Question:
If $y = \sqrt{\left(\frac{1+\cos 2\theta}{1 -\cos 2\theta}\right)} $ , then $\frac{dy}{d\theta} $ at $\theta =\frac{3\pi}{4}$ is:
If $y = \sqrt{\left(\frac{1+\cos 2\theta}{1 -\cos 2\theta}\right)} $ , then $\frac{dy}{d\theta} $ at $\theta =\frac{3\pi}{4}$ is:
Updated On: Jul 6, 2022
- -2
- 2
- $\pm$
- none of these
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The Correct Option is A
Solution and Explanation
$y = \sqrt{\left(\frac{1+\cos 2\theta}{1 -\cos 2\theta}\right)} $ ,
$\Rightarrow \:\: y = \sqrt{ \frac{2 \:\cos^2 \theta}{2 \: \sin^2\theta}} = \sqrt{\cot^2 \: \theta}$
$\Rightarrow \:\: y = \cot \: \theta$
Differentiate w.r.t. $' \theta '$, we get
$\frac{dy }{d \theta} = - cosec^2 \theta$
Now , $ \left( \frac{dy }{d \theta}\right)_{\theta = \frac{3 \pi}{4}}$
$ = -cosec^2 \left( \frac{3 \pi}{4} \right)$
$ = -cosec^2 \left(\pi - \frac{ \pi}{4} \right)$
$ = -cosec^2 \frac{ \pi}{4}$
$ = - 2 \left( \because \sin \frac{ \pi}{4} = \frac{ 1}{\sqrt{2}} \right) $
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Notes on Continuity and differentiability
Concepts Used:
Continuity & Differentiability
Definition of Differentiability
f(x) is said to be differentiable at the point x = a, if the derivative f ‘(a) be at every point in its domain. It is given by
Definition of Continuity
Mathematically, a function is said to be continuous at a point x = a, if
It is implicit that if the left-hand limit (L.H.L), right-hand limit (R.H.L), and the value of the function at x=a exist and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is unspecified or does not exist, then we say that the function is discontinuous.