Question:
The subtangent at $x = \pi/2$ on the curve $y = x \sin x $ is
The subtangent at $x = \pi/2$ on the curve $y = x \sin x $ is
Updated On: Apr 1, 2026
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- $\pi/2$
- none of these
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The Correct Option is C
Solution and Explanation
$y=x\,sin\,x$
$\Rightarrow \frac{dy}{dx}=x\,cos\,x+sin\,x\cdot1=x\,cos\,x+sin\,x$
At $x=\frac{\pi}{2}, \frac{dy}{dx}=0+sin \frac{\pi}{2}=1$
and $y=\frac{\pi}{2} sin \frac{\pi}{2} =\frac{\pi}{2}$
Now subtangent $=\left|\frac{y}{y_{1}}\right|_{at\,x=\pi/2}=\frac{\pi/2}{1}=\frac{\pi}{2}$
$\Rightarrow \frac{dy}{dx}=x\,cos\,x+sin\,x\cdot1=x\,cos\,x+sin\,x$
At $x=\frac{\pi}{2}, \frac{dy}{dx}=0+sin \frac{\pi}{2}=1$
and $y=\frac{\pi}{2} sin \frac{\pi}{2} =\frac{\pi}{2}$
Now subtangent $=\left|\frac{y}{y_{1}}\right|_{at\,x=\pi/2}=\frac{\pi/2}{1}=\frac{\pi}{2}$
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Application of Derivatives
Various Applications of Derivatives-
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\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)
This is also known to be as the Average Rate of Change.
Increasing and Decreasing Function:
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
- If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
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- The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.
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