Question:
The angle between the curve \(2y = e^{-x/2}\) and the y-axis is \(\tan^{-1}(k)\). Then \(k =\)
The angle between the curve \(2y = e^{-x/2}\) and the y-axis is \(\tan^{-1}(k)\). Then \(k =\)
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Angle with y-axis can be computed from slope of tangent using \(\tan \theta = \frac{1}{m}\) and taking positive magnitude.
Updated On: Jul 18, 2026
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The Correct Option is D
Solution and Explanation
Step 1: Differentiate the curve.
Given \(2y = e^{-x/2} \implies y = \frac{1}{2} e^{-x/2}\)
\(\frac{dy}{dx} = \frac{-1}{4} e^{-x/2}\)
Step 2: Slope of tangent at point of intersection with y-axis.
At y-axis, \(x = 0\): slope \(m = -1/4\)
Step 3: Angle with y-axis.
Slope with y-axis: \(\tan \phi = 1/m = 1/(-1/4) = -4\)
Step 4: Take positive magnitude.
\(\tan^{-1}(k) = |\tan^{-1}(-4)| \implies k = 4\)
Step 5: Final conclusion.
Hence, \[ \boxed{k = 4} \]
Given \(2y = e^{-x/2} \implies y = \frac{1}{2} e^{-x/2}\)
\(\frac{dy}{dx} = \frac{-1}{4} e^{-x/2}\)
Step 2: Slope of tangent at point of intersection with y-axis.
At y-axis, \(x = 0\): slope \(m = -1/4\)
Step 3: Angle with y-axis.
Slope with y-axis: \(\tan \phi = 1/m = 1/(-1/4) = -4\)
Step 4: Take positive magnitude.
\(\tan^{-1}(k) = |\tan^{-1}(-4)| \implies k = 4\)
Step 5: Final conclusion.
Hence, \[ \boxed{k = 4} \]
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