1 (d) 0

If \(y=\sin (x / 2) \mid[1 /\\{\cos (x / 2) \cos x\\}]+[1 /\\{\cos x \cos (3 x / 2)\\}]\) \(+[1 /\\{\cos (3 \mathrm{x} / 2) \cos 2 \mathrm{x}\\}] \mid\) then \((\mathrm{dy} / \mathrm{dx})_{\mathrm{x}=(\pi / 2)}=\) (a) \((3 / 2)\) (b) \((1 / 2)\) (c) \(-1\) (d) 1

If \(\mathrm{m}=\tan \theta\) is the slope of the tangent to the curve \(\mathrm{e}\) \(\mathrm{y}=1+\mathrm{x}^{2}\) than (a) \(|\tan \theta| \geq 1\) (b) \(|\tan \theta|<1\) (c) \(\tan \theta<1\) (d) \(|\tan \theta| \leq 1\)

The point on the curve \(\mathrm{y}=(\mathrm{x}-2)(\mathrm{x}-3)\) at which the tangent makes an angle of \(225^{\circ}\) with positive direction of \(\mathrm{x}\) -axis has co-ordinates (a) \((0,3)\) (b) \((3,0)\) (c) \((-3,0)\) (d) \((0,-3)\)
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